Question

Why is

int *a = new int[10];

written as

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

instead of

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

?

The way I see it, in the second block of code you’re saying a, which was a pointer variable, is now an array. In the third block of code you’re saying the thing a points to is now an array. Why does the second make more sense than the third?

Answer 1

new int[10] returns a pointer to the first element in the array (which is of type int*). It does not return a pointer to the array (which would be of type int(*)[10]).

a = new int[10] means make a point to the first element of the dynamically allocated array. *a is not a pointer at all. It is the object pointed to by the pointer a (which is of type int).

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Note that if you actually had a named array object, the syntax would still be the same:

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

Why? In C++, in most cases, whenever you use an array, it is implicitly converted to a pointer to its initial element. So here, int* a = x is the same as int* a = &x[0];.

(There are several cases where the array-to-pointer decay does not occur, most notably when the array is the operand of the & or sizeof operators; they allow you to get the address of the array and the size of the array, respectively.)