Why does my strCopy implementation work?
c++
Solution
Not just under the hood… both `dst` and `src`, despite appearances, actually are pointers! The `[]` syntax in parameter lists is syntactic sugar (or syntactic pepper really) but it's lying to you; these are `char* dst` and `char const* src` for reals.
8.3.5/5 [dcl.fct] Functions:
After determining the type of each parameter, any parameter of type “array of T” or of function type T is adjusted to be “pointer to T”.
Problem
Here is an implementation of strCopy ``` void strcopy2(char *dst, char const *src){ while ((*dst++ = *src++)) ; } ``` Our professor asked us to reproduce this code without using pointers, so I came up with the following function: ``` void strcopy(char dst[], char const src[]){ size_t i = 0; while (dst[i] = src[i++]) ; } ``` It works well, but I realised, that under the hood the function must still be using pointers, as we nowhere return any value. In other words, I though the last function would use pass by value but this is obviously not the case. So what is happening under water, and is there actually any difference between the two methods?