滑动选项卡布局上的添加按钮

Question

我想在我的标签布局上添加自定义按钮。此展开按钮将展开我当前的标签视图。我已经实现了谷歌滑动标签中给出的所有功能。

Answer 1

你可以在FrameLayout的帮助下完成它。

#include "stdafx.h"
#include <functional>
#include <iostream>

using namespace std::placeholders;

class Observable
{
public:
    void Register(std::function<void()> Callback)
    {

    }
    std::function<void()> OnStateChanged;

    void ChangeState()
    {
        std::cout << "DevicioClient::ChangeState" << std::endl;
        OnStateChanged();
    }
};

class Observer
{
    Observable client;

    void OnStateChanged()
    {
        std::cout << "DevicioConsumer::OnStateChanged" << std::endl;
    }
public:
    void Run()
    {
    // does work
        client.OnStateChanged = []{
            std::cout << "client.OnStateChanged" << std::endl;
        };
    // doesn't work
    //  OnStateChanged = std::bind(&Observable::OnStateChanged, this);

        client.ChangeState();
    }
};
int _tmain(int argc, _TCHAR* argv[])
{
    Observer consumer;
    consumer.Run();

    getchar();
    return 0;
}

Answer 2

在TabLayout旁边添加按钮是否有意义？管理它（只是一个alignParentRight按钮和tabLayout to_leftOf）要容易得多，结果也是一样的。