問:有5個(gè)耗時(shí)任務(wù),每次最多執(zhí)行兩次,如何實(shí)現(xiàn)?
1. 創(chuàng)建信號量
/*
創(chuàng)建信號量為2
*/
dispatch_semaphore_t semaphore = dispatch_semaphore_create(2);
2. 創(chuàng)建隊(duì)列
dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);
3.開始異步執(zhí)行
信號量用法解釋:
- 信號量的數(shù)量初始化,最好就是使用要執(zhí)行任務(wù)的多少來定,如題是兩次,則設(shè)置為2
- 每次執(zhí)行一個(gè)任務(wù),調(diào)用方法dispatch_semaphore_wait,讓信號量減1
- 每次結(jié)束一個(gè)任務(wù),調(diào)用方法dispatch_semaphore_signal,讓信號量+1
- 信號量做了這些加減操作,如果信號量小于1,則任務(wù)不會執(zhí)行,如果大于設(shè)置的信號量,則會多執(zhí)行任務(wù)。所以,dispatch_semaphore_wait和dispatch_semaphore_signal應(yīng)該是成對出現(xiàn),才能保證信號量的設(shè)定
注意:
1.可以看出,如果設(shè)置信號量為1,就等同于同步執(zhí)行的效果
2.也有一些用法是設(shè)置信號量為0的,進(jìn)入任務(wù)的時(shí)候signal,離開任務(wù)的時(shí)候wait。
dispatch_async(queue, ^{
dispatch_semaphore_wait(semaphore, DISPATCH_TIME_FOREVER);
NSLog(@"1任務(wù)開始執(zhí)行");
[NSThread sleepForTimeInterval:3];
NSLog(@"1任務(wù)結(jié)束執(zhí)行");
dispatch_semaphore_signal(semaphore);
});
dispatch_async(queue, ^{
dispatch_semaphore_wait(semaphore, DISPATCH_TIME_FOREVER);
NSLog(@"2任務(wù)開始執(zhí)行");
[NSThread sleepForTimeInterval:3];
NSLog(@"2任務(wù)結(jié)束執(zhí)行");
dispatch_semaphore_signal(semaphore);
});
dispatch_async(queue, ^{
dispatch_semaphore_wait(semaphore, DISPATCH_TIME_FOREVER);
NSLog(@"3任務(wù)開始執(zhí)行");
[NSThread sleepForTimeInterval:3];
NSLog(@"3任務(wù)結(jié)束執(zhí)行");
dispatch_semaphore_signal(semaphore);
});
dispatch_async(queue, ^{
dispatch_semaphore_wait(semaphore, DISPATCH_TIME_FOREVER);
NSLog(@"4任務(wù)開始執(zhí)行");
[NSThread sleepForTimeInterval:3];
NSLog(@"4任務(wù)結(jié)束執(zhí)行");
dispatch_semaphore_signal(semaphore);
});
dispatch_async(queue, ^{
dispatch_semaphore_wait(semaphore, DISPATCH_TIME_FOREVER);
NSLog(@"5任務(wù)開始執(zhí)行");
[NSThread sleepForTimeInterval:3];
NSLog(@"5任務(wù)結(jié)束執(zhí)行");
dispatch_semaphore_signal(semaphore);
});
zzZZZ```