Function passing arguments in reverse

Question

Here is my function:

void abc(char  *def, unsigned int w, unsigned int x, unsigned int y, unsigned int z)
{
   printf("val 1 : %d\n", w);
   printf("val 2 : %d\n", x);
   printf("val 3 : %d\n", y);
   printf("val 4 : %d\n", z);
}

and here is where I call this function:

unsigned int exp[4] = { 1, 2, 3, 4 };
unsigned short count = 0;
abc(anyarray, exp[count++], exp[count++], exp[count++], exp[count++]);

and here is the output that I expect:

val1 : 1
val2 : 2
val3 : 3
val4 : 4

but what I get is completely reverse of it:

val1 : 4
val2 : 3
val3 : 2
val4 : 1

I don't know why? Any help would be appreciated.

Answer 1

From standard docs, 5.4

Except where noted, the order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is unspecified58) Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored. The requirements of this paragraph shall be met for each allowable ordering of the subexpressions of a full expression; otherwise the behavior is undefined.

An example from the Standard docs itself,

i = v[i ++]; / / the behavior is undefined

And it is for the very same reason that

abc(anyarray, exp[count++], exp[count++], exp[count++], exp[count++]); is undefined..

Answer 2

You should not use the ++ operator, operating on the same variable, more than once in the same statement. The order in which the operation will be performed is not defined.

Try:

abc(anyarray, exp[count], exp[count+1], exp[count+2], exp[count+3]);  
count += 4;