在不使用typedef的情况下声明块方法参数

Question

是否可以在不使用typedef的情况下在Objective-C中指定方法块参数？它必须像函数指针一样,但是如果不使用中间的typedef,我就无法获得胜利的语法:

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

只有以上编译,所有这些都失败了:

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

我不记得我尝试过的其他组合.

Answer 1

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

Answer 2

这就是它的方式,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

另一个例子(这个问题得益于多个):

@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
}];