XCODE将方法作为参数传递

Question

我不想将方法作为参数传递给另一个方法,因此它知道在结束运行时调用的方法.可能吗？

[self bringJSON:(NSString *)_passedValua:(NSObject *)anotherMethod];

Answer 1

正如@Daniel在评论中提到的,你可以使用选择器.基本方案如下:

// Method declaration - method accept selector as parameter
- (void) bringJSON:(NSString *)_passedValua toMethod:(SEL)anotherMethod];

// Method call - create selector from method name (mind the colon in selector - it is required if your method takes 1 parameter) 
[self bringJSON:jsonString toMethod:@selector(methodName:)];

// Implementation
- (void) bringJSON:(NSString *)_passedValua toMethod:(SEL)anotherMethod]{
   ...
   if ([target respondsToSelector:anotherMethod])
      [target performSelector:anotherMethod withObject:_passedValua];
}