不使用 typedef 声明块方法参数

Question

是否可以在 Objective-C 中不使用 typedef 来指定方法块参数？它一定是，就像函数指针一样，但如果不使用中间 typedef，我就无法找到获胜的语法：

typedef BOOL (^PredicateBlock_t)(int);
- (void) myMethodTakingPredicate:(PredicateBlock_t)predicate

只有上面的编译，所有这些都失败：

-  (void) myMethodTakingPredicate:( BOOL(^block)(int) ) predicate
-  (void) myMethodTakingPredicate:BOOL (^predicate)(int)

而且我不记得我尝试过哪些其他组合。

Answer 1

- ( void )myMethodTakingPredicate: ( BOOL ( ^ )( int ) )predicate

Answer 2

事情是这样的，例如...

[self smartBlocks:@"Pen" youSmart:^(NSString *response) {
        NSLog(@"Response:%@", response);
    }];


- (void)smartBlocks:(NSString *)yo youSmart:(void (^) (NSString *response))handler {
    if ([yo compare:@"Pen"] == NSOrderedSame) {
        handler(@"Ink");
    }
    if ([yo compare:@"Pencil"] == NSOrderedSame) {
        handler(@"led");
    }
}

Answer 3

http://fuckingblocksyntax.com

作为方法参数：

- (void)someMethodThatTakesABlock:(returnType (^)(parameterTypes))blockName;

Answer 4

再举个例子（这个问题受益多）：

@implementation CallbackAsyncClass {
void (^_loginCallback) (NSDictionary *response);
}
// …


- (void)loginWithCallback:(void (^) (NSDictionary *response))handler {
    // Do something async / call URL
    _loginCallback = Block_copy(handler);
    // response will come to the following method (how is left to the reader) …
}

- (void)parseLoginResponse {
    // Receive and parse response, then make callback

   _loginCallback(response);
   Block_release(_loginCallback);
   _loginCallback = nil;
}


// this is how we make the call:
[instanceOfCallbackAsyncClass loginWithCallback:^(NSDictionary *response) {
   // respond to result
}];

Answer 5

更清楚了！

[self sumOfX:5 withY:6 willGiveYou:^(NSInteger sum) {
    NSLog(@"Sum would be %d", sum);
}];

- (void) sumOfX:(NSInteger)x withY:(NSInteger)y willGiveYou:(void (^) (NSInteger sum)) handler {
    handler((x + y));
}