Home  Forums  Reviews  Guides  Newsgroups  Register  Search 
Thread Tools 
somenath 


 
pete
Guest
Posts: n/a

somenath wrote:
> > Hi All, > I am trying to undestand "Type Conversions" from K&R book.I am not > able to understand the > bellow mentioned text > "Conversion rules are more complicated when unsigned operands are > involved. The problem > is that comparisons between signed and unsigned values are machine > dependent, because > they depend on the sizes of the various integer types. For example, > suppose that int is 16 bits > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > int, is promoted to a > signed long. But 1L > 1UL because 1L is promoted to unsigned long > and thus appears > to be a large positive number" > To understand it i wrote a small code > > #include<stdio.h> > int main(void) > { > signed long sl = 5; > unsigned long ul= 5; > if (sl>ul) > { > printf("\n 5 is greater than 5\n"); > } > else > { > printf("5 is less than 5 \n"); > } > return 0; > } > > OP is > 5 is greater than 5 > Could anybody help me to understand the conversion rules when > unsigned operands are involved ? "rank" is the missing word here. To compare two expressions, they must be converted to the same type. The rule for comparing two exressions is simple, and signed or unsigned types do not form any exceptions to this rule: The lower ranking type is converted to the higher ranking type. The above text about comparisons being machine dependent because of sizes, is wrong. For example, suppose that int and long are both 32 bits. Then 1L < 1U, because 1U, which is an unsigned int, is *converted* (not promoted) to a signed long, and it has absolutely nothing to do with whether or not int and long are the same size. "Integer promotions" on the other hand, which are distinct from the kind of conversions performed on an operand of a relational operator, are dependent upon whether or not type int can represent all the positive values of the lower ranking type. N869 6.3.1 Arithmetic operands 6.3.1.1 Boolean, characters, and integers [#1] Every integer type has an integer conversion rank defined as follows:  No two signed integer types shall have the same rank, even if they have the same representation.  The rank of a signed integer type shall be greater than the rank of any signed integer type with less precision.  The rank of long long int shall be greater than the rank of long int, which shall be greater than the rank of int, which shall be greater than the rank of short int, which shall be greater than the rank of signed char.  The rank of any unsigned integer type shall equal the rank of the corresponding signed integer type, if any.  The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.  The rank of char shall equal the rank of signed char and unsigned char.  The rank of _Bool shall be less than the rank of all other standard integer types.  The rank of any enumerated type shall equal the rank of the compatible integer type (see 6.7.2.2).  The rank of any extended signed integer type relative to another extended signed integer type with the same precision is implementationdefined, but still subject to the other rules for determining the integer conversion rank.  For all integer types T1, T2, and T3, if T1 has greater rank than T2 and T2 has greater rank than T3, then T1 has greater rank than T3. /* BEGIN new.c */ #include<stdio.h> #include<limits.h> int main(void) { signed long sl = 5; unsigned u = 5; if (sl > u) { puts("5 is greater than 5"); } else { puts("5 is less than 5"); } printf("LONG_MAX is %ld\n", LONG_MAX); printf("UINT_MAX is %u \n", UINT_MAX); return 0; } /* END new.c */ On my machine the output of new.c is: 5 is greater than 5 LONG_MAX is 2147483647 UINT_MAX is 4294967295 The unsigned type is converted to the long signed type, even though on my machine, they have the same size and number of bits.  pete 




