Objective-C：typedef为一个块，在方法声明中使用它。我该如何实现？

Just trying to get a handle on blocks. I get the concept. They're like function pointers, but they're actually objects; you can declare a block variable and assign it a block value; call it like a function; they get "frozen in time," for lack of a term, when they get executed, etc. I've created a few blocks and run them successfully, in a few different formats, but when it comes to using them in a method--either with a typedef or without--I'm having a lot of trouble. For example, here's an object interface I created, just to get a handle on the syntax. I have almost no idea how to implement it.

只是想试试块。我明白了。它们就像函数指针,但它们实际上是对象;你可以声明一个块变量并为它赋值一个块值;把它称为功能;他们得到“及时冻结”,缺乏一个术语,当他们被执行时等等。我已经创建了几个块并以几种不同的格式成功运行它们,但是当它在方法中使用时 - - 使用typedef或不使用 - 我遇到了很多麻烦。例如,这是我创建的对象接口,只是为了获得语法句柄。我几乎不知道如何实现它。

// AnObject.h

#import <Foundation/Foundation.h>

// The idea with the block and the method below is for the block to take
// an int, multiply it by 3, and return a "tripled" int.  The method
// will then repeat: this process however many times the user wants via
// the howManyTimes parameter and return that value in the form of an int.

typedef int (^triple)(int);

@interface AnObject : NSObject
{
    int num;
}

-(int)repeat:(int)howManyTimes withBlock:(triple)someBlock;

@end

Here's what I have for an implementation, so far:

到目前为止,这是我实现的内容:

#import "AnObject.h"

@implementation AnObject

@synthesize num;

-(int)repeat:(int)howManyTimes withBlock:(triple)someBlock {
    for (int i = 0; i <= howManyTimes; i++) {
        // What the heck am I supposed to put here?  I'm baffled by the
        // syntax over and over again.
    }
}

@end

I know I'm not addressing the instance variable yet. Again, this is a rough draft, just trying to get a handle on how blocks work. Am I even declaring this method right? I'm reading Big Nerd Ranch's Objective-C Programming, Mike Clark's article on blocks from Pragmatic Studio, and several SO threads. Can't find anything relevant. Thanks.

我知道我还没有解决实例变量。再次,这是一个粗略的草案,只是试图了解块如何工作。我甚至宣布这种方法对吗?我正在阅读Big Nerd Ranch的Objective-C编程,Mike Clark关于Pragmatic Studio中的块的文章,以及几个SO线程。找不到任何相关的东西。谢谢。

EDIT: XCode 4.3.2, if it matters.

编辑:XCode 4.3.2,如果重要的话。

FURTHER EDIT: Ok. Using BJ's (slightly modified) example, I think I've come up with a really complicated way of multiplying 5 by 3. :)

进一步编辑:好的。使用BJ(稍加修改)的例子,我想我已经提出了一种将5乘以3的非常复杂的方法。:)

// BJ's implementation:

-(int)repeat:(int)howManyTimes withBlock:(Triple)someBlock {

    int blockReturnValue;

    for (int i = 0; i <= howManyTimes; i++) {
        blockReturnValue = someBlock(i);
    }
    return blockReturnValue;
}

Main:

...
    @autoreleasepool
    {
        AnObject *obj = [[AnObject alloc] init];

        NSLog(@"%d", [obj repeat: 5 withBlock: ^ (int number) {
            return number * 3;
        }]);

    }
    return 0;
...

And the output is:

输出是:

15

Now, it's kicking back 15, because the block I defined as an argument is run only once, right? It multiplies "number," which is 5 in this case, by 3 and freezes that answer, right? I'm sure I just created a completely useless method, and I don't yet understand how to utilize the benefits/features of a block. Am I correct?

现在,它正在回击15,因为我定义为参数的块只运行一次,对吧?它将“数字”(在这种情况下为5)乘以3并冻结该答案,对吧?我确定我刚刚创建了一个完全没用的方法,我还不知道如何利用块的优点/功能。我对么?

/********************* UPDATE *********************/

/ *********************更新********************* /

UPDATE: I understand what you're saying, CRD. Just a correction though, for any new programmers who might be reading this