objective-c - Objective-C以非阻塞方式实现超时

Question

我正在调用一个外部异步函数，该函数应在完成后立即调用回调。
但是，由于该函数是外部函数，因此我不控制其实现，并且我想以5秒钟为例设置超时，并考虑如果在这5天内未传递给该外部异步函数的回调调用了超时的操作秒。

我目前发现的唯一方法是使当前线程进入休眠状态，从而实际上阻塞了该线程。
这是一个例子:
+(void)myFuncWithCompletion:(void (^ _Nonnull)(BOOL))completion{
BOOL timedOut = NO;
BOOL __block finishedAsyncCall = NO;
[someObj someAsyncMethod {
// completion callback
finishedAsyncCall = YES;
if (!timedOut) {
completion(YES);
}
}];
// This is the logic I want to fix. My goal is to make something similar but non-blocking.
long timeoutInSeconds = 5;
long startTime = [[NSDate date] timeIntervalSince1970];
long currTime = [[NSDate date] timeIntervalSince1970];
while (!finishedAsyncCall && startTime + timeoutInSeconds > currTime) {
[NSThread sleepForTimeInterval:0];
currTime = [[NSDate date] timeIntervalSince1970];
}
if (!finishedAsyncCall) {
timedOut = YES;
completion(NO);
}
}

Answer 1

您可以使用dispatch_after代替-[NSThread sleepForTimeInterval:]

double delayInSeconds = 5.0;
dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(delayInSeconds * NSEC_PER_SEC)); // 1 
dispatch_after(popTime, dispatch_get_main_queue(), ^(void){ // 2 
    if (!finishedAsyncCall ) {
        timedOut = YES;
        completion(NO);
    } 
});