0

我试图OpenFileDialog在下面的代码中使用,但编译器显示异常System::NullReferenceException,一旦我将代码放在trycatch块之间,不会引发异常,但现在显示对话框!?

try 
{ 
    if(openFileDialog1->ShowDialog() == System::Windows::Forms::DialogResult::OK)
    { 
         this->textBox18->Text=openFileDialog1->FileName->ToString();  

    } 
} 
catch(System::NullReferenceException^ e){ e->Message;}
4

1 回答 1

2

在使用它之前,您必须创建OpenFileDialog该类的实例。在此代码段中:

private: System::Windows::Forms::OpenFileDialog^ openFileDialog1; 

您只是声明了一个类型的变量OpenFileDialog(用 初始化null),但您需要先实例化它才能使用它

OpenFileDialog^ openFileDialog1 = gcnew OpenFileDialog;
于 2012-09-16T10:53:54.993 回答