std::list为空时调用pop_front的访问越界问题

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#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <list>
#include <unistd.h>
#include <assert.h>

using namespace std;

void Test() // failed
{
    std::list<int *> list;
    int n = 1;
    std::cout << "n address:" << static_cast<void *>(&n) << std::endl;

    while (1)
    {
        list.push_back(&n);
        std::cout << "size:" << list.size() << " front:" << static_cast<void *>(list.front()) << std::endl;
        list.pop_front();
        std::cout << "size:" << list.size() << " front:" << static_cast<void *>(list.front()) << std::endl;

        if (list.empty())
            assert(list.front() == NULL); // 这里会断言失败，若list中为指针时在pop_front前一定要先检查下是否为空，否则会导致访问越界
        usleep(1000*500);
    }
}

void Test2() // pass
{
    std::list<int> list2;
    int n = 1;

    while (1)
    {
        list2.push_back(n);
        std::cout << "size:" << list2.size() << " front:" << list2.front() << std::endl;
        list2.pop_front();
        std::cout << "size:" << list2.size() << " front:" << list2.front() << std::endl;

        if (list2.empty())
            assert(list2.front() == 0); // 这里断言成功
        usleep(1000*500);
    }
}

    int
main( int argc, char **argv )
{
    Test2();
    Test();

    return 0;
}

作者：帅得不敢出门 c++哈哈堂:31843264