QtTaskTree::Iterator Class
class QtTaskTree::Iterator用作 For 元素内部迭代器的基类。更多内容...
| 头文件： | #include <qtasktree.h> |
| CMake： | find_package(Qt6 REQUIRED COMPONENTS TaskTree) target_link_libraries(mytarget PRIVATE Qt6::TaskTree) |
| qmake： | QT += tasktree |
| 自： | Qt 6.11 |
| 由以下版本继承： | QtTaskTree::ForeverIterator、QtTaskTree::ListIterator 、QtTaskTree::RepeatIterator 以及QtTaskTree::UntilIterator |
| 状态： | 技术预览 |
该类处于技术预览阶段，内容可能会有变动。
注意：此类中的所有函数均为可重入函数。
公共函数
另请参阅 For 、ForeverIterator 、RepeatIterator 、UntilIterator 和ListIterator 。
成员函数文档
[constexpr noexcept default] Iterator::Iterator(QtTaskTree::Iterator &&other)
通过Move构造一个Iterator 的实例。
qsizetype Iterator::iteration() const
返回在以下结构中Do 主体内当前正在执行的处理程序的迭代索引：For (Iterator) >> 。请仅在位于食谱Do 主体内的任何GroupItem 元素的处理程序主体中调用此函数，否则可能会导致程序崩溃。请确保已将Iterator 传递给For 元素。
使用示例：
constQList<std::chrono::seconds>timeouts={5s, 1s, 3s};
constListIterator iterator(timeouts);
const autoonSetup= [iterator](std::chrono::milliseconds &timeout) {
timeout= *iterator;
qDebug() << "Starting" << iterator.iteration() << "iteration with timeout"
<< *iterator << "秒。";
};
const autoonDone= [iterator]{
qDebug() << "Finished" << iterator.iteration() << "iteration with timeout"
<< *迭代器 << "秒。";
};
constGroup sequentialRecipe=For(迭代器)>>Do {
QTimeoutTask(onSetup,onDone)
};
constGroup parallelRecipe=For(iterator)>>Do {
parallel,
QTimeoutTask(onSetup,onDone)
};执行sequentialRecipe 时的输出结果为：
Starting 0 iteration with timeout 5s seconds.
Finished 0 iteration with timeout 5s seconds.
Starting 1 iteration with timeout 1s seconds.
Finished 1 iteration with timeout 1s seconds.
Starting 2 iteration with timeout 3s seconds.
Finished 2 iteration with timeout 3s seconds.在顺序模式下，done 处理程序中迭代索引的顺序将得到保证。
执行parallelRecipe 时的输出结果为：
Starting 0 iteration with timeout 5s seconds.
Starting 1 iteration with timeout 1s seconds.
Starting 2 iteration with timeout 3s seconds.
Finished 1 iteration with timeout 1s seconds.
Finished 2 iteration with timeout 3s seconds.
Finished 0 iteration with timeout 5s seconds.在并行模式下，done处理程序中迭代索引的顺序无法保证保持不变，且取决于任务完成的顺序。并行Do 主体中done处理程序返回的迭代索引与原始迭代中对应setup处理程序的索引相匹配，因此后续done处理程序中迭代索引的顺序可能并非递增。
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