About the problems encountered when using the timer class to stop with a button (why does the QPushButton (for the first time) need to be clicked twice to respond?)

 Why? QPushButton（ for the first time ） It takes two clicks to respond ？

I found a problem when learning the timer class （ I wonder if ？）.
ask ： When clicking pause 1 when , Mouse to double-click , The timer will stop , see Qt Assistant pair clicked Obviously not

translate ： When the button is activated ( namely , Press and release when the mouse cursor is inside the button ), When you type a shortcut key , Or when click() or animateClick() When called , This signal is sent . It is worth noting that , If the setDown()、setChecked() or toggle(), This signal will not be sent .

understand ： Shouldn't it be a mouse click ？

#include "widget.h"
#include "ui_widget.h"

Widget::Widget(QWidget *parent) :
    QWidget(parent),
    ui(new Ui::Widget)
{
    
    ui->setupUi(this);
    // Create a timer object 
    timer1 = new QTimer(this);
    timer2 = new QTimer(this);
    // Object on 
    timer1->start(500);
    timer2->start(1000);
    int num1=0;
    int num2=0;
    // When the timer is turned on, a timeout The signal 
    connect(timer1,&QTimer::timeout,[=]()mutable{
    
        ui->label->setText(QString::number(num1++));
    });

    connect(timer2,&QTimer::timeout,[=]()mutable{
    
        ui->label_2->setText(QString::number(num2++));
    });
}

Widget::~Widget()
{
    
    delete ui;
}


void Widget::on_pushButton_clicked()// Pause 1
{
    
    connect(ui->pushButton, &QPushButton::clicked,this,[=](){
    
        timer1->stop();// Click twice to stop  ？？？
    });
}

void Widget::on_pushButton_2_clicked()// continue 1
{
    
    connect(ui->pushButton_2, &QPushButton::clicked,this,[=](){
    
        timer1->start(500);
    });
}

** attach ：** This is my in windows Up operation Qt（5.7.0） When I met

** reflection ：** This sum doubleClicked What's the difference ？
This signal is emitted when the button is interactively double clicked by the user.
When the user interactively double clicks the button , Will send this signal .

Now record .

** solve ：** The above problem is caused by turning to the slot , If you write the code of clicking the button into the constructor, there will be no problem .