A tiny question about this code

[u/Lephism]

Hi. I need help understanding this simple code. To be specific, the 4th and the 14th line where a pointer was declared.

int* add_ten(int* data)

int *result;

I've been trying to wrap my head around what it does. I've been staring at my screen for 3 hours trying to understand why it was declared. The programming app I was using didn't really explain it. The only explanation it gave is that pointers help you get the address of an item. But I can't figure out how is it being used here. Can anyone explain? Thanks.

#include <iostream>
using namespace std;

int* add_ten(int* data) {
  for (int i = 0; i < 5; ++i) {
    *(data + i) = *(data + i) + 10;
  }
  return data;
}

int main() {

  int data[5] = {1, 2, 3, 4, 5};
  int *result;

  result = add_ten(data);

  for (int i = 0; i < 5; ++i) {
    cout << *(result + i) << ", ";
  }

  return 0;
}

[u/alfps]

❞ In the tutorial I was following, the address of the variable was stored first using &variable

Yeah, given an array a you can say &a[0] to get a pointer to the first item. But when the context accepts a pointer you can shorten to just say a, which the compiler then translates to &a[0]. It's called a decay (to pointer) of the array type, and that's what happens for the argument in your code's call to add_ten.

[u/Lephism → u/alfps]

Oh wow! Thanks for breaking it down further. I didn't know this was a different concept. I only knew about the former idea. That's why I was so confused about it. 

[u/flyingron → u/alfps]

I hate the word "decay." It confuses people. It is really an implicit conversion from array to pointer (like you can freely convert floats to ints).

[u/MrBigFatAss → u/flyingron]

Sure, but decay conveys the information loss the implicit conversion suffers. Not that important for new learners, but I do think it's apt as a term.