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起初，我的思路是，先取得Layout的items数量， 然后通过索引来移除每一个items，代码如下：

    QHBoxLayout * hly = new QHBoxLayout;

    for(int i = 0; i < 5; i++)
    {
        QPushButton * btn = new QPushButton;
        hly->addWidget(btn);
    }

    int hlyCount = hly->count();
    qDebug()<<"hlyCount =="<<hlyCount;

    for(int i = 0; i < hlyCount; i++)
    {
        QLayoutItem * item = hly->itemAt(i);
        hly->removeItem(item);
        qDebug()<<"remove item "<<i;
    }

    hlyCount = hly->count();
    qDebug()<<"hlyCount =="<<hlyCount;

而输出结果有些意外：

11:21:42: Starting E:\Qt_projcet\play\build-play-Desktop_Qt_5_12_2_MinGW_64_bit-Debug\debug\play.exe...
hlyCount == 5
remove item  0
remove item  1
remove item  2
remove item  3
remove item  4
hlyCount == 2
11:21:58: E:/Qt_projcet/play/build-play-Desktop_Qt_5_12_2_MinGW_64_bit-Debug/debug/play.exe exited with code 0

查看Qt帮助手册有解释，itemAt()有三点值得关注：

1-If there is no such item, the function must return 0
1-如果子项不存在，返回零

2-Items are numbered consecutively from 0
2-item的排序从零开始

3-If an item is deleted, other items will be renumbered.
3-如果子项被删除，其他子项将被从新排序

哇~看到这里，知道了答案。下面看一下更改后的代码：

    QHBoxLayout * hly = new QHBoxLayout;

    for(int i = 0; i < 5; i++)
    {
        QPushButton * btn = new QPushButton;
        hly->addWidget(btn);
    }

    int hlyCount = hly->count();
    qDebug()<<"hlyCount =="<<hlyCount;

    for(int i = hlyCount - 1; i >= 0 ; i--)    //###改动：items编号，从大到小遍历
    {
        QLayoutItem * item = hly->itemAt(i);
        if (item != nullptr)                  //###改动：判断子项是否存在
            hly->removeItem(item);
        qDebug()<<"remove item "<<i;
    }

    hlyCount = hly->count();
    qDebug()<<"hlyCount =="<<hlyCount;

运行结果：

11:44:34: Starting E:\Qt_projcet\play\build-play-Desktop_Qt_5_12_2_MinGW_64_bit-Debug\debug\play.exe...
hlyCount == 5
remove item  4
remove item  3
remove item  2
remove item  1
remove item  0
hlyCount == 0
11:44:43: E:/Qt_projcet/play/build-play-Desktop_Qt_5_12_2_MinGW_64_bit-Debug/debug/play.exe exited with code 0

ok了