返回指向新建数组的指针

Question

指向数组的指针声明为Type (*p)[N];.例如,

int a[5] = { 1, 2, 3, 4, 5 };
int(*ptr_a)[5] = &a;
for (int i = 0; i < 5; ++i){
    cout << (*ptr_a)[i] << endl;
}

将ouptut五个整数a.

如何转换new int[5]为类型int (*p)[5].

例如,当我编写一个返回指向新数组的指针的函数时,以下代码不会编译.

int (*f(int x))[5] {
    int *a = new int[5];
    return a; // Error: return value type does not match the function type.
}

它产生:

error: cannot convert ‘int*’ to ‘int (*)[5]’

Answer 1

您可以使用:

int (*a)[5] = new int[1][5];

例:

#include <iostream>

int main()
{
   int (*a)[5] = new int[1][5];
   for ( int i = 0; i < 5; ++i )
   {
      (*a)[i] = 10*i;
      std::cout << (*a)[i] << std::endl;
   }
   delete [] a;
}

输出:

0
10
20
30
40