标签：++ 运行 ons i++ mat add 判断 从零开始 hbox
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
标签：++ 运行 ons i++ mat add 判断 从零开始 hbox
原文地址：https://www.cnblogs.com/azbane/p/11617138.html
