«

Union un
{ int I;
char c[2];
}

main()
{
union un x;
x.c[0]=10;
x.c[1]=1;
printf("%d",.i);
}


What is the output results? Please explain in detail under.

On 29 Mar 2007 23:24:56 -0700, "??????" <> wrote:

>Union un
>{ int I;
>char c[2];
>}
>
>main()
>{
> union un x;
> x.c[0]=10;
> x.c[1]=1;
> printf("%d",.i);
>}
>
>
>What is the output results? Please explain in detail under.

That code does not even compile with my compilers and should not
compile with yours. You should at least make an effort to insure that
your code compiles before submitting it to this newsgroup. Otherwise,
it can be difficult for anyone here to provide help to you.

Best regards
--
jay

On 3月30日, 下�3时46分, "Joachim Schmitz" <nospam.schm...@hp.com> wrote:
> "�õ³¬·²" <wangchao...@gmail.com> schrieb im Newsbeitragnews:1175235895.997556.161250@d57g2000h sg.googlegroups.com...
> #include <stdio.h>> Union un
> union un
> > { int I;
> > char c[2];
> > }
> };
>
> > main()
> int main(void)
> > {
> > union un x;
> > x.c[0]=10;
> > x.c[1]=1;
> > printf("%d",.i);
>
> printf("%d\n",x.I);
> return 0;> }
>
Thank you very much!I have something wrong with the procedure, as you
corrected by the right .

#include <stdio.h>
union un
{
int I;
char c[2];
}


int main(void)
{
union un x;
x.c[0]=10;
x.c[1]=1;
printf("%d",.i);

printf("%d\n",x.I);
return 0;
}

> > What is the output results? Please explain in detail under.
>
> Lots of compiler errors and some warnings too, without the modifications
> mentioned above. After having fixed these, on one of my machine the output
> was 167838688, on another it was 1073807626.
> Heavily depends in the implementation and machine architecture (big
> endian/little endian)
>
> Bye, Jojo

However, I still do not know how the compiler theory, Why the digital
output ? Depends what the outcome is different for each output?

"Íõ³¬·²" <> schrieb im Newsbeitrag
news: oups.com...
#include <stdio.h>
> Union un
union un
> { int I;
> char c[2];
> }
};
>
> main()
int main(void)
> {
> union un x;
> x.c[0]=10;
> x.c[1]=1;
> printf("%d",.i);
printf("%d\n",x.I);
return 0;
> }
>
>
> What is the output results? Please explain in detail under.
Lots of compiler errors and some warnings too, without the modifications
mentioned above. After having fixed these, on one of my machine the output
was 167838688, on another it was 1073807626.
Heavily depends in the implementation and machine architecture (big
endian/little endian)

Bye, Jojo

??? said:

> Union un
> { int I;
> char c[2];
> }
>
> main()
> {
> union un x;
> x.c[0]=10;
> x.c[1]=1;
> printf("%d",.i);
> }
>
>
> What is the output results?

It would be undefined because you're missing a header, except that it
won't compile because you're missing an x. And when you've added the x,
you still have to change i to I or I to i (either will do), and U to u.

Once you've done all that, the answer is "it depends".

All in all, I wouldn't bother if I were you.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

On 30 Mar 2007 00:02:33 -0700, "???" <> wrote:

>On 3?30?, ??3?46?, "Joachim Schmitz" <nospam.schm...@hp.com> wrote:
>> "Íõ³¬·²" <wangchao...@gmail.com> schrieb im Newsbeitragnews:1175235895.997556.161250@d57g2000h sg.googlegroups.com...
>> #include <stdio.h>> Union un
>> union un
>> > { int I;
>> > char c[2];
>> > }
>> };
>>
>> > main()
>> int main(void)
>> > {
>> > union un x;
>> > x.c[0]=10;
>> > x.c[1]=1;
>> > printf("%d",.i);
>>
>> printf("%d\n",x.I);
>> return 0;> }
>>
>Thank you very much!I have something wrong with the procedure, as you
>corrected by the right .

Your code still does not compile!

>#include <stdio.h>
>union un
>{
>int I;
>char c[2];
> }

Change that to:

};

>
>
>int main(void)
>{
> union un x;
> x.c[0]=10;
> x.c[1]=1;
> printf("%d",.i);

That statement won't compile and requires a diagnostic. Just delete
it, since your following statement is correct.

>
> printf("%d\n",x.I);
> return 0;
>}

I notice that you changed your posted name from "?????" to "???". I
have a feeling that if you change your posted name to simply "?", your
code will compile

All kidding aside, I'll tell you this ... in my 15+ years of
programming in C, I have never--not once--legitimately used a union,
because I've never needed to legitimately use one.

I've dealt with unions declared by compilers, which most compilers
inevitably seem to use (and I can live with that), and I've dealt with
fellow programmers who've declared unions (IMHO, unnecessarily). But
I've never found the need to legitimately use unions myself.

My advice to you is to understand unions, but to never use them in
your own code. In your example, I can tell you that you will not get
the results you expect on a lot of implementations, such as those in
which sizeof(int) != 2 (many if not most) and on those in which the
Endianness is Big when yours is Little or vice versa.

Best regards
--
jay

> Union un
> { int I;
> char c[2];
> }
>

After "fixing" your code to make it compiled:

- int 4 bytes
- char 1 byte, char[2] - 2 bytes

so, after union is created it is filled with trash. Then you assign a values
but only to "char" components, the other two bytes of "int" component are
still filled with "trash". So final answer is "you will get unspecified
number".

"Marcin" <> schrieb im Newsbeitrag
news:euij31$880$...
>> Union un
>> { int I;
>> char c[2];
>> }
>>
>
> After "fixing" your code to make it compiled:
>
> - int 4 bytes
> - char 1 byte, char[2] - 2 bytes
>
> so, after union is created it is filled with trash. Then you assign a
> values
> but only to "char" components, the other two bytes of "int" component are
> still filled with "trash". So final answer is "you will get unspecified
> number".
Ah, yes, I missed that these other bytes still may be uninitialized... (on
an implementation with sizeof(int) != 2)

but
union un x;
x.I = 0;
x.c[0]=10;
x.c[1]=1;
printf("%d",x.I);

would still give some implementation/architecture dependant output.

Bye, Jojo

Marcin said:

<snip>

> After "fixing" [the] code to make it compiled:
>
> - int 4 bytes
> - char 1 byte, char[2] - 2 bytes

What makes you think int is 4 bytes? It might be only one byte, or it
might be two bytes, or it might even be ninety-seven bytes.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

"Richard Heathfield" <> ha scritto nel messaggio
news...
> What makes you think int is 4 bytes? It might be only one byte, or it
> might be two bytes, or it might even be ninety-seven bytes.

It can't be one byte.
One byte is defined as the size of an unsigned char, i.e. CHAR_BIT bits.
All 2^CHAR_BIT = UCHAR_MAX + 1 possible sequences of CHAR_BIT bits must be
valid unsigned char representations. Also, CHAR_BIT is required to be at
least 8.

An int must be able to represent all possible values of unsigned char, so
INT_MAX >= UCHAR_MAX.
Also, it must be able to represent all negative numbers down to -INT_MAX.
So, it must be able to represent at least 2*UCHAR_MAX - 1 values. To do
that, it must have at least ceil(log2(2*UCHAR_MAX - 1)) = CHAR_BIT + 1 bits.

Since any object must occupy an integer number of bytes, an int must have at
least 2 bytes.