How are arguments evaluated in a function call?

Question

Consider this code:

void res(int a,int n)
{
    printf("%d %d, ",a,n); 
}

void main(void) 
{
    int i; 
    for(i=0;i<5;i++)
        res(i++,i);
    //prints 0 1, 2 3, 4 5

    for(i=0;i<5;i++)
        res(i,i++);
    //prints 1 0, 3 2, 5 4
}

Looking at the conclusion, it seems that the arguments are not evaluated right to left every time. What exactly is going on here?

Answer 1

The evaluation order of the arguments in the function call is not specified. The compiler can evaluate them depending on what order it can decide.

From the standard C99 6.5.2.2/10 "Functional Calls / Semantics":

The order of evaluation of the function pointer, the actual arguments and the subexpressions in the actual arguments are not defined, but there is a sequence point before the actual call.

If you need to provide a specific order, using time series is the usual workaround:

int i; 
for(i=0;i<5;i++) {
    int tmp = i;
    int tmp2 = i++;

    res(tmp2,tmp);
}

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. , , . (6.5/2 "" )

Answer 2

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