What does f(x++) mean?

[becte]

What does the standard say about f(x++) ?
1. x is incremented and then the that value is
used in function f, i.e. x++; f(x);
2. the value of x (before ++) is send to f and
after returning. x is increased, i.e f(x); x++;
3. undefined behavior, it is different for different compilers

I encountered the problem when I was porting code (not written by me)
from AIX to Linux. Visual age and gcc gave different results (2 and 1
respectively), so I had to manually change f(x++); to f(x); x++
in a lot of places because f(x++) was frequently used.

[becte]

"Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote in message news:<Pi******* *************** *************@u nix44.andrew.cm u.edu>...

This argument by itself is okay; but in combination with
the other arguments, which also try to use the value of
'index', it invokes undefined behavior. IOW, it's perfectly
well-defined and okay to write

foo(x++)

but it's entirely wrong and bad to write

foo(x, x++)

or

foo(x++, x)

or any combination of x++ and x without a sequence point
in between them. The ',' separator between function arguments
is *not* a sequence point, even though the comma operator (also
',', but in a different context) *is* a sequence point.

This answers my question.
I wrote a small test program:

#include <stdio.h>

int i = 10;

int f(int a, int b)
{
printf ("a = %d, b = %d, Global i = %d\n", a, b, i);
return 0;
}

int main()
{
f(i, i--);
return 0;
}

In AIX (xlC) I get
a = 10, b = 10, Global i = 9
In Linux (gcc) I get
a = 9, b = 10, Global i = 9

[Peter Nilsson]

"becte" <he************ ***@hotmail.com > wrote in message
news:b4******** *************** ***@posting.goo gle.com...

"Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote in message news:<Pi******* *************** *************@u nix44.andrew.cm u.edu>...

This argument by itself is okay; but in combination with
the other arguments, which also try to use the value of
'index', it invokes undefined behavior. IOW, it's perfectly
well-defined and okay to write

foo(x++)

but it's entirely wrong and bad to write ^^^^^^^^^^^^^^^ ^^^^ ^^^^^^^^
foo(x, x++) ^^^^^^^^^^^

f(i, i--) is no different.

or

foo(x++, x)

or any combination of x++ and x without a sequence point
in between them. The ',' separator between function arguments
is *not* a sequence point, even though the comma operator (also
',', but in a different context) *is* a sequence point.

This answers my question.
I wrote a small test program:

You can't test undefined behaviour.
#include <stdio.h>

int i = 10;

int f(int a, int b)
{
printf ("a = %d, b = %d, Global i = %d\n", a, b, i);
return 0;
}

int main()
{
f(i, i--);
The behaviour of the line is not defined by C. That doesn't merely mean that
the result is not specified, it means that *anything* can happen, including
a subsequent formatting of your harddisk.
return 0;
}

In AIX (xlC) I get
a = 10, b = 10, Global i = 9
In Linux (gcc) I get
a = 9, b = 10, Global i = 9

Your test program is flawed, so it can't demonstrate anything.

In any case, the mere observation that the same program generates different
results on two different implementations does not prove the program is in
error.

#include <stdio.h>

int f(void) { puts("f()"); return 42; }
int g(void) { puts("g()"); return 42; }

int main(void)
{
f() + g();
return 0;
}

This is a legitimate C program which can output one of two possible results.
But the language doesn't specify which one.

Expressions like (i + i++) and foo(i, i++) fall in a much worse category of
behaviour: undefined.

--
Peter