Question

有些人可以解释为什么发生这种情况? 我写道:

(1) 窗口管理类别,作为单一州实施,有<条码>(Instance()方法,其定义如下:

static WindowManager * instance_;
//...
WindowManager * WindowManager::instance_ = 0;
WindowManager & WindowManager::Instance()
{
    if (!instance_)
        instance_ = new WindowManager();
    return *instance_;
}

2) aWindowManager:createWindow 回归提及新建立的窗口的方法,如:

Window & WindowManager::createWindow()
{
    windows_.push_back(Window());
    return windows_.at(windows_.size() - 1);
}

3)a 在窗口内打印信息的方法

在我的主要方案中,我撰写了以下文件:

ui::Window & win1 = ui::WindowManager::Instance().createWindow();
ui::Window & win2 = ui::WindowManager::Instance().createWindow();
win1.print("First window");
win2.print("Second window");

This does not work! Only the second call to print is executed (for win2). However, if I change the order, like so:

ui::Window & win1 = ui::WindowManager::Instance().createWindow();
win1.print("First window");
ui::Window & win2 = ui::WindowManager::Instance().createWindow();
win2.print("Second window");

那么,所有事情都是按预期运作的。如果任何人能够说明这种情况,将非常感谢任何帮助。

Answer 1

在这里,你有迅速的例子说明问题:

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

struct T
{
    int id;
    T(int id) : id(id)
       { cout << "created " << id << endl; }
    T(T const& t) : id(t.id)
       { cout << "copy: " << t.id << endl; }
    void print(char const* m)
       { cout << id << ": " << m << endl; }
};

vector<T> ts;

T& create(int id)
{
    ts.push_back(T(id));
    return ts.at(ts.size() - 1);
}

int main()
{
    // Uncomment these lines and compare results
    //int const max_windows = 10;
    //ts.reserve(max_windows);

    T& t1 = create(1);
    T& t2 = create(2);
    t1.print("t1");
    t2.print("t2");
}

Compile and run as it is and see what is printed to stdout. Then uncomment reserve() call, compile and run once again and compare with previous output.

问题是<代码>std:vector:push_back造成数据内部重新分配。这使你在重新定位之前获得的病媒要素的所有参考资料、点或探测器都失效。

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