Question

我对obj c比较新，所以对于我的学校工作，我需要在（相同）函数中执行三行之间添加延迟。是否有任何有利的选择？

line 1: [executing first operation];

line 2: Delay  /* I need to introduce delay here */

line 3: [executing second operation];

line 4: Delay  /* I need to introduce delay here again */

line 5: [executing second operation];

在此先感谢，任何帮助将不胜感激。

Answer 1

所以我找到的解决方案是：

[operation 1]
    double delayInSeconds = 9.3;
    dispatch_time_t popTime2 = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(delayInSeconds * NSEC_PER_SEC));
    dispatch_after(popTime, dispatch_get_main_queue(), ^(void){
        [operation 2];

double delayInSeconds2 = 9.3;
dispatch_time_t popTime2 = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(delayInSeconds2 * NSEC_PER_SEC));
dispatch_after(popTime2, dispatch_get_main_queue(), ^(void){
   [operation 3];

 });

 });

这是正确的方法吗？如果是的话，无论如何我都不需要在操作2的延迟中嵌套操作3

Answer 2

你可以通过NSRunLoop方法实现目标 - （void）performSelector：（SEL）aSelector withObject :( nullable id）anArgument afterDelay：（NSTimeInterval）delay;

[executing first operation];
[self performSelector:@selector(2nd_Operation) withObject:nil 
 afterDelay:1st_Delay_In_Seconds];
[self performSelector:@selector(3rd_Operation) withObject:nil 
 afterDelay:2nd_Delay_In_Seconds];

请注意，如果您将上述代码放在一个方法中，则两个延迟时间几乎同时开始。如果你想在0.5s延迟后调用2nd_operation然后在另外0.5s延迟后调用3rd_operation，那么2nd_Delay_Time应该是0.5s + 2nd_operation执行时间+0.5s。如果由于某种原因需要准确的延迟，我建议将最后一行放在2nd_operation的完整块中。

在操作之间增加延迟（obj c）

2 个答案: