What is the type of an int array?

Question

I am reading about arrays in C++. I tried the following code:

int main()
{
   int a[10];
   int *p;
   p = &a;
}

I got compiler error:

pointers.cpp:10:6: error: cannot convert ‘int (*)[10]’ to ‘int*’ in assignment
p = &a;

In order to understand the array type to be able to assign to a pointer I tried this code:

int main()
{
   int a[10];
   int *r[10];
   r = a;
}

Compilation error:

: error: incompatible types in assignment of ‘int [10]’ to ‘int* [10]’
  r = a;

Then I tried this:

int main()
{
   int a[10];
   int *r[10];
   r = &a;
}

Compilation error:

  error: incompatible types in assignment of ‘int (*)[10]’ to ‘int* [10]’
  r = &a;

What is the type int (*)[10] ?

@dandan78 The question is about C++, not about C. Arrays don't work exactly the same in the two languages. — , Dec 19 '14 at 12:32
Just don't try with C++, if you don't know what you are doing. Your house could fall. — nbro, Dec 19 '14 at 12:39

score 3 · Accepted Answer · answered Dec 19 '14 at 12:31

int (*)[10] is a pointer to an array of 10 ints. It points at the entire array. In your first code, when you do &a, you are getting the address of the entire array, which has this type. If you wanted to store that, you'd have to write:

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

In the code that you tried, int *r[10] is instead an array of pointers (rather than a pointer to an array).

An int* on the other hand, points at a single int. You could, for example, take the address of one of your array elements, &a[0], and it would be an int*.

int *p = &a[0];

In fact, the name of an array often undergoes an implicit conversion called array-to-pointer conversion in which it becomes a pointer to the first element in the array. We typically say that the array decays to a pointer to its first element.

int *p = a; // Here, a is decaying

score 2 · Answer 2 · answered Dec 19 '14 at 12:31

2

The right way to go about this is to pass the address of the first element:

int a[10];
int *p;
p = &a[0];

The others can be accessed by incrementing p:

p++;

answered Dec 19 '14 at 12:31

karlphillip

92,053
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Mike Seymour · Answer 3 · 2014-12-19T12:39:07.767

What is the type of an int array?

There's a different type for each size. For example, the type of an array of ten is int[10]. There's also an incomplete type int[], representing an array of unknown size.

What is the type int (*)[10] ?

That's a pointer to an array of size 10.

In your first snippet, &a is a pointer to the array, and you can't assign that to a pointer to a single int, since int and int[10] are incompatible types. In the other examples, r is an array of pointers, and you can't assign anything to an array.

You could use either of these:

int *p = a;      // implicit array-to-pointer conversion
int *p = &a[0];  // explicitly take address of element

to assign a pointer to the first element of the array. Or if you want a pointer to the array (perserving the size in its type), that's

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

The extra () indicate that this is a pointer to an array, not an array of pointers.

score 1 · Answer 4 · answered Dec 19 '14 at 12:34

1

You simply have to do

int a[10];
int *p;
p = a;

assign p the adress of array a; i-e address of its first element.

answered Dec 19 '14 at 12:34

Ali Kazmi

1,460
9
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score 0 · Answer 5 · answered Dec 19 '14 at 12:33

0

As the error says, its int (*)[10], so you should use int (*r)[10] instead of int *r[10]. The extra () are the difference between a pointer to an array of 10 ints and an array of 10 pointers to ints.

answered Dec 19 '14 at 12:33

Zaskar

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3
10

What is the type of an int array?

5 Answers5