在哪里放置动态变量的删除运算符

Question

我试图找出在下面的程序中将delete pointerstatement 放在哪里。我想清除pointer指向的内存空间，以避免内存泄漏。似乎无论我把它放在哪里，我都会收到一条错误消息：

main(8282,0x7fff95d823c0) malloc: *** error for object 0x7fff582d3960: pointer being freed was not allocated
*** set a breakpoint in malloc_error_break to debug

我不确定如何解决这个问题。任何帮助表示赞赏。

完整代码：

#include <iostream>
#include <vector>
using namespace std;

int main() {

    //initialize vector
    vector<int> historyValues;

    //initialize pointer and int variable
    int *pointer;
    pointer = new int;
    pointer = 0;
    int currentValue;

    //make pointer point to the address of currentValue
    pointer = &currentValue;

    //increment pointer by 1 for a total of 10 times.
    //since pointer is pointing at currentValue, currentValue should change also.
    //push back the current value of currentValue into the vector.

    for (int i = 0; i < 10; i++) {

        *pointer += 1;
        historyValues.push_back(currentValue);
    }

    //print final results
    cout << "currentValue: " << currentValue << endl;
    cout << "*pointer: " << *pointer << endl;

    cout << "History of integers stored in currentValue: ";

    for (int i = 0; i < historyValues.size(); i++) {

        cout << historyValues[i] << " ";
    }

    cout << endl;
    cout << "Program finished" << endl;

    return 0;
}

Answer 1

在这个程序中pointer 指代可以释放的动态内存块的唯一位置是在pointer = new int; 和pointer = 0; 之间如果你在这两行之间移动delete pointer，你会没事的。

那么改变的结果是：

pointer = new int;
delete pointer;
pointer = 0;

但是，您最好删除所有这三行并以 pointer = &currentValue; 开头，因为您的代码从不使用动态分配的 int。

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另外，您的评论“增量指针”不正确。您正在递增指针的目标，而不是指针。