在同一个指针上调用new []两次而不调用delete []会导致内存泄漏？ [英] Does calling new [] twice on the same pointer without calling delete [] in between cause a memory leak?

问题描述

我听说你应该通常删除，当你使用新，但当我运行一个简单的测试程序（下面），它似乎没有什么影响我放在arraySize或numLoops 。这是否会导致内存泄漏？

I've heard that you should usually "delete" whenever you use "new", yet when I run a simple test program (below), it doesn't seem to make a difference which numbers I put for arraySize or numLoops. Does this cause a memory leak?

#include <iostream>

int main()
{
    double *array;
    const int arraySize = 100000;
    const int numLoops = 100000;

    for (int i = 0; i < numLoops; i++)
    {
        // do I need to call "delete [] array;" here?
        array = new double[arraySize];
    }

    int x;
    std::cin >> x; // pause the program to view memory consumption

    delete [] array;

    return 0;
}

推荐答案

不，

每次调用 new 或 new [] ，一些存储器被分配，并且给出该存储器的地址（以便能够使用它）。每个内存最终必须 delete d（或 delete [] d）。

Each time you call new or new[], some memory is allocated, and you are given the address of that memory (in order to be able to use it). Each piece of memory must eventually be deleted (or delete[]d).

您正在将该地址存储在 array 中，但是在下一次迭代时立即覆盖它。因此，你没有办法 delete - 你分配的所有内存块。因此，您有内存泄漏。

You are storing that address in array, but then immediately overwriting it on the next iteration. Therefore you have no way of delete-ing all of pieces of memory that you've been allocated. Therefore you have a memory leak.

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