objective-c - 如何创建一个返回 block 的 objective-c 方法

Question

-(NSMutableArray *)sortArrayByProminent:(NSArray *)arrayObject
{
    NSArray * array = [arrayObject sortedArrayUsingComparator:^(id obj1, id obj2) {
        Business * objj1=obj1;
        Business * objj2=obj2;
        NSUInteger prom1=[objj1 .prominent intValue];
        NSUInteger prom2=[objj2 .prominent intValue];
        if (prom1 > prom2) {
            return NSOrderedAscending;
        }
        if (prom1 < prom2) {
            return NSOrderedDescending;
        }
        return NSOrderedSame;
    }];

    NSMutableArray *arrayHasBeenSorted = [NSMutableArray arrayWithArray:array];

    return arrayHasBeenSorted;
}

所以基本上我有这个 block 用于对数组进行排序。

现在我想编写一个返回该 block 的方法。

我该怎么做？

我试过了

+ (NSComparator)(^)(id obj1, id obj2)
{
    (NSComparator)(^ block)(id obj1, id obj2) = {...}
    return block;
}

我们就说它还行不通吧。

Answer 1

像这样返回一个 block 的方法签名应该是

+(NSInteger (^)(id, id))comparitorBlock {
    ....
}

这分解为:

+(NSInteger (^)(id, id))comparitorBlock;
^^    ^      ^  ^   ^  ^       ^
ab    c      d  e   e  b       f

a = Static Method
b = Return type parenthesis for the method[just like +(void)...]
c = Return type of the block
d = Indicates this is a block (no need for block names, it's just a type, not an instance)
e = Set of parameters, again no names needed
f = Name of method to call to obtain said block

更新:在您的特定情况下，NSComparator 已经是 block 类型。它的定义是:

typedef NSComparisonResult (^NSComparator)(id obj1, id obj2);

因此，您需要做的就是返回此 typedef:

+ (NSComparator)comparator {
   ....
}