About Pointers and arrays in C++

arrays, c, c++, pointers

Solution

Firstly, `a` is an array of 10 `int`s. That's the easy part.

`p1` is an "pointer to `int`". You are assigning to it the value of `&a[0]`. This takes the address of the first element of `a`. So `p1` now points to the first element of `a`.

`p2` is also an "pointer to `int`". You are assigning `a` directly to it. In this case, a standard conversion must take place called array-to-pointer conversion. Basically, an array can be converted to a pointer to its first element. You are assigning the result of this conversion to `p2`. So `p2` is also a pointer to the first element of `a`.

`p3` is a "pointer to array of 10 `int`". You are taking the address of the array `a` and assigning it to this pointer. So now this pointer points at the array itself (not the first element of it).

You might think "well the first element has the same address as the array, so what's the difference?" In fact, you'll notice the difference when you try to increment the pointer. Incrementing either `p1` or `p2` will give you a pointer to the second element of the array. Incrementing `p3` will give you a pointer to the next array of 10 `int`s (which doesn't actually exist).

┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
│ int │ int │ int │ int │ int │ int │ int │ int │ int │ int │ 
└─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘
   ^
   └── p1, p2, p3

So if you start off with the all pointing as you have described and then increment them, you get:

┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬┄
│ int │ int │ int │ int │ int │ int │ int │ int │ int │ int │ 
└─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴┄
         ^                                                     ^
         └── p1, p2                                            └── p3

Problem

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