Question

我有3个依赖信号，我想将它们的值组合成一个对象。我想出了两个选择。

选项1：

 + (RACSignal *)createObject {
    RACSignal *paramsSignal = [[[self class] createObject1] flattenMap:^RACStream *(NSString *object1) {
        return [[[self class] createObject2:object1] flattenMap:^RACStream *(NSString *object2) {
            return [[[self class] createObject3:object2] flattenMap:^RACStream *(NSString *object3) {
                return [RACSignal return:RACTuplePack(object1, object2, object3)];
            }];
        }];
    }];

    return [paramsSignal map:^id(RACTuple *tuple) {
        return [[CombinedObject alloc] initWithO1: tuple.first O2: tuple.second O3: tuple.third];
    }];
}

我不喜欢所有的嵌套和封闭。

选项2：

+ (RACSignal *)createObject {
    RACSignal *paramsSignal = [[[[self class] createObject1] flattenMap:^id(NSManagedObjectModel *object1) {
        return [RACSignal combineLatest:@[[RACSignal return:object1], [[self class] createObject2:object1]]];
    }] flattenMap:^RACStream *(RACTuple *tuple) {
        return [RACSignal combineLatest:@[[RACSignal return:tuple.first], [RACSignal return:tuple.second], [[self class] createObject3:tuple.second]]];
    }];


    return [paramsSignal map:^id(RACTuple *tuple) {
        return [[CombinedObject alloc] initWithO1: tuple.first O2: tuple.second O3: tuple.third];
    }];
}

没有嵌套或闭包，但是过多的元组和对象正在通过每个信号......

所以我想知道是否有更优雅的解决方案我不存在。

感谢。

Answer 1

堆栈式方法怎么样？

- (NSArray *) createPhases {
    return @[@selector(createObject1:), @selector(createObject2:), @selector(createObject3:)];
}

...


+ (RACSignal *)createObject {

    NSMutableArray *stack = [NSMutableArray new];
    for (SEL phase: [self createPhases]) {
        stack = [[[self class] performSelector:phase withObject:stack] flattenMap:^RACStream *(NSString *append) {
            return [stack arrayByAddingArray:@[append]]; // or with some other manipulations
        }];
    }


    return [self withObjects:stack map:^id(NSArray *stack) {
        return [[CombinedObject alloc] initWithObjects:stack];
    }];
}

- (...) createObject1:(NSArray *)chain {
    // here chain will be empty, actually (@[])

    return ... // create object 1
}

- (...) createObject2:(NSArray *)chain {
    // here chain will be @[object1]

    return [... ... withObject1:[chain lastObject]]; // create object 2
}

- (...) createObject3:(NSArray *)chain {
    // here chain will be @[object1, object2]

    return [... ... withObject2:[chain lastObject]]; // create object 3
}

链接相关信号并组合它们的值

1 个答案: