doubt regarding sizeof operator

Hi All,

I am not able to understand the behavior of the bellow mentioned
program.

#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is

Size = 0

But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?

But if I rewrite the code as mentioned bellow the output is 4

#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}

Size = 4

Could you please help me to understand the problem?

Regards
Somenath

somenath wrote:

Quote:

>
Hi All,
>
I am not able to understand the behavior of the bellow mentioned
program.
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
>
Size = 0
>
But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?
>

The expression "sizeof 1<0" is false, hence the zero. sizeof binds
tighter than '<', so you have written (sizeof 1) < 0.

--
Ian Collins.

On 10 30 , 4 44 , somenath <somenath...@gmail.comwrote:

Quote:

Hi All,
>
I am not able to understand the behavior of the bellow mentioned
program.
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;}
>
The output of the program is
>
Size = 0
>
But my understanding is sizeof returns the size of the type .In this
case should it not returns sizeof (int) ?
>
But if I rewrite the code as mentioned bellow the output is 4
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
>
}
>
Size = 4
>
Could you please help me to understand the problem?
>
Regards
Somenath

printf("\n Size = %d \n",sizeof (1<0));//sizeof(bool)
printf("\n Size = %d \n",sizeof 1<0);// equal to ((sizeof (int))<0)
of course zero --false--

somenath <somenathpal@gmail.comwrites:

Quote:

I am not able to understand the behavior of the bellow mentioned
program.
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
>
Size = 0

Which is exactly what it should be, as Ian Collins explained.

[...]

Quote:

But if I rewrite the code as mentioned bellow the output is 4
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}
>
Size = 4

(1<0) yields a value of type int (with the value 0), so sizeof (1<0)
yields sizeof(int). Apparently that's 4 on your system. But you
print the value using a "%d" format, which is correct for type int,
but not for type size_t.

For a small value like this, a reasonable way to print it is to
convert it to int:

printf("\n Size = %d \n",(int)sizeof (1<0));

--
Keith Thompson (The_Other_Keith) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Oct 30, 11:12 am, Keith Thompson <ks...@mib.orgwrote:

Quote:

(1<0) yields a value of type int (with the value 0), so sizeof (1<0)
yields sizeof(int). Apparently that's 4 on your system. But you
print the value using a "%d" format, which is correct for type int,
but not for type size_t.
>
For a small value like this, a reasonable way to print it is to
convert it to int:
>
printf("\n Size = %d \n",(int)sizeof (1<0));

The only way is this

#if __STDC_VERSION__ >= 199901L
printf("%zu\n", sizeof (int));
#else
printf("%u\n", (unsigned)sizeof (int));
#endif

"somenath" <somenathpal@gmail.coma écrit dans le message de news:
1193733892.491643.315010@y27g2000pre.googlegroups. com...

Quote:

>
I am not able to understand the behavior of the bellow mentioned
program.
>
#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}

This looks like a typo in a program that prints the size of floating point
numbers:

"Charlie Gordon" <news@chqrlie.orga écrit dans le message de news:
47272d17$0$5451$426a74cc@news.free.fr...

Quote:

"somenath" <somenathpal@gmail.coma écrit dans le message de news:
1193733892.491643.315010@y27g2000pre.googlegroups. com...

Quote:

>>
>I am not able to understand the behavior of the bellow mentioned
>program.
>>
>#include<stdio.h>
>int main(void)
>{
> printf("\n Size = %d \n",sizeof 1<0);
> return 0;
>}

>
This looks like a typo in a program that prints the size of floating point
numbers: (OOPS, my fingers slipped)

#include<stdio.h>
int main(void)
{
printf("\n Size = %d \n",sizeof 1.0);
return 0;
}

Note that sizeof 1.0, which is the same as sizeof(double), should be cast to
(int) for printf's %d format specifier. The format specifier for size_t is
%zu, but it is c99 specific.

--
Chqrlie.

vipvipvipvipvip.ru@gmail.com writes:

Quote:

The only way is this
>
#if __STDC_VERSION__ >= 199901L
printf("%zu\n", sizeof (int));
#else
printf("%u\n", (unsigned)sizeof (int));
#endif

I would recommend "%lu" and unsigned long for the second case.
--
char a[]="\n .CJacehknorstu";int putchar(int);int main(void){unsigned long b[]
={0x67dffdff,0x9aa9aa6a,0xa77ffda9,0x7da6aa6a,0xa6 7f6aaa,0xaa9aa9f6,0x11f6},*p
=b,i=24;for(;p+=!*p;*p/=4)switch(0[p]&3)case 0:{return 0;for(p--;i--;i--)case+
2:{i++;if(i)break;else default:continue;if(0)case 1:putchar(a[i&15]);break;}}}

vipvipvipvipvip.ru@gmail.com writes:

Quote:

On Oct 30, 11:12 am, Keith Thompson <ks...@mib.orgwrote:

Quote:

>(1<0) yields a value of type int (with the value 0), so sizeof (1<0)
>yields sizeof(int). Apparently that's 4 on your system. But you
>print the value using a "%d" format, which is correct for type int,
>but not for type size_t.
>>
>For a small value like this, a reasonable way to print it is to
>convert it to int:
>>
> printf("\n Size = %d \n",(int)sizeof (1<0));

>
The only way is this
>
#if __STDC_VERSION__ >= 199901L
printf("%zu\n", sizeof (int));
#else
printf("%u\n", (unsigned)sizeof (int));
#endif

Why is that the only way?

It's safe to assume that sizeof(int) does not exceed INT_MAX (even
though it's not absolutely guaranteed by the standard), so for this
particular case, converting to int and using "%d" is not unreasonable.

If you want to be a bit safer, convert to unsigned long and use "%lu".

"%zu" is the correct format for C99 -- but in practice, there's a
significant risk that the compiler will claim to conform to C99, but
the runtime library won't support "%zu". "%lu" will work properly for
all sizes up to 2**32-1, and perhaps more.

--
Keith Thompson (The_Other_Keith) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

somenath wrote:

Quote:

>
#include<stdio.h>
int main(void) {
printf("\n Size = %d \n",sizeof 1<0);
return 0;
}
The output of the program is
Size = 0

Here "sizeof 1" returns size of an int, which is not < 0, so the
comparison function returns 0, which is then printed. Fully
parenthized the operation is "((sizeof 1) < 0)"

Quote:

>
But my understanding is sizeof returns the size of the type .In
this case should it not returns sizeof (int) ?
>
But if I rewrite the code as mentioned bellow the output is 4
>
#include<stdio.h>
int main(void) {
printf("\n Size = %d \n",sizeof (1<0));
return 0;
}
>
Size = 4

Here "1 < 0" returns an int (which happens to be 0, and doesn't
matter). So sizeof (1 < 0) returns the sizeof an int, which
happens to be 4 on your system. Fully parenthized the operation is
"(sizeof (1 < 0))".

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

On Tue, 30 Oct 2007 09:03:33 +0000, qiooeer wrote:

Quote:

printf("\n Size = %d \n",sizeof (1<0));//sizeof(bool)

(1<0) has type int.
%d is for signed int, not size_t.
--
Army1987 (Replace "NOSPAM" with "email")
A hamburger is better than nothing.
Nothing is better than eternal happiness.
Therefore, a hamburger is better than eternal happiness.

On Oct 30, 11:15 pm, "Charlie Gordon" <n...@chqrlie.orgwrote:

Quote:

That's true, but it will not handle huge variables under Win64 where %zu
might not be supported either. Should we use %llu ?

You cannot have an object whose size SIZE_MAX.
%zu will work everywhere, anywhere as long as it is supported.
Else you cast to unsigned.

<vipvipvipvipvip.ru@gmail.coma écrit dans le message de news:
1193781055.450457.290450@o80g2000hse.googlegroups. com...

Quote:

On Oct 30, 11:15 pm, "Charlie Gordon" <n...@chqrlie.orgwrote:

Quote:

>"Keith Thompson" <kst-u@mib.orga écrit dans le message de news:

ln8x5k1j9e.fsf@nuthaus.mib.org...

Quote:

>That's true, but it will not handle huge variables under Win64 where %zu
>might not be supported either. Should we use %llu ?

You cannot have an object whose size SIZE_MAX.
%zu will work everywhere, anywhere as long as it is supported.
Else you cast to unsigned.

Why did you snip the relevant part by Keith Thomson below:

Quote:

>>"%zu" is the correct format for C99 -- but in practice, there's a
>>significant risk that the compiler will claim to conform to C99, but
>>the runtime library won't support "%zu". "%lu" will work properly for
>>all sizes up to 2**32-1, and perhaps more.

On the win64 architecture, size_t is 64 bits, while both unsigned and
unsigned long are 32 bits.
One can have objects larger than 4 GB on thosw systems for which %lu and
(unsigned long)sizeof(object) will fail to print the correct size. VC++ is
not c99 conforming, as %zu might not be supported. So the only way to print
a size on win64 is to cast to (unsigned long long) and use the %llu format.

--
Chqrlie.

"Charlie Gordon" <news@chqrlie.orgwrites:

Quote:

One can have objects larger than 4 GB on thosw systems for which %lu and
(unsigned long)sizeof(object) will fail to print the correct size. VC++ is
not c99 conforming, as %zu might not be supported. So the only way to print
a size on win64 is to cast to (unsigned long long) and use the %llu format.

If VC++ is not C99 conforming, then it may not have unsigned long
long type at all, and %llu may also not be supported.
--
"Some people *are* arrogant, and others read the FAQ."
--Chris Dollin

Keith Thompson <kst-u@mib.orgwrote:

Quote:

"%zu" is the correct format for C99 -- but in practice, there's a
significant risk that the compiler will claim to conform to C99, but
the runtime library won't support "%zu".

Possibly, but in that case, you have a broken implementation. %zu should
be among the least of your concerns. I'd advice upgrading to a correct,
well-built C99 implementation immediately. If that's not possible,
reverting to a correct, well-built C89 implementation would be
preferable over struggling with a chimaera.

Richard

"Ben Pfaff" <blp@cs.stanford.edua écrit dans le message de news:
87zly0usev.fsf@blp.benpfaff.org...

Quote:

"Charlie Gordon" <news@chqrlie.orgwrites:
>

Quote:

>One can have objects larger than 4 GB on thosw systems for which %lu and
>(unsigned long)sizeof(object) will fail to print the correct size. VC++
>is
>not c99 conforming, as %zu might not be supported. So the only way to
>print
>a size on win64 is to cast to (unsigned long long) and use the %llu
>format.

>
If VC++ is not C99 conforming, then it may not have unsigned long
long type at all, and %llu may also not be supported.

Can anyone with VC++ experience what is the current status on c99
conformance and long long support ?

--
Chqrlie.

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