为什么将DispatchQueue延迟函数放在另一个DispatchQueue中对Swift没有影响

问题描述

预期行为:

对于i = 0，在0秒后调用print语句.

For i = 0, print statement called after 0 second .

对于i = 1，在1.1秒后调用打印语句

For i = 1, print statement called after 1.1 seconds

对于i = 2，在2.2秒后调用打印语句

For i = 2, print statement called after 2.2 seconds

实际行为:

分别在0、1、2、3秒之后调用的打印语句，即内部延迟功能被忽略.

Print statement called after 0, 1, 2, 3 seconds respectively i.e. inner delay function is ignored.

那为什么会有差异呢?

So why the discrepancy?

    for i in 0...3 {

        DispatchQueue.main.asyncAfter(deadline: .now() + .seconds(i), execute: {

            DispatchQueue.main.asyncAfter(deadline: .now() + .seconds(i/10), execute: {
                print("function was called")
            })
        })
    }

推荐答案

有很多方法可以做到这一点.这里是3:

There are a number of ways you could do this. Here are 3:

使用计时器每秒重复一次块

var i = 0
Timer.scheduledTimer(withTimeInterval: 1, repeats: true) { timer in
    i += 1
    print("hello \(i)")
    if i == 5 {
        timer.invalidate()
    }
}

调度多个异步任务

for i in 1...5 {
    DispatchQueue.main.asyncAfter(deadline: .now() + .seconds(i)) {
        print("hello \(i)")
    }
}

注意:这会将所有任务预先排入队列，如果有大量任务，则可能会使队列溢出.

Note: This queues all of the tasks up front and could overflow a queue if there were a large number of them.

在后台运行循环并进入睡眠状态，然后切换到前台进行打印

DispatchQueue.global().async {
    for i in 1...5 {
        sleep(1)
        DispatchQueue.main.async {
            print("hello \(i)")
        }
    }
}

注意:如果循环中的工作要花费大量时间，这会稍微漂移(相隔1秒).

Note: This will drift a little (not be 1 second apart) if the work in the loop takes a significant amount of time.

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