-(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用于排序数组的块。
现在我想编写一个返回该块的方法。
我该怎么做?
我试过
+ (NSComparator)(^)(id obj1, id obj2)
{
(NSComparator)(^ block)(id obj1, id obj2) = {...}
return block;
}
我们只是说它不起作用。
答案 0 :(得分:52)
返回像这样的块的方法签名应该是
+(NSInteger (^)(id, id))comparitorBlock {
....
}
这会分解为:
+(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
已经是块类型。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 {
....
}