如何在Objective-C中实现链式语法?
天蓬老师
天蓬老师 2017-04-24 09:11:13
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如何在Objective-C中实现链式语法?

例如:

someObject.a.and.b.someMethod(5)

这里主要指的是点链式语法,不同于常见的[[[[someObject a] and] b] someMethod:5] 中括号链式语法

如何设计?

天蓬老师
天蓬老师

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membalas semua(1)
Peter_Zhu

Masalah ini telah diselesaikan melalui Stackoverflow. Alamat pautan: http://stackoverflow.com/questions/27739106/how-to-implement-chainable-syntax-with-objective-c/27740550#27740550

Contoh:

@class ClassB;
@interface ClassA : NSObject

//1. we define some the block properties
@property(nonatomic, readonly) ClassA *(^aaa)(BOOL enable);
@property(nonatomic, readonly) ClassA *(^bbb)(NSString* str);
@property(nonatomic, readonly) ClassB *(^ccc)(NSString* str);

@implement ClassA

//2. we implement these blocks, and remember the type of return value, it's important to chain next block

- (ClassA *(^)(BOOL))aaa
{
    return ^(BOOL enable) {
        //code
        if (enable) {
            NSLog(@"ClassA yes");
        } else {
            NSLog(@"ClassA no");
        }
        return self;
    }
}

- (ClassA *(^)(NSString *))bbb
{
    return ^(NSString *str)) {
        //code
        NSLog(@"%@", str);
        return self;
    }
}

// Here returns a instance  which is kind of ClassB, then we can chain ClassB's block.
// See below .ccc(@"Objective-C").ddd(NO)
- (ClassB * (^)(NSString *))ccc
{
    return ^(NSString *str) {
        //code
        NSLog(@"%@", str);
        ClassB* b = [[ClassB alloc] initWithString:ccc];
        return b;
    }
}

//------------------------------------------
@interface ClassB : NSObject
@property(nonatomic, readonly) ClassB *(^ddd)(BOOL enable);

- (id)initWithString:(NSString *)str;

@implement ClassB

- (ClassB *(^)(BOOL))ddd
{
    return ^(BOOL enable) {
        //code
        if (enable) {
            NSLog(@"ClassB yes");
        } else {
            NSLog(@"ClassB no");
        }
        return self;
    }
}

// At last, we can do it like this------------------------------------------
id a = [ClassA new];
a.aaa(YES).bbb(@"HelloWorld!").ccc(@"Objective-C").ddd(NO)

Jika anda mempunyai pendapat atau idea yang lebih baik, perbincangan akan diteruskan :]

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