pete 


 
somenath
Guest
Posts: n/a

On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote:
> somenath wrote: > > > Hi All, > > I am trying to undestand "Type Conversions" from K&R book.I am not > > able to understand the > > bellow mentioned text > > "Conversion rules are more complicated when unsigned operands are > > involved. The problem > > is that comparisons between signed and unsigned values are machine > > dependent, because > > they depend on the sizes of the various integer types. For example, > > suppose that int is 16 bits > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > int, is promoted to a > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > and thus appears > > to be a large positive number" > > To understand it i wrote a small code > > > #include<stdio.h> > > int main(void) > > { > > signed long sl = 5; > > unsigned long ul= 5; > > if (sl>ul) > > { > > printf("\n 5 is greater than 5\n"); > > } > > else > > { > > printf("5 is less than 5 \n"); > > } > > return 0; > > } > > > OP is > > 5 is greater than 5 > > Could anybody help me to understand the conversion rules when > > unsigned operands are involved ? > > "rank" is the missing word here. > To compare two expressions, they must be converted to the same type. > The rule for comparing two exressions is simple, > and signed or unsigned types do not form any exceptions to this rule: > The lower ranking type is converted to the higher ranking type. > > The above text about comparisons > being machine dependent because of sizes, is wrong. > For example, suppose that int and long are both 32 bits. > Then 1L < 1U, because 1U, which is an unsigned int, > is *converted* (not promoted) to a signed long, > and it has absolutely nothing to do with whether or not > int and long are the same size. > > "Integer promotions" on the other hand, which are distinct > from the kind of conversions performed on an operand > of a relational operator, are dependent upon whether > or not type int can represent all the positive values > of the lower ranking type. > > N869 > 6.3.1 Arithmetic operands > 6.3.1.1 Boolean, characters, and integers > [#1] Every integer type has an integer conversion rank > defined as follows: >  No two signed integer types shall have the same rank, > even if they have the same representation. >  The rank of a signed integer type shall be greater than > the rank of any signed integer type with less > precision. >  The rank of long long int shall be greater than the > rank of long int, which shall be greater than the rank > of int, which shall be greater than the rank of short > int, which shall be greater than the rank of signed > char. >  The rank of any unsigned integer type shall equal the > rank of the corresponding signed integer type, if any. >  The rank of any standard integer type shall be greater > than the rank of any extended integer type with the > same width. >  The rank of char shall equal the rank of signed char > and unsigned char. >  The rank of _Bool shall be less than the rank of all > other standard integer types. >  The rank of any enumerated type shall equal the rank of > the compatible integer type (see 6.7.2.2). >  The rank of any extended signed integer type relative > to another extended signed integer type with the same > precision is implementationdefined, but still subject > to the other rules for determining the integer > conversion rank. >  For all integer types T1, T2, and T3, if T1 has greater > rank than T2 and T2 has greater rank than T3, then T1 > has greater rank than T3. > > /* BEGIN new.c */ > > #include<stdio.h> > #include<limits.h> > > int main(void) > { > signed long sl = 5; > unsigned u = 5; > > if (sl > u) { > puts("5 is greater than 5"); > } else { > puts("5 is less than 5"); > } > printf("LONG_MAX is %ld\n", LONG_MAX); > printf("UINT_MAX is %u \n", UINT_MAX); > return 0; > > } > > /* END new.c */ > > On my machine the output of new.c is: > 5 is greater than 5 > LONG_MAX is 2147483647 > UINT_MAX is 4294967295 > > The unsigned type is converted to the long signed type, > even though on my machine, > they have the same size and number of bits. > Hi Pete, Many thanks for this information. But i am still not able to understand the comparision. As par my signed long has higher rank than unsigned int . so "u" will be converted to Signed long .Does it mean the value of "u" which is 5 will be converted to 5 ? Even then how "if (sl > u)" is true ? Regards, Somenath 




