Question

Execution of this simple code:

int  foo(int* a){
    cout <<"a="<<a;
    *a=1;
    cout <<", *a="<<*a<<endl;
    return 0;}

int main () {
    int* ptr;
    ptr=new int[2];
    ptr[0]=0;
    ptr[1]=0;

    cout<< foo(ptr) <<" "<< ptr <<" *ptr="<< *ptr <<endl;
    cout<< foo(ptr) <<" "<< ptr <<" *ptr="<< *ptr <<endl;

    return 0;}

Leads to (linux):

a=0x939f008, *a=1
0 0x939f008 *ptr=0
a=0x939f008, *a=1
0 0x939f008 *ptr=1

Please explain why *ptr=0 in the second line, but not in the fourth; could it be, that “things” are “fetched” to cout from right to left? Than – how does it really work (step-by-step at runtime)?

Answer 1

The order of evaluation of arguments to a function is Unspecified as per the C++ Standard.
It may be:

• Left to Right or
• Right to Left or
• Any other order

One of my previous answer here, explains this in depth and detail.