conditional expression

[chandanlinster]

consider the following program:

#include <stdio.h>
#include <stdlib.h>

int
main(void)
{
int i;
double f;

printf("size = %u", sizeof((0)?f:i) ); /* 0 is zero*/

putchar('\n');
exit(0);
}

output:
size = 8

since 'i' is supposed to be an "int", on my 32-bit system I expected
the size to 4 bytes. Why does this happen?

[Marc Boyer]

Re: conditional expression

Le 06-09-2006, chandanlinster <chandanlinster @gmail.coma écrit :

consider the following program:
>
#include <stdio.h>
#include <stdlib.h>
>
int
main(void)
{
int i;
double f;
>
printf("size = %u", sizeof((0)?f:i) ); /* 0 is zero*/
>
putchar('\n');
exit(0);
}
>
output:
size = 8
>
since 'i' is supposed to be an "int", on my 32-bit system I expected
the size to 4 bytes. Why does this happen?

Because 0?f:i is an expression, it must have a type, staticaly.
In this case, the type is certainly 'double'. If you want details,
you should have a look on the 'implicit conversion' chapter in
a good book.

Marc Boyer

[lovecreatesbea...@gmail.com]

Re: conditional expression


chandanlinster wrote:

consider the following program:
>
#include <stdio.h>
#include <stdlib.h>
>
int
main(void)
{
int i;
double f;
>
printf("size = %u", sizeof((0)?f:i) ); /* 0 is zero*/
>
putchar('\n');
exit(0);
}
>
output:
size = 8
>
since 'i' is supposed to be an "int", on my 32-bit system I expected
the size to 4 bytes. Why does this happen?

This is the same result on my computer. The operand of the sizeof is
(0) ? f : i which has the type same as f. "The C programming language,
2nd", sec 2.11 has an example on this.

[SM Ryan]

Re: conditional expression

"chandanlinster " <chandanlinster @gmail.comwrote :
# consider the following program:
#
# #include <stdio.h>
# #include <stdlib.h>
#
# int
# main(void)
# {
# int i;
# double f;
#
# printf("size = %u", sizeof((0)?f:i) ); /* 0 is zero*/

Even if the compiler optimise the expression, the type
of the conditional is double and the integer value
must be converted to double.

--
SM Ryan http://www.rawbw.com/~wyrmwif/
GERBILS
GERBILS
GERBILS

[Peter Shaggy Haywood]

Re: conditional expression

Groovy hepcat chandanlinster was jivin' on 6 Sep 2006 02:00:18 -0700
in comp.lang.c.
conditional expression's a cool scene! Dig it!

> printf("size = %u", sizeof((0)?f:i) ); /* 0 is zero*/

In addition to what else you've been told, this causes undefined
behaviour. You have a %u conversion specifier with a size_t argument.
size_t is an implementation defined unsigned integer type. Thus, it
might be unsigned int, but it might not. It's entirely possible that
it is some other unsigned integer type.
The usual way of handling this is well known (See question 12.9b in
the FAQ.), and has been posted in this newsgroup recently.

--

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