somenath 
pete
Guest
Posts: n/a

somenath wrote:
> > On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote: > > somenath wrote: > > > > > Hi All, > > > I am trying to undestand "Type Conversions" from K&R book.I am not > > > able to understand the > > > bellow mentioned text > > > "Conversion rules are more complicated when unsigned operands are > > > involved. The problem > > > is that comparisons between signed and unsigned values are machine > > > dependent, because > > > they depend on the sizes of the various integer types. For example, > > > suppose that int is 16 bits > > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > > int, is promoted to a > > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > > and thus appears > > > to be a large positive number" > > > To understand it i wrote a small code > > > > > #include<stdio.h> > > > int main(void) > > > { > > > signed long sl = 5; > > > unsigned long ul= 5; > > > if (sl>ul) > > > { > > > printf("\n 5 is greater than 5\n"); > > > } > > > else > > > { > > > printf("5 is less than 5 \n"); > > > } > > > return 0; > > > } > > > > > OP is > > > 5 is greater than 5 > > > Could anybody help me to understand the conversion rules when > > > unsigned operands are involved ? > > > > "rank" is the missing word here. > > To compare two expressions, they must be converted to the same type. > > The rule for comparing two exressions is simple, > > and signed or unsigned types do not form any exceptions to this rule: > > The lower ranking type is converted to the higher ranking type. > > > > The above text about comparisons > > being machine dependent because of sizes, is wrong. > > For example, suppose that int and long are both 32 bits. > > Then 1L < 1U, because 1U, which is an unsigned int, > > is *converted* (not promoted) to a signed long, > > and it has absolutely nothing to do with whether or not > > int and long are the same size. > > > > "Integer promotions" on the other hand, which are distinct > > from the kind of conversions performed on an operand > > of a relational operator, are dependent upon whether > > or not type int can represent all the positive values > > of the lower ranking type. > > > > N869 > > 6.3.1 Arithmetic operands > > 6.3.1.1 Boolean, characters, and integers > > [#1] Every integer type has an integer conversion rank > > defined as follows: > >  No two signed integer types shall have the same rank, > > even if they have the same representation. > >  The rank of a signed integer type shall be greater than > > the rank of any signed integer type with less > > precision. > >  The rank of long long int shall be greater than the > > rank of long int, which shall be greater than the rank > > of int, which shall be greater than the rank of short > > int, which shall be greater than the rank of signed > > char. > >  The rank of any unsigned integer type shall equal the > > rank of the corresponding signed integer type, if any. > >  The rank of any standard integer type shall be greater > > than the rank of any extended integer type with the > > same width. > >  The rank of char shall equal the rank of signed char > > and unsigned char. > >  The rank of _Bool shall be less than the rank of all > > other standard integer types. > >  The rank of any enumerated type shall equal the rank of > > the compatible integer type (see 6.7.2.2). > >  The rank of any extended signed integer type relative > > to another extended signed integer type with the same > > precision is implementationdefined, but still subject > > to the other rules for determining the integer > > conversion rank. > >  For all integer types T1, T2, and T3, if T1 has greater > > rank than T2 and T2 has greater rank than T3, then T1 > > has greater rank than T3. > > > > /* BEGIN new.c */ > > > > #include<stdio.h> > > #include<limits.h> > > > > int main(void) > > { > > signed long sl = 5; > > unsigned u = 5; > > > > if (sl > u) { > > puts("5 is greater than 5"); > > } else { > > puts("5 is less than 5"); > > } > > printf("LONG_MAX is %ld\n", LONG_MAX); > > printf("UINT_MAX is %u \n", UINT_MAX); > > return 0; > > > > } > > > > /* END new.c */ > > > > On my machine the output of new.c is: > > 5 is greater than 5 > > LONG_MAX is 2147483647 > > UINT_MAX is 4294967295 > > > > The unsigned type is converted to the long signed type, > > even though on my machine, > > they have the same size and number of bits. > > > > Hi Pete, > Many thanks for this information. But i am still not able to > understand the comparision. > As par my signed long has higher rank than unsigned int . so "u" will > be converted to > Signed long .Does it mean the value of "u" which is 5 will be > converted to 5 ? > Even then how "if (sl > u)" is true ? The answer is simple: I was wrong. The rules are more complicated. Both sl and u get converted to type long unsigned in the above expression on my machine. The reason is because on my machine, signed long can't represent the entire range of unsigned values. The conversion in question, is called "the usual arithmetic conversions". N869 6.3.1.8 Usual arithmetic conversions [#1] Many operators that expect operands of arithmetic type cause conversions and yield result types in a similar way. The purpose is to determine a common real type for the operands and result. For the specified operands, each operand is converted, without change of type domain, to a type whose corresponding real type is the common real type. Unless explicitly stated otherwise, the common real type is also the corresponding real type of the result, whose type domain is the type domain of the operands if they are the same, and complex otherwise. This pattern is called the usual arithmetic conversions: First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double. Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double. Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float. Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands: If both operands have the same type, then no further conversion is needed. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank. Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type. Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.  pete 




