STL deque pop_front()是否会自动回收内存? [英] Will STL deque pop_front() automatically recycle memory?

问题描述

我有一个程序，其中收集一些数据并将它们临时存储在双端队列中

I have a program in which I collect some data and store them temporarily in a deque

    typedef vector<float> floatVector;
    ...
    floatVector * currRecord;
    deque<floatVector *> data;
    ...
    ...

    for (...)
    {
        ...
        currRecord = new floatVector(10); 
        data.push_back(currRecord);
    }

稍后，我想将数据保存到文件

Later, I want to save data to file

    while (data.size() > 0) 
    {
        for (int i=0; i < 10; i++) 
        {
            fprintf(fPtr, "%lf\t", data[0]->at(i) );
        }
    fprintf(fPtr,"\n");
    data.pop_front();
    }

所以，我的问题是，该程序会导致内存泄漏吗?我使用new运算符为每个currRecord向量分配内存. deque pop_front函数是否会自动回收内存?还是我需要放

So, my question is, will this program cause a memory leak? I use new operator to allocate memory for each currRecord vector. Will the deque pop_front function automatically recycle memory? Or do I need to put

    delete [] data[0]

之前

    data.pop_front();

?另外，如果data是向量而不是双端队列，那么一切都一样吗?谢谢！

? Also, if data is a vector instead of a deque, will everything be the same? Thanks!

推荐答案

您有一个std::deque指针，每个指针都拥有一个资源(内存).调用pop_front()将从容器中删除一个指针，但不会释放该指针拥有的内存.由于您使用new分配了内存，因此您还必须调用delete.如果容器是std::vector，则情况不变.

You have a std::deque of pointers and each pointer owns a resource (memory). Calling pop_front() will remove a pointer from the container but it doesn't release the memory the pointer owns. Since you allocate the memory with new you must also call delete. The situation is unchanged if the container is a std::vector.

如果更改为std::deque<floatvector>或类似std::shared_ptr的智能指针容器，则可以避免内存泄漏.

You could avoid memory leaks if you changed to a std::deque<floatvector> or a container of smart pointers like std::shared_ptr.

请注意，调用new时并没有使用[]，因此请使用不带方括号的普通delete.

Note that you didn't use [] when you called new so use plain delete without square brackets.

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