当我在 C++ 中擦除我的向量时，我遇到了双重错误

Question

如下代码。

我想把向量中的元素移到后面。

eg: [(1),2,3,4] -> [2,3,4,(1)]

但是，它会导致双重释放问题。这段代码的逻辑很简单。

我想我误用了擦除功能。是真的吗？谁能告诉我详情吗？

感谢您的阅读。

这是输出：

*** Error in '/home/ubuntu/workspace/hello-cpp-world.cc.o': double free or corruption (out): 0x00000000016ffca0 ***

这是代码片段：

#include <iostream>
#include <vector>

int main() {
    std::vector<int*> targets;

    int* a = new int;
    *a = 1;
    int* b = new int;
    *b = 2;
    targets.push_back(a);
    targets.push_back(b);
    int i =0;
    for (std::vector<int*>::iterator obj1 = targets.begin(); obj1 != targets.end(); i++)
    {
        if(i==0)
        {
            int* tmp = *obj1;
            targets.push_back(tmp);
            targets.erase(obj1);
        }
        else obj1++;
    }

}

Answer 1

调用 push_back 或 erase 到 std::vector 可能会使迭代器无效。使用索引更容易。

#include <iostream>
#include <vector>

int main() {
    std::vector<int*> targets;

    int* a = new int;
    *a = 1;
    int* b = new int;
    *b = 2;
    targets.push_back(a);
    targets.push_back(b);
    int i =0;
    for(size_t obj1 = 0; obj1 < targets.size(); i++)
    {
        if(i==0)
        {
            int* tmp = targets[obj1];
            targets.push_back(tmp);
            targets.erase(targets.begin() + obj1);
        }
        else obj1++;
    }

}

因为 obj1 除了在 i==0 时递增之外没有其他用途，你可以写得更简单

#include <iostream>
#include <vector>

int main() {
    std::vector<int*> targets;

    int* a = new int;
    *a = 1;
    int* b = new int;
    *b = 2;
    targets.push_back(a);
    targets.push_back(b);

    int* tmp = targets[0];
    targets.push_back(tmp);
    targets.erase(targets.begin());

}

当我在 C++ 中擦除我的向量时，我遇到了双重错误

I meets double free error when I erase my vector in c++

c++

vector

std

erase

double-free