pete 
somenath
Guest
Posts: n/a

On Jul 25, 8:35 am, pete <(EMail Removed)> wrote:
> somenath wrote: > > > On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote: > > > somenath wrote: > > > > > Hi All, > > > > I am trying to undestand "Type Conversions" from K&R book.I am not > > > > able to understand the > > > > bellow mentioned text > > > > "Conversion rules are more complicated when unsigned operands are > > > > involved. The problem > > > > is that comparisons between signed and unsigned values are machine > > > > dependent, because > > > > they depend on the sizes of the various integer types. For example, > > > > suppose that int is 16 bits > > > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > > > int, is promoted to a > > > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > > > and thus appears > > > > to be a large positive number" > > > > To understand it i wrote a small code > > > > > #include<stdio.h> > > > > int main(void) > > > > { > > > > signed long sl = 5; > > > > unsigned long ul= 5; > > > > if (sl>ul) > > > > { > > > > printf("\n 5 is greater than 5\n"); > > > > } > > > > else > > > > { > > > > printf("5 is less than 5 \n"); > > > > } > > > > return 0; > > > > } > > > > > OP is > > > > 5 is greater than 5 > > > > Could anybody help me to understand the conversion rules when > > > > unsigned operands are involved ? > > > > "rank" is the missing word here. > > > To compare two expressions, they must be converted to the same type. > > > The rule for comparing two exressions is simple, > > > and signed or unsigned types do not form any exceptions to this rule: > > > The lower ranking type is converted to the higher ranking type. > > > > The above text about comparisons > > > being machine dependent because of sizes, is wrong. > > > For example, suppose that int and long are both 32 bits. > > > Then 1L < 1U, because 1U, which is an unsigned int, > > > is *converted* (not promoted) to a signed long, > > > and it has absolutely nothing to do with whether or not > > > int and long are the same size. > > > > "Integer promotions" on the other hand, which are distinct > > > from the kind of conversions performed on an operand > > > of a relational operator, are dependent upon whether > > > or not type int can represent all the positive values > > > of the lower ranking type. > > > > N869 > > > 6.3.1 Arithmetic operands > > > 6.3.1.1 Boolean, characters, and integers > > > [#1] Every integer type has an integer conversion rank > > > defined as follows: > > >  No two signed integer types shall have the same rank, > > > even if they have the same representation. > > >  The rank of a signed integer type shall be greater than > > > the rank of any signed integer type with less > > > precision. > > >  The rank of long long int shall be greater than the > > > rank of long int, which shall be greater than the rank > > > of int, which shall be greater than the rank of short > > > int, which shall be greater than the rank of signed > > > char. > > >  The rank of any unsigned integer type shall equal the > > > rank of the corresponding signed integer type, if any. > > >  The rank of any standard integer type shall be greater > > > than the rank of any extended integer type with the > > > same width. > > >  The rank of char shall equal the rank of signed char > > > and unsigned char. > > >  The rank of _Bool shall be less than the rank of all > > > other standard integer types. > > >  The rank of any enumerated type shall equal the rank of > > > the compatible integer type (see 6.7.2.2). > > >  The rank of any extended signed integer type relative > > > to another extended signed integer type with the same > > > precision is implementationdefined, but still subject > > > to the other rules for determining the integer > > > conversion rank. > > >  For all integer types T1, T2, and T3, if T1 has greater > > > rank than T2 and T2 has greater rank than T3, then T1 > > > has greater rank than T3. > > > > /* BEGIN new.c */ > > > > #include<stdio.h> > > > #include<limits.h> > > > > int main(void) > > > { > > > signed long sl = 5; > > > unsigned u = 5; > > > > if (sl > u) { > > > puts("5 is greater than 5"); > > > } else { > > > puts("5 is less than 5"); > > > } > > > printf("LONG_MAX is %ld\n", LONG_MAX); > > > printf("UINT_MAX is %u \n", UINT_MAX); > > > return 0; > > > > } > > > > /* END new.c */ > > > > On my machine the output of new.c is: > > > 5 is greater than 5 > > > LONG_MAX is 2147483647 > > > UINT_MAX is 4294967295 > > > > The unsigned type is converted to the long signed type, > > > even though on my machine, > > > they have the same size and number of bits. > > > Hi Pete, > > Many thanks for this information. But i am still not able to > > understand the comparision. > > As par my signed long has higher rank than unsigned int . so "u" will > > be converted to > > Signed long .Does it mean the value of "u" which is 5 will be > > converted to 5 ? > > Even then how "if (sl > u)" is true ? > > The answer is simple: I was wrong. > The rules are more complicated. > Both sl and u get converted to type long unsigned in the But does it mean 5 will be converted to 5 ? Even then how "if (sl > u)" is true ? 




