char * and int *
 

why following's output is noting in C program

int a=5;
printf("\n%s\n",(char *)(&a));

and
following lines give correct output

printf("\n%d\n",*(int *)(&a));

Re: char * and int *
 

gyan wrote:
> why following's output is noting in C program
>
> int a=5;
> printf("\n%s\n",(char *)(&a));
Read up on the conversion specifiers a bit.
Specifically, from the man page of printf(), I'll quote (partly) a
couple that
might be of interest to you:
s The char * argument is expected to be a pointer to an array
of
character type (pointer to a string). Characters from the
array
are written up to (but not including) a terminating NUL
charac-
ter;...

p The void * pointer argument is printed in hexadecimal (as
if by
`%#x' or `%#lx').
So `s' is not the correct thing to use when you are trying to print out
the
address (provided, I understand you correctly), rather p is.
>
> and
> following lines give correct output
>
> printf("\n%d\n",*(int *)(&a));
is the same as
printf("\n%d\n",a);
and prints 5 with a newline before and after the number.The cast is
superfluous.

Re: char * and int *
 

gyan wrote:

> why following's output is noting in C program
>
> int a=5;
> printf("\n%s\n",(char *)(&a));

`%s` wants a string, ie a char* pointer to a null-terminated
sequence of characters. `a`, whose address you have taken and
then forcibly (but reversibly) converted to `char*`, isn't
any sequence of characters at all, so whatever you get out
isn't C's responsibility any more - it will be an accident
of the implementation. (Or even a deliberate feature, but
usually not.)

*As it happens*, the underlying implementation may not care.
There are some bytes in the object `a`; it may treat them
as characters and print until it finds a zero byte. If you
machine is big-endian, or the character with value 5 is
invisible, you'll get what you got - no visible characters
displayed.

> and
> following lines give correct output
>
> printf("\n%d\n",*(int *)(&a));

Because you ask to print an int, and what you give it is

*(int *)(&a)

which is

*(&a)

because the cast is redundant, &a already being int*, and

*&a === a

by the co-definitions of * and &.

--
Chris "electric hedgehog" Dollin
It's called *extreme* programming, not *stupid* programming.

Re: char * and int *
 

I agree with you all, but i am playing little bit.
see i am working on
SunOS au1646s 5.8 Generic_117000-01 sun4u sparc SUNW,Sun-Fire-15000

here int takes 4 bytes.
Now i have prgram:
int a=6;
char *tmp;
tmp=(char *)(&a);
*(tmp+4)='\0';
printf("\n%s\n",tmp+3);

Now i run it through debugger dbx.

according to output:
p *(tmp) = '\0'
p *(tmp+1) = '\0'
p *(tmp+2) = '\0'
p -f%d *(tmp+3) = 6
p *(tmp+3) = '\0'

So when i write
printf("\n%s\n",tmp+3);
i hope that program should see 6 at that adress
(tmp+3) and '\0' at next address.
Now %s should print 6, since i am passing a char * who stores a char array
type having 6 at oth byte and '\0' at 1st byte.

Re: char * and int *
 

> why following's output is noting in C program
>
> int a=5;
> printf("\n%s\n",(char *)(&a));

Odds are you're on a big endian endian machine that read the upper 8
value bits (which are 0 in this case) of |a| and printed nothing
because it treated it as a 0 length NUL terminated string.

> and
> following lines give correct output
>
> printf("\n%d\n",*(int *)(&a));

*(int*)&a is the same thing as a

Re: char * and int *
 

On Wed, 06 Jul 2005 18:54:36 -0400, "gyan" <singhg4@anz.com> wrote in
comp.lang.c:

> I agree with you all, but i am playing little bit.
> see i am working on
> SunOS au1646s 5.8 Generic_117000-01 sun4u sparc SUNW,Sun-Fire-15000
>
> here int takes 4 bytes.
> Now i have prgram:
> int a=6;
> char *tmp;
> tmp=(char *)(&a);
> *(tmp+4)='\0';

If sizeof(int) is four or less on your platform, the line above
invokes undefined behavior. And if sizeof(int) is not 4 on your
platform, it makes no sense at all.

> printf("\n%s\n",tmp+3);

The problem is that once you have produced undefined behavior, and you
have, the rest of the program is completely meaningless as far as the
C standard, and therefore this newsgroup, are concerned.

>
> Now i run it through debugger dbx.

Debuggers are completely off topic here.

>
> So when i write
> printf("\n%s\n",tmp+3);
> i hope that program should see 6 at that adress
> (tmp+3) and '\0' at next address.
> Now %s should print 6, since i am passing a char * who stores a char array
> type having 6 at oth byte and '\0' at 1st byte.

No it shouldn't, you are mistaken. The character '6' in your
implementation's character set will print "6". The binary value 0x6
will not.

--
Jack Klein
Home: http://JK-Technology.Com
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