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About Pointers and arrays in C++

I would like to ask a question about pointers and arrays in C++.

int a[10];

int *p1; p1 = &a[0];

int *p2; p2 = a;

int (*p3)[10]; p3 = &a;

What are the differences between p1, p2 and p3? They are very confusing.

like image 902
redcap Avatar asked Dec 09 '22 16:12

redcap


2 Answers

Firstly, a is an array of 10 ints. 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 ints (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
like image 174
Joseph Mansfield Avatar answered Dec 11 '22 05:12

Joseph Mansfield


p1 points to the first element of a.

p2 causes a to decay into a pointer to the first element, so it also points to the first element of a.

p3 is read as a pointer to an array of ten integers. It points to a itself, and incrementing it will move it ahead by sizeof(int) * 10 instead of sizeof(int).

like image 33
chris Avatar answered Dec 11 '22 06:12

chris