somenath 
somenath
Guest
Posts: n/a

On Jul 25, 8:35 am, pete <(EMail Removed)> wrote:
> somenath wrote: > > > On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote: > > > somenath wrote: > > > > > Hi All, > > > > I am trying to undestand "Type Conversions" from K&R book.I am not > > > > able to understand the > > > > bellow mentioned text > > > > "Conversion rules are more complicated when unsigned operands are > > > > involved. The problem > > > > is that comparisons between signed and unsigned values are machine > > > > dependent, because > > > > they depend on the sizes of the various integer types. For example, > > > > suppose that int is 16 bits > > > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > > > int, is promoted to a > > > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > > > and thus appears > > > > to be a large positive number" > > > > To understand it i wrote a small code > > > > > #include<stdio.h> > > > > int main(void) > > > > { > > > > signed long sl = 5; > > > > unsigned long ul= 5; > > > > if (sl>ul) > > > > { > > > > printf("\n 5 is greater than 5\n"); > > > > } > > > > else > > > > { > > > > printf("5 is less than 5 \n"); > > > > } > > > > return 0; > > > > } > > > > > OP is > > > > 5 is greater than 5 > > > > Could anybody help me to understand the conversion rules when > > > > unsigned operands are involved ? > > > > "rank" is the missing word here. > > > To compare two expressions, they must be converted to the same type. > > > The rule for comparing two exressions is simple, > > > and signed or unsigned types do not form any exceptions to this rule: > > > The lower ranking type is converted to the higher ranking type. > > > > The above text about comparisons > > > being machine dependent because of sizes, is wrong. > > > For example, suppose that int and long are both 32 bits. > > > Then 1L < 1U, because 1U, which is an unsigned int, > > > is *converted* (not promoted) to a signed long, > > > and it has absolutely nothing to do with whether or not > > > int and long are the same size. > > > > "Integer promotions" on the other hand, which are distinct > > > from the kind of conversions performed on an operand > > > of a relational operator, are dependent upon whether > > > or not type int can represent all the positive values > > > of the lower ranking type. > > > > N869 > > > 6.3.1 Arithmetic operands > > > 6.3.1.1 Boolean, characters, and integers > > > [#1] Every integer type has an integer conversion rank > > > defined as follows: > > >  No two signed integer types shall have the same rank, > > > even if they have the same representation. > > >  The rank of a signed integer type shall be greater than > > > the rank of any signed integer type with less > > > precision. > > >  The rank of long long int shall be greater than the > > > rank of long int, which shall be greater than the rank > > > of int, which shall be greater than the rank of short > > > int, which shall be greater than the rank of signed > > > char. > > >  The rank of any unsigned integer type shall equal the > > > rank of the corresponding signed integer type, if any. > > >  The rank of any standard integer type shall be greater > > > than the rank of any extended integer type with the > > > same width. > > >  The rank of char shall equal the rank of signed char > > > and unsigned char. > > >  The rank of _Bool shall be less than the rank of all > > > other standard integer types. > > >  The rank of any enumerated type shall equal the rank of > > > the compatible integer type (see 6.7.2.2). > > >  The rank of any extended signed integer type relative > > > to another extended signed integer type with the same > > > precision is implementationdefined, but still subject > > > to the other rules for determining the integer > > > conversion rank. > > >  For all integer types T1, T2, and T3, if T1 has greater > > > rank than T2 and T2 has greater rank than T3, then T1 > > > has greater rank than T3. > > > > /* BEGIN new.c */ > > > > #include<stdio.h> > > > #include<limits.h> > > > > int main(void) > > > { > > > signed long sl = 5; > > > unsigned u = 5; > > > > if (sl > u) { > > > puts("5 is greater than 5"); > > > } else { > > > puts("5 is less than 5"); > > > } > > > printf("LONG_MAX is %ld\n", LONG_MAX); > > > printf("UINT_MAX is %u \n", UINT_MAX); > > > return 0; > > > > } > > > > /* END new.c */ > > > > On my machine the output of new.c is: > > > 5 is greater than 5 > > > LONG_MAX is 2147483647 > > > UINT_MAX is 4294967295 > > > > The unsigned type is converted to the long signed type, > > > even though on my machine, > > > they have the same size and number of bits. > > > Hi Pete, > > Many thanks for this information. But i am still not able to > > understand the comparision. > > As par my signed long has higher rank than unsigned int . so "u" will > > be converted to > > Signed long .Does it mean the value of "u" which is 5 will be > > converted to 5 ? > > Even then how "if (sl > u)" is true ? > > The answer is simple: I was wrong. > The rules are more complicated. > Both sl and u get converted to type long unsigned in the But does it mean 5 will be converted to 5 ? Even then how "if (sl > u)" is true ? 




