Parameter evaluation order before a function calling in C

c parameters evaluation
corto picture corto · Dec 17, 2008 · Viewed 44.1k times · Source

Can it be assumed a evaluation order of the function parameters when calling it in C ? According to the following program, it seems that there is not a particular order when I executed it. 

#include <stdio.h>

int main()
{
   int a[] = {1, 2, 3};
   int * pa; 

   pa = &a[0];
   printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa), *(pa++),*(++pa));
   /* Result: a[0] = 3  a[1] = 2    a[2] = 2 */

   pa = &a[0];
   printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(pa),*(++pa));
   /* Result: a[0] = 2  a[1] = 2     a[2] = 2 */

   pa = &a[0];
   printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(++pa), *(pa));
   /* a[0] = 2  a[1] = 2 a[2] = 1 */

}

Answer

Grant Wagner picture Grant Wagner · Dec 17, 2008

No, function parameters are not evaluated in a defined order in C.

See Martin York's answers to What are all the common undefined behaviour that c++ programmer should know about?.

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