Objective-C传递块作为参数

Question

我怎样才能传递Block到Function/ Method？

我尝试- (void)someFunc:(__Block)someBlock无济于事.

即.什么类型的Block？

Answer 1

块的类型取决于其参数及其返回类型.在一般情况下,块类型的声明方式与函数指针类型相同,但替换为*a ^.将块传递给方法的一种方法如下:

- (void)iterateWidgets:(void (^)(id, int))iteratorBlock;

但正如你所看到的那样,这很麻烦.您可以改为使用a typedef来使块类型更清晰:

typedef void (^ IteratorBlock)(id, int);

然后将该块传递给类似的方法:

- (void)iterateWidgets:(IteratorBlock)iteratorBlock;

Answer 2

这个问题最简单的解释是遵循以下模板:

1.阻止作为方法参数

模板

- (void)aMethodWithBlock:(returnType (^)(parameters))blockName {
        // your code
}

例

-(void) saveWithCompletionBlock: (void (^)(NSArray *elements, NSError *error))completionBlock{
        // your code
}

案件的其他用途:

2.作为财产阻止

模板

@property (nonatomic, copy) returnType (^blockName)(parameters);

例

@property (nonatomic,copy)void (^completionBlock)(NSArray *array, NSError *error);

3.阻止作为方法参数

模板

[anObject aMethodWithBlock: ^returnType (parameters) {
    // your code
}];

例

[self saveWithCompletionBlock:^(NSArray *array, NSError *error) {
    // your code
}];

4.阻止作为局部变量

模板

returnType (^blockName)(parameters) = ^returnType(parameters) {
    // your code
};

例

void (^completionBlock) (NSArray *array, NSError *error) = ^void(NSArray *array, NSError *error){
    // your code
};

5.阻止作为typedef

模板

typedef returnType (^typeName)(parameters);

typeName blockName = ^(parameters) {
    // your code
}

例

typedef void(^completionBlock)(NSArray *array, NSError *error);

completionBlock didComplete = ^(NSArray *array, NSError *error){
    // your code
};

Answer 3

这可能会有所帮助:

- (void)someFunc:(void(^)(void))someBlock;

Answer 4

您可以这样做,将块作为块参数传递:

//creating a block named "completion" that will take no arguments and will return void
void(^completion)() = ^() {
    NSLog(@"bbb");
};

//creating a block namd "block" that will take a block as argument and will return void
void(^block)(void(^completion)()) = ^(void(^completion)()) {
    NSLog(@"aaa");
    completion();
};

//invoking block "block" with block "completion" as argument
block(completion);

Answer 5

在下面的示例中使用с函数传递块的另一种方法.我创建了在后台和主队列中执行任何操作的函数.

blocks.h文件

void performInBackground(void(^block)(void));
void performOnMainQueue(void(^block)(void));

blocks.m文件

#import "blocks.h"

void performInBackground(void(^block)(void)) {
    if (nil == block) {
        return;
    }

    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_BACKGROUND, 0), block);
}

void performOnMainQueue(void(^block)(void)) {
    if (nil == block) {
        return;
    }

    dispatch_async(dispatch_get_main_queue(), block);
}

比必要时导入blocks.h并调用它:

- (void)loadInBackground {

    performInBackground(^{

        NSLog(@"Loading something in background");
        //loading code

        performOnMainQueue(^{
            //completion hadler code on main queue
        });
    });
}

Answer 6

如果适用于您,您还可以将块设置为简单属性:

@property (nonatomic, copy) void (^didFinishEditingHandler)(float rating, NSString *reviewString);

确保块属性是"复制"!

当然你也可以使用typedef:

typedef void (^SimpleBlock)(id);

@property (nonatomic, copy) SimpleBlock someActionHandler;

Answer 7

您还可以使用常用的c函数语法调用或调用块

-(void)iterateWidgets:(IteratorBlock)iteratorBlock{

    iteratorBlock(someId, someInt);
}

有关块的更多信息

http://developer.apple.com/library/ios/#documentation/cocoa/Conceptual/Blocks/Articles/bxGettingStarted.html#//apple_ref/doc/uid/TP40007502-CH7-SW1