somenath 
pete
Guest
Posts: n/a

somenath wrote:
> > On Jul 25, 8:35 am, pete <(EMail Removed)> wrote: > > somenath wrote: > > > > > On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote: > > > > somenath wrote: > > > > > > > Hi All, > > > > > I am trying to undestand "Type Conversions" from K&R book.I am not > > > > > able to understand the > > > > > bellow mentioned text > > > > > "Conversion rules are more complicated when unsigned operands are > > > > > involved. The problem > > > > > is that comparisons between signed and unsigned values are machine > > > > > dependent, because > > > > > they depend on the sizes of the various integer types. For example, > > > > > suppose that int is 16 bits > > > > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > > > > int, is promoted to a > > > > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > > > > and thus appears > > > > > to be a large positive number" > > > > > To understand it i wrote a small code > > > > > > > #include<stdio.h> > > > > > int main(void) > > > > > { > > > > > signed long sl = 5; > > > > > unsigned long ul= 5; > > > > > if (sl>ul) > > > > > { > > > > > printf("\n 5 is greater than 5\n"); > > > > > } > > > > > else > > > > > { > > > > > printf("5 is less than 5 \n"); > > > > > } > > > > > return 0; > > > > > } > > > > > > > OP is > > > > > 5 is greater than 5 > > > > > Could anybody help me to understand the conversion rules when > > > > > unsigned operands are involved ? > > > > > > "rank" is the missing word here. > > > > To compare two expressions, they must be converted to the same type. > > > > The rule for comparing two exressions is simple, > > > > and signed or unsigned types do not form any exceptions to this rule: > > > > The lower ranking type is converted to the higher ranking type. > > > > > > The above text about comparisons > > > > being machine dependent because of sizes, is wrong. > > > > For example, suppose that int and long are both 32 bits. > > > > Then 1L < 1U, because 1U, which is an unsigned int, > > > > is *converted* (not promoted) to a signed long, > > > > and it has absolutely nothing to do with whether or not > > > > int and long are the same size. > > > > > > "Integer promotions" on the other hand, which are distinct > > > > from the kind of conversions performed on an operand > > > > of a relational operator, are dependent upon whether > > > > or not type int can represent all the positive values > > > > of the lower ranking type. > > > > > > N869 > > > > 6.3.1 Arithmetic operands > > > > 6.3.1.1 Boolean, characters, and integers > > > > [#1] Every integer type has an integer conversion rank > > > > defined as follows: > > > >  No two signed integer types shall have the same rank, > > > > even if they have the same representation. > > > >  The rank of a signed integer type shall be greater than > > > > the rank of any signed integer type with less > > > > precision. > > > >  The rank of long long int shall be greater than the > > > > rank of long int, which shall be greater than the rank > > > > of int, which shall be greater than the rank of short > > > > int, which shall be greater than the rank of signed > > > > char. > > > >  The rank of any unsigned integer type shall equal the > > > > rank of the corresponding signed integer type, if any. > > > >  The rank of any standard integer type shall be greater > > > > than the rank of any extended integer type with the > > > > same width. > > > >  The rank of char shall equal the rank of signed char > > > > and unsigned char. > > > >  The rank of _Bool shall be less than the rank of all > > > > other standard integer types. > > > >  The rank of any enumerated type shall equal the rank of > > > > the compatible integer type (see 6.7.2.2). > > > >  The rank of any extended signed integer type relative > > > > to another extended signed integer type with the same > > > > precision is implementationdefined, but still subject > > > > to the other rules for determining the integer > > > > conversion rank. > > > >  For all integer types T1, T2, and T3, if T1 has greater > > > > rank than T2 and T2 has greater rank than T3, then T1 > > > > has greater rank than T3. > > > > > > /* BEGIN new.c */ > > > > > > #include<stdio.h> > > > > #include<limits.h> > > > > > > int main(void) > > > > { > > > > signed long sl = 5; > > > > unsigned u = 5; > > > > > > if (sl > u) { > > > > puts("5 is greater than 5"); > > > > } else { > > > > puts("5 is less than 5"); > > > > } > > > > printf("LONG_MAX is %ld\n", LONG_MAX); > > > > printf("UINT_MAX is %u \n", UINT_MAX); > > > > return 0; > > > > > > } > > > > > > /* END new.c */ > > > > > > On my machine the output of new.c is: > > > > 5 is greater than 5 > > > > LONG_MAX is 2147483647 > > > > UINT_MAX is 4294967295 > > > > > > The unsigned type is converted to the long signed type, > > > > even though on my machine, > > > > they have the same size and number of bits. > > > > > Hi Pete, > > > Many thanks for this information. But i am still not able to > > > understand the comparision. > > > As par my signed > > > long has higher rank than unsigned int . so "u" will > > > be converted to > > > Signed long .Does it mean the value of "u" which is 5 will be > > > converted to 5 ? > > > Even then how "if (sl > u)" is true ? > > > > The answer is simple: I was wrong. > > The rules are more complicated. > > Both sl and u get converted to type long unsigned in the > > But does it mean 5 will be converted to 5 ? No. The converted expression is equal to ((long unsigned)5). > Even then how "if > (sl > u)" is true ? The rule for converting a negative value to an unsigned integer type is to add (1 + UNSIGNED_INTEGER_TYPE_MAX) repeatedly to the negative value until it is not negative any more. 5 gets converted to 4294967291LU  pete 




