Portable?

[Gibby Koldenhof]

Hiya,

Quick question:

/* test.c, compile with gcc -ansi -pedantic -Wall test.c -o test */

int foo(int a, int b, int c, int d) { return 0; }
int bar(int a, int b, int c, int d) { return 1; }

int main(void)
{
int a = 1;

a = (a ? foo : bar)(0, 1, 2, 3);

return a;
}

Looks conforming to me [the (a ? foo : bar)] ... right?

Cheers,
Gibby Koldenhof

[Martin Johansen]

"Gibby Koldenhof" <flux@__*********@nebule.com> wrote in message
news:42**********************@news.wanadoo.nl...

Hiya,

Quick question:

/* test.c, compile with gcc -ansi -pedantic -Wall test.c -o test */

int foo(int a, int b, int c, int d) { return 0; }
int bar(int a, int b, int c, int d) { return 1; }

int main(void)
{
int a = 1;

a = (a ? foo : bar)(0, 1, 2, 3);

return a;
}

Looks conforming to me [the (a ? foo : bar)] ... right?

Cheers,
Gibby Koldenhof

Yes, I think this is OK, C is beautifull.

I know this is not:

(a ? foo : bar) = NULL;

[Peter Nilsson]

Gibby Koldenhof wrote:

/* test.c, compile with gcc -ansi -pedantic -Wall test.c -o test */

int foo(int a, int b, int c, int d) { return 0; }
int bar(int a, int b, int c, int d) { return 1; }

int main(void)
{
int a = 1;

a = (a ? foo : bar)(0, 1, 2, 3);

return a;
}

Looks conforming to me [the (a ? foo : bar)] ... right?

It's 'conforming', but that doesn't mean much! More importantly,
it is 'strictly conforming'. Although, without any output, this
is often true of programs which would otherwise only be
'conforming'.

But in answer to your real question, you can portably use
(compatible) function pointers as operands to the conditional
operator.

--
Peter