indexing an integer pointer in C++

Question

I'm a beginner in C++ and trying to get my head around every new concept I come across by writting as many various little programs (programmlets) as possible. So I've just concocted the following piece of code:

#include <iostream>
using namespace std;

int main(){
int inumbers[] = {1 ,2 , 3, 4, 5};
int *p;
int i;

p = inumbers;

for(i = 0; p[i]; i++) cout << p[i] << '\n';

return 0;

}

And I fail to understand a seemingly simple thing: how does a compiler "know" when to stop incrementing the loop variable "i"? Surprisingly, the code does work the way it was supposed to.

Answer 1

The compiler does not know when to stop.

for(i = 0; p[i]; i++)

will only stop when p[i] is 0. You just got lucky and p[5] happened to be 0. There is nothing saying that it has to be so and it could keep going like this example. As soon as i >= array size you are accessing memory the array does not own and you enter the world of undefined behavior.

Also if you had a 0 in the array like

int inumbers[] = {1 ,2 , 0, 4, 5};

then the program will only print

1
2

And it will not visit the rest of the array.

When dealing with an array or a standard container you can use a ranged based for loop like

for (const auto & e : inumbers)
    std::cout << e << '\n';

To iterate through it's contents. Then const & is not really needed in this case as we are dealing with an int but it is a good habit to get into as passing by reference avoids copying and making it const prevents you from accidentally modifying the element when doing a read only operation.

When dealing with a pointer you need to provide a size like

for (int i = 0; i < size_of_pointed_to_array, i++)
    std::cout << p[i] << '\n';

Answer 2

You got lucky. By random chance, it appears that there was a zero in memory after the last element in your inumbers array.

Your program is not actually correct. Instead, you might want to limit your loop by counting array members:

for(i = 0; i < sizeof(inumbers)/sizeof(int); i++) cout << p[i] << '\n';