pete 
somenath
Guest
Posts: n/a

On Jul 25, 3:35 pm, pete <(EMail Removed)> wrote:
> somenath wrote: > > > On Jul 25, 8:35 am, pete <(EMail Removed)> wrote: > > > somenath wrote: > > > > > On Jul 23, 12:37 pm, pete <(EMail Removed)> wrote: > > > > > somenath wrote: > > > > > > > Hi All, > > > > > > I am trying to undestand "Type Conversions" from K&R book.I am not > > > > > > able to understand the > > > > > > bellow mentioned text > > > > > > "Conversion rules are more complicated when unsigned operands are > > > > > > involved. The problem > > > > > > is that comparisons between signed and unsigned values are machine > > > > > > dependent, because > > > > > > they depend on the sizes of the various integer types. For example, > > > > > > suppose that int is 16 bits > > > > > > and long is 32 bits. Then 1L < 1U, because 1U, which is an unsigned > > > > > > int, is promoted to a > > > > > > signed long. But 1L > 1UL because 1L is promoted to unsigned long > > > > > > and thus appears > > > > > > to be a large positive number" > > > > > > To understand it i wrote a small code > > > > > > > #include<stdio.h> > > > > > > int main(void) > > > > > > { > > > > > > signed long sl = 5; > > > > > > unsigned long ul= 5; > > > > > > if (sl>ul) > > > > > > { > > > > > > printf("\n 5 is greater than 5\n"); > > > > > > } > > > > > > else > > > > > > { > > > > > > printf("5 is less than 5 \n"); > > > > > > } > > > > > > return 0; > > > > > > } > > > > > > > OP is > > > > > > 5 is greater than 5 > > > > > > Could anybody help me to understand the conversion rules when > > > > > > unsigned operands are involved ? > > > > > > "rank" is the missing word here. > > > > > To compare two expressions, they must be converted to the same type. > > > > > The rule for comparing two exressions is simple, > > > > > and signed or unsigned types do not form any exceptions to this rule: > > > > > The lower ranking type is converted to the higher ranking type. > > > > > > The above text about comparisons > > > > > being machine dependent because of sizes, is wrong. > > > > > For example, suppose that int and long are both 32 bits. > > > > > Then 1L < 1U, because 1U, which is an unsigned int, > > > > > is *converted* (not promoted) to a signed long, > > > > > and it has absolutely nothing to do with whether or not > > > > > int and long are the same size. > > > > > > "Integer promotions" on the other hand, which are distinct > > > > > from the kind of conversions performed on an operand > > > > > of a relational operator, are dependent upon whether > > > > > or not type int can represent all the positive values > > > > > of the lower ranking type. > > > > > > N869 > > > > > 6.3.1 Arithmetic operands > > > > > 6.3.1.1 Boolean, characters, and integers > > > > > [#1] Every integer type has an integer conversion rank > > > > > defined as follows: > > > > >  No two signed integer types shall have the same rank, > > > > > even if they have the same representation. > > > > >  The rank of a signed integer type shall be greater than > > > > > the rank of any signed integer type with less > > > > > precision. > > > > >  The rank of long long int shall be greater than the > > > > > rank of long int, which shall be greater than the rank > > > > > of int, which shall be greater than the rank of short > > > > > int, which shall be greater than the rank of signed > > > > > char. > > > > >  The rank of any unsigned integer type shall equal the > > > > > rank of the corresponding signed integer type, if any. > > > > >  The rank of any standard integer type shall be greater > > > > > than the rank of any extended integer type with the > > > > > same width. > > > > >  The rank of char shall equal the rank of signed char > > > > > and unsigned char. > > > > >  The rank of _Bool shall be less than the rank of all > > > > > other standard integer types. > > > > >  The rank of any enumerated type shall equal the rank of > > > > > the compatible integer type (see 6.7.2.2). > > > > >  The rank of any extended signed integer type relative > > > > > to another extended signed integer type with the same > > > > > precision is implementationdefined, but still subject > > > > > to the other rules for determining the integer > > > > > conversion rank. > > > > >  For all integer types T1, T2, and T3, if T1 has greater > > > > > rank than T2 and T2 has greater rank than T3, then T1 > > > > > has greater rank than T3. > > > > > > /* BEGIN new.c */ > > > > > > #include<stdio.h> > > > > > #include<limits.h> > > > > > > int main(void) > > > > > { > > > > > signed long sl = 5; > > > > > unsigned u = 5; > > > > > > if (sl > u) { > > > > > puts("5 is greater than 5"); > > > > > } else { > > > > > puts("5 is less than 5"); > > > > > } > > > > > printf("LONG_MAX is %ld\n", LONG_MAX); > > > > > printf("UINT_MAX is %u \n", UINT_MAX); > > > > > return 0; > > > > > > } > > > > > > /* END new.c */ > > > > > > On my machine the output of new.c is: > > > > > 5 is greater than 5 > > > > > LONG_MAX is 2147483647 > > > > > UINT_MAX is 4294967295 > > > > > > The unsigned type is converted to the long signed type, > > > > > even though on my machine, > > > > > they have the same size and number of bits. > > > > > Hi Pete, > > > > Many thanks for this information. But i am still not able to > > > > understand the comparision. > > > > As par my signed > > > > long has higher rank than unsigned int . so "u" will > > > > be converted to > > > > Signed long .Does it mean the value of "u" which is 5 will be > > > > converted to 5 ? > > > > Even then how "if (sl > u)" is true ? > > > > The answer is simple: I was wrong. > > > The rules are more complicated. > > > Both sl and u get converted to type long unsigned in the > > > But does it mean 5 will be converted to 5 ? > > No. The converted expression is equal to ((long unsigned)5). > > > Even then how "if > > (sl > u)" is true ? > > The rule for converting a negative value to an unsigned integer type > is to add (1 + UNSIGNED_INTEGER_TYPE_MAX) repeatedly > to the negative value until it is not negative any more. > > 5 gets converted to 4294967291LU But is this rules applies to largest negative number such as 2147483648 ? then what will be the result when it is converted to unsigned integer type? 




somenath 
pete
Guest
Posts: n/a

somenath wrote:
> > On Jul 25, 3:35 pm, pete <(EMail Removed)> wrote: > > The rule for converting a negative value to an unsigned integer type > > is to add (1 + UNSIGNED_INTEGER_TYPE_MAX) repeatedly > > to the negative value until it is not negative any more. > > > > 5 gets converted to 4294967291LU > > But is this rules applies to largest negative number such as > 2147483648 ? I have some shocking news for you. On some implementations of C, like mine and probably yours, 2147483648 is a positive quantity. (214748364 is the result of the unary minus operator operating on a value of 2147483648. On my machine, 2147483648 excedes LONG_MAX, so it is converted to long unsigned. 2147483648 equals 2147483648, on my C implementation. INT_MIN probably expands to (2147483647  1) on your machine, like it does on mine. > then what will be the result when it is converted to > unsigned integer type? This is the output on my machine: /* BEGIN new.c output */ 2147483648 is greater than zero. (2147483647  1) is less than zero. 2147483648 == 2147483648 (long unsigned)5 is 4294967291 (long unsigned)2147483648 is 2147483648 LONG_MAX is 2147483647 UINT_MAX is 4294967295 INT_MIN expands to (2147483647  1) /* END new.c output */ /* BEGIN new.c */ #include<stdio.h> #include<limits.h> #define str(s) # s #define xstr(s) str(s) int main(void) { puts("/* BEGIN new.c output */\n"); if (2147483648 > 0) { puts("2147483648 is greater than zero.\n"); } if (0 > (2147483647  1)) { puts("(2147483647  1) is less than zero.\n"); } if (2147483648 == 214748364 { puts("2147483648 == 2147483648\n"); } printf("(long unsigned)5 is %lu\n", (long unsigned)5); printf("(long unsigned)2147483648 is %lu\n\n", (long unsigned)214748364; printf("LONG_MAX is %ld\n", LONG_MAX); printf("UINT_MAX is %u \n", UINT_MAX); puts("INT_MIN expands to " xstr(INT_MIN)); puts("\n/* END new.c output */"); return 0; } /* END new.c */  pete 




pete 
pete
Guest
Posts: n/a

pete wrote:
> > somenath wrote: > > > > On Jul 25, 3:35 pm, pete <(EMail Removed)> wrote: > > > > The rule for converting a negative value to an unsigned integer type > > > is to add (1 + UNSIGNED_INTEGER_TYPE_MAX) repeatedly > > > to the negative value until it is not negative any more. > > > > > > 5 gets converted to 4294967291LU > > > > But is this rules applies to largest negative number such as > > 2147483648 ? > > I have some shocking news for you. > On some implementations of C, like mine and probably yours, > 2147483648 is a positive quantity. > (214748364 is the result of the unary minus operator > operating on a value of 2147483648. > On my machine, 2147483648 excedes LONG_MAX, > so it is converted to long unsigned. Not actually converted; The original type of 2147483648, is long unsigned on my machine. > 2147483648 equals 2147483648, on my C implementation. > INT_MIN probably expands to (2147483647  1) > on your machine, like it does on mine. > > > then what will be the result when it is converted to > > unsigned integer type? > > This is the output on my machine: > > /* BEGIN new.c output */ > > 2147483648 is greater than zero. > > (2147483647  1) is less than zero. > > 2147483648 == 2147483648 > > (long unsigned)5 is 4294967291 > (long unsigned)2147483648 is 2147483648 > > LONG_MAX is 2147483647 > UINT_MAX is 4294967295 > INT_MIN expands to (2147483647  1) > > /* END new.c output */ > > /* BEGIN new.c */ > > #include<stdio.h> > #include<limits.h> > > #define str(s) # s > #define xstr(s) str(s) > > int main(void) > { > puts("/* BEGIN new.c output */\n"); > if (2147483648 > 0) { > puts("2147483648 is greater than zero.\n"); > } > if (0 > (2147483647  1)) { > puts("(2147483647  1) is less than zero.\n"); > } > if (2147483648 == 214748364 { > puts("2147483648 == 2147483648\n"); > } > printf("(long unsigned)5 is %lu\n", > (long unsigned)5); > printf("(long unsigned)2147483648 is %lu\n\n", > (long unsigned)214748364; > printf("LONG_MAX is %ld\n", LONG_MAX); > printf("UINT_MAX is %u \n", UINT_MAX); > puts("INT_MIN expands to " xstr(INT_MIN)); > puts("\n/* END new.c output */"); > return 0; > } > > /* END new.c */ > >  > pete  pete 




pete 


 
Thread Tools  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
(int) > (unsigned) > (int) or (unsigned) > (int) > (unsigned):I'll loose something?  pozz  C Programming  12  03202011 11:32 PM 
Type conversion (to unsigned) can not have aggregate operand.  Olaf  VHDL  4  06112007 06:32 PM 
What happens when type conversion between signed and unsigned happens?  NM  C++  6  09202006 05:39 PM 
Explicit unsigned/signed conversion: ANSI/ISO rules?  Ken Tough  C Programming  4  06082004 04:38 PM 
silly unsigned/signed byte conversion question  aspa  Java  9  10292003 10:37 PM 
Powered by vBulletin®. Copyright ©2000  2014, vBulletin Solutions, Inc..
SEO by vBSEO ©2010, Crawlability, Inc. 