Re: Implementation-defined behaviour

Dann Corbit wrote:
> 
> --- Module:   foo.c (C)
>                             _
>     printf("%d %d\n", n++, n);
> foo.c(8) : Warning 564: variable 'n' depends on order of evaluation

I wonder a bit about this one. Let's consider a simpler version:


#include <stdio.h>


int somefunc(int x)
{
	printf("%d\n", x);
	
	return x;
}


int main(void)
{
	int x= 1;
	
	somefunc(x++);	
	
	return 0;
}



Above, is it well defined that it will be printed "1"?



#include <stdio.h>


int somefunc(int x)
{
	printf("%d\n", x);
	
	return x;
}


int main(void)
{
	int x= 1;
	
	somefunc( (x++, somefunc(x)) );	
	
	return 0;
}


In the above, the expression (x++, somefunc(x)) is evaluated to 2, so I
assume it is guaranteed that it will print "2".