删除二维数组C ++ [英] Delete 2D array C++

问题描述

是释放一个二维数组，这两种方法相似？

Are these two methods of freeing a 2D array similar?

int** M = new int*[5];

for (int i = 0; i < 5; ++i)
    M[i] = new int[3];

for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 3; ++j) {
        M[i][j] = i + j;
    }
}

删除我：

for (int i = 0; i < 5; ++i)
    delete [] M[i];
delete [] M;

和删除二：

delete [] *M;
delete [] M;

是这两个code相同呢？

Are these two code equivalent?

推荐答案

删除[] * M; 相同删除[]中号[0] ，所以它是不等同于删除所有 M [I] 在一个循环中，因为只有第一个将被删除。循环是为了避免内存泄漏的正确方法。

delete [] *M; is the same as delete [] M[0], so it is not equivalent to deleting all M[i] in a loop because only the first one would be deleted. The loop is the correct way to avoid memory leaks.

或者更好，使用的std ::矢量代替手工分配，你将不必担心删除指针。

Or better, use std::vector instead of manual allocations and you won't need to worry about deleting pointers.

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