Question

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 1

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?.