问题
I'm trying to understand how to create a dynamic array of pointers in C++.
I understand that new returns a pointer to the allocated block of memory and int*[10] is an array of pointers to int. But why to you assign it to a int**? I'm struggling to understand that.
int **arr = new int*[10];
回答1:
According to the C++ Standard (4.2 Array-to-pointer conversion)
1 An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array.
So for example if you have an array like this
int a[] = { 1, 2, 3, 4, 5 };
then in this declaration
int *p = a;
the array designator used as the initializer is implicitly converted to pointer to its first element.
So in general if you have array
T a[N];
then in expressions with rare exceptions it is converted to pointer to its first element of the type T *.
In this declaration
int **arr = new int*[10];
the initializer is an array elements of which has the type int *. You can introduce a typedef or an alias declaration
typedef int * T;
or
using T = int *;
So you can write
T * arr = new T[10];
that is the pointer arr points to the first element of the dynamically allocated array. As the elements of the array has the type int * then the type of the pointer to an element of the array is int **.
That is the operator new returns pointer to the first element of the dynamically allocated array.
来源：https://stackoverflow.com/questions/47925124/c-dynamic-array-of-pointers