修改函数中的2d数组 [英] modifying 2d array in a function

问题描述

所以我写了一个函数，将一个二维数组90度，我已经告诉IRC，我不能通过引用传递二维数组（例如void test（char A [] [10]& ）），我只是应该通过我的数组通常的方式，但是当我这样做，这个函数不会改变实际的数组。那么如何在函数中修改我的原始数组呢？

So I've written a function that turns a 2D array 90 degrees and I've been told on IRC that I can't pass 2D array by reference (for example void test(char A[][10]&)) and that I just should pass my array the usual way, however when I do that, this function doesn't change the actual array. So how do I modify my original array in a function ?

void one(char A[][10], int N)
{
    char B [10][10];
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            B[j][N-i-1] = A[i][j];
    A = B;
}

推荐答案

A = B; 不会将数组B的元素永久复制到A.永久更改为 A 元素的无效分配。 A 保留函数返回时的原始值。你需要做一个成员明智的副本永久复制 B 的元素到 A 。

A = B ; doesn't copy the elements of array B to A permanently. It's an invalid assignment to change to elements of A permanently. A retains it's original values upon function return. You need to do a member wise copy to permanently copy the elements of B to A.

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