在特定的多线程情况下，我感到困惑，无法找到对此情况的明确解释。在下面的代码中，两个自定义线程正在安全地编写+读取线程，但主线程也正在同时读取。所以这是我的问题：我也必须互斥读取功能吗？还是绝对不可能使应用程序崩溃，例如可能是由于矢量中先前删除的指针引起的？希望你们能帮助我，谢谢！

#include <thread>
#include <mutex>
#include <iostream>
#include <vector>

int g_i = 0;
std::vector<int> test;
std::mutex g_i_mutex;  // protects g_i

void safe_increment()
{
    std::lock_guard<std::mutex> lock(g_i_mutex);
    ++g_i;
    test.resize(test.size() + 1,2);
    std::cout << std::this_thread::get_id() << ": " << g_i << '\n';

    for (std::vector<int>::const_iterator i = test.begin(); i != test.end(); ++i)
        std::cout << std::this_thread::get_id() << " thread vector: " << *i << '\n';
    // g_i_mutex is automatically released when lock
    // goes out of scope
}


void request_threadedvar()
{
    for (std::vector<int>::const_iterator i = test.begin(); i != test.end(); ++i)
        std::cout << std::this_thread::get_id() << " request threaded vector: " << *i << '\n';
}

int main()
{
    std::cout << "main: " << g_i << '\n';
    test.resize(test.size() + 1,1);
    for (std::vector<int>::const_iterator i = test.begin(); i != test.end(); ++i)
        std::cout << "main vector: " << *i << '\n';

    std::thread t1(safe_increment);
    request_threadedvar();
    std::thread t2(safe_increment);

    t1.join();
    t2.join();

    std::cout << "main: " << g_i << '\n';
    for (std::vector<int>::const_iterator i = test.begin(); i != test.end(); ++i)
        std::cout << "main vector: " << *i << '\n';

}