Re: 'C' Operators precedence

Richard B. Johnson (
Wed, 20 May 1998 14:48:42 -0400 (EDT)

On Wed, 20 May 1998, Matthew Kirkwood wrote:

> On Wed, 20 May 1998, Richard B. Johnson wrote:
> And this _without_ optimisation. gcc is right to do this; if a function
> clobbers registers, then let it do so when the variables don't contain
> anything interersting anyway.
> Your point was..?

Sigh. It has nothing to do with clobbering registers. It has to do with
the fact that the function(s) and/or macros, anything that closes (), will
all be 'evaluated' first. To evaluate a function, requires that any
parameters passed to the function be stabilized by a sequence-point, and
the function be called. To evaluate a macro, requires only that it be
expanded as text. However, the () for passed parameters to a macro, if
it exists, requires that the contents of () be evaluated before the
macro is expanded, because () has the highest presedence of any 'C'

#include <stdio.h>

static int a, b, c, d, e;
static int f()
return a;
int main()
b = 1;
c = 2;
d = 3;
e = 4;
printf("%d\n", f() + b + c + d + e );
printf("%d\n", b + f() + c + d + e );
printf("%d\n", b + c + f() + d + e );
printf("%d\n", b + c + d + e + f() );
return 0;


Compile this with -S.
This will show that the function is always called first, regardless
of where it is first seen by the compiler. It must do this because
of the reasons previously stated.

Dick Johnson
Penguin : Linux version 2.1.101 on an i586 machine (66.15 BogoMips).
Warning : It's hard to remain at the trailing edge of technology.

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