ios - 为什么这个连接不起作用？

Question

也许我误解了concat？为什么这段代码不输出“B”我只得到一个输出“A”不应该像这样连接:

RACSignal *signal = [RACSignal createSignal:^ RACDisposable * (id<RACSubscriber>  subscriber) {
    [subscriber sendNext:@"A"];
    return nil;
}];

signal = [signal concat:[RACSignal createSignal:^ RACDisposable * (id<RACSubscriber> subscriber) {
    [subscriber sendNext:@"B"];
    return nil;
}] ];


signal= [signal doNext:^(id x){
    NSLog(@"doNext  %@",x);
}];

signal= [signal flattenMap:^RACStream *(id value) {
    NSLog(@"flattenMap %@", value);
   return [RACSignal return:value];
}];

signal=[signal doNext:^(id x){
    NSLog(@"doNext after flattenMap %@",x);
}];


[signal subscribeNext:^(id x) {
    NSLog(@"subscribeNext  %@", x);
}];

Answer 1

您的第一个信号永远不会完成。如果你在调用 [subscriber sendNext:@"A"] 之后调用 [subscriber sendCompleted]，那么你会发现 -concat: 做了什么你期待的。