使用下拉列表小部件时如何 return 函数值

Question

我在 Ubuntu。当我在下拉列表中单击时，我需要 return 一个函数，但不明白该怎么做。例如：

...
dropdownList.set_active_text("Choose");
dropdownList.signal_changed().connect(sigc::mem_fun(*this, &usb_boot::showing));
std::cout << "var = " << var << std::endl;
...

void usb_boot::showing(){

Gtk::MessageDialog dialogue(*this, dropdownList.get_active_text());
dialog.set_secondary_text("Choose list");
dialog.run();
std::cout << "You choose :\n" << dropdownList.get_active_row_number() << " " << dropdownList.get_active_text() << std::endl;

add return here ?

如何return dropdownList.get_active_text() 和 dropdownList.get_active_row_number() 到变量？

Answer 1

如果我对你的问题的理解正确，你想在处理程序范围之外重用你从处理程序（即 showing）中获得的一些值。为此，我建议使用 lambda 表达式并通过引用捕获变量。例如：

#include <iostream>
#include <gtkmm.h>

int main(int argc, char **argv)
{
    auto app = Gtk::Application::create(argc, argv, "so.question.q63850579");
    
    Gtk::Window window;

    Gtk::ComboBoxText combo;
    combo.append("option 1");
    combo.append("option 2");
    combo.append("option 3");

    // These are your variables, which you want to set to the values the user will choose. They are defined
    // outise the handler:
    std::string choice;
    int row;

    // Here you set the handler. The variables 'choice', 'row' and 'combo' are all passed by reference to
    // the handler (notice the '&'):
    combo.signal_changed().connect([&choice, &row, &combo](){
        choice = combo.get_active_text();
        row = combo.get_active_row_number(); 

        // Inside the scope of the handler, we can see the variable content changing everytime the user
        // changes a value:
        std::cout << "Your current selection is item #" << row << ", which is: " << choice << std::endl;
    });

    window.add(combo);
    window.show_all();

    int returnCode = app->run(window);

    // When the window closes, the variables are read again, but this time from outside the handler's
    // scope. This is possible because they were references:
    std::cout << "Your final selection was item #" << row << ", which is: " << choice << std::endl;

    return returnCode;
}

由于您使用 sigc::mem_fun，另一种选择是将内容保存在 class 的成员变量中，该成员变量表示 this 指向的对象。我个人更喜欢 lambda 表达式，因为这些值可能与 class.

无关

使用下拉列表小部件时如何 return 函数值

How to return a function value when using a dropdown list widget

c++

widget

gtkmm3