C ++从函数返回多维数组 [英] C++ Returning multidimension array from function

问题描述

如何返回隐藏在私有字段中的多维数组？

How do I return a multidimensional array hidden in a private field?

class Myclass {
private:
 int myarray[5][5];
public:
 int **get_array();
};

........

int **Myclass::get_array() {
 return myarray;
}

无法转换 int（*）[5] 5] 到 int ** 在返回test.cpp / Polky / src line 73 C / C ++ Problem

cannot convert int (*)[5][5] to int** in return test.cpp /Polky/src line 73 C/C++ Problem

推荐答案

二维数组不会衰减为指向int的指针。它衰减到指向int数组的指针 - 也就是说，只有第一维衰减为指针。指针不指向int指针，当以指针的大小递增时，指向5个整数的数组。

A two-dimensional array does not decay to a pointer to pointer to ints. It decays to a pointer to arrays of ints - that is, only the first dimension decays to a pointer. The pointer does not point to int pointers, which when incremented advance by the size of a pointer, but to arrays of 5 integers.

class Myclass {
private:
    int myarray[5][5];
public:
    typedef int (*pointer_to_arrays)[5]; //typedefs can make things more readable with such awkward types

    pointer_to_arrays get_array() {return myarray;}
};

int main()
{
    Myclass o;
    int (*a)[5] = o.get_array();
    //or
    Myclass::pointer_to_arrays b = o.get_array();
}

指向指针的指针（ int ** <

A pointer to pointer (int**) is used when each subarray is allocated separately (that is, you originally have an array of pointers)

int* p[5];
for (int i = 0; i != 5; ++i) {
    p[i] = new int[5];
}

这里我们有一个5个指针数组，每个指针指向第一个项目一个单独的内存块，共有6个不同的内存块。

Here we have an array of five pointers, each pointing to the first item in a separate memory block, altogether 6 distinct memory blocks.

在二维数组中，您将获得一个连续的内存块：

In a two-dimensional array you get a single contiguous block of memory:

int arr[5][5]; //a single block of 5 * 5 * sizeof(int) bytes

这些东西的布局完全不同，因此这些东西不能以同样的方式返回和传递。

You should see that the memory layout of these things are completely different, and therefore these things cannot be returned and passed the same way.

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