Qt智能指针信号槽连接问题

正常new的指针，信号槽如下写法肯定没问题的

  

   

    

     

    
    

     

      ContHeightPropertyDialog  *dia = 
      new 
      ContHeightPropertyDialog(
      this);
     
    
   

    

     

    
    

     

      connect(dia, &ContHeightPropertyDialog::execClicked, 
      this, [ = ]()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     

      qDebug() << 
      "execClicked";
     
    
   

    

     

    
    

     

      });
     
    
   

    

     

    
    

     

      int code = dia->
      exec();
     
    
  

使用智能指针是否也能向上面一样信号连接了，经过测试发现了几个坑

错误1：connect参数1，直接写成智能指针变量

代码写成如下

  

   

    

     

    
    

     

      QScopedPointer<ContHeightPropertyDialog> dia(new ContHeightPropertyDialog(this));
     
    
   

    

     

    
    

     

      connect(dia, &ContHeightPropertyDialog::execClicked, 
      this, [ = ]()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
	
      qDebug() << 
      "execClicked";
     
    
   

    

     

    
    

     

      });
     
    
   

    

     

    
    

     

      int code = dia->
      exec();
     
    
  

这时你将出现如下报错

error: C2664: “QMetaObject::Connection QObject::connect(const QObject *,const char *,const char *,Qt::ConnectionType) const”: 无法将参数 1 从“QScopedPointer<ContHeightPropertyDialog,QScopedPointerDeleter<T>>”转换为“const ProcessPropertyEditor *”
with
[
    T=ContHeightPropertyDialog
]

 正确的写法：使用.data()

  

   

    

     

    
    

     

      QScopedPointer<ContHeightPropertyDialog> dia(new ContHeightPropertyDialog(this));
     
    
   

    

     

    
    

     

      connect(dia.
      data(), &ContHeightPropertyDialog::execClicked, 
      this, [ = ]()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
	
      qDebug() << 
      "execClicked";
     
    
   

    

     

    
    

     

      });
     
    
   

    

     

    
    

     

      int code = dia->
      exec();
     
    
  

错误2：connect参数4，lamba表达式采用传值的方式，调用public 函数

代码写成如下

  

   

    

     

    
    

     

      QScopedPointer<ContHeightPropertyDialog> dia(new ContHeightPropertyDialog(this));
     
    
   

    

     

    
    

     

      connect(dia.
      data(), &ContHeightPropertyDialog::execClicked, 
      this, [ = ]()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     

      qDebug() << 
      "execClicked";
     
    
   

    

     

    
    

     

      dia->
      showResult(
      QStringLiteral(
      "高度差: 11"));
     
    
   

    

     

    
    

     

      });
     
    
  

这时你将出现如下报错，如果是直接new的对象，这么写肯定是没问题的

error: C2248: “QScopedPointer<ContHeightPropertyDialog,QScopedPointerDeleter<T>>::QScopedPointer”: 无法访问 private 成员(在“QScopedPointer<ContHeightPropertyDialog,QScopedPointerDeleter<T>>”类中声明)
with
[
    T=ContHeightPropertyDialog
]

 正确的写法：lambda表达式改为传引用

  

   

    

     

    
    

     

      QScopedPointer<ContHeightPropertyDialog> dia(new ContHeightPropertyDialog(this));
     
    
   

    

     

    
    

     

      connect(dia, &ContHeightPropertyDialog::execClicked, 
      this, [ & ]()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     

      qDebug() << 
      "execClicked";
     
    
   

    

     

    
    

     

      dia->
      showResult(
      QStringLiteral(
      "高度差: 11"));
     
    
   

    

     

    
    

     

      });