doubt regarding sizeof operator

[somenath]

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

[Ian Collins]

somenath wrote:

>
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.

[qiooeer]

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

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--

[Keith Thompson]

somenath <so*********@gm ail.comwrites:

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.

[...]

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)size of (1<0));

--
Keith Thompson (The_Other_Keit h) ks***@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"

[vipvipvipvipvip.ru]

On Oct 30, 11:12 am, Keith Thompson <[e-mail] rote:

(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)size of (1<0));

The only way is this

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

[Charlie Gordon]

"somenath" <so*********@gm ail.coma écrit dans le message de news:
11************* *********@y27g2 00...legr oups.com...

>
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]

"Charlie Gordon" <ne**@chqrlie.o rga écrit dans le message de news:
47************* *********@news. free.fr...

"somenath" <so*********@gm ail.coma écrit dans le message de news:
11************* *********@y27g2 00...legr oups.com...

>>
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.

[Ben Pfaff]

vi************* ***@gmail.com writes:

The only way is this

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

I would recommend "%lu" and unsigned long for the second case.
--
char a[]="\n .CJacehknorstu" ;int putchar(int);in t main(void){unsi gned long b[]
={0x67dffdff,0x 9aa9aa6a,0xa77f fda9,0x7da6aa6a ,0xa67f6aaa,0xa a9aa9f6,0x11f6} ,*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)bre ak;else default:continu e;if(0)case 1:putchar(a[i&15]);break;}}}

[Keith Thompson]

vi************* ***@gmail.com writes:

On Oct 30, 11:12 am, Keith Thompson <[e-mail] rote:

>(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)size of (1<0));

The only way is this

#if __STDC_VERSION_ _ >= 199901L
printf("%zu\n", sizeof (int));
#else
printf("%u\n", (unsigned)sizeo f (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_Keit h) ks***@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"

[CBFalconer]

somenath wrote:

>
#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)"

>
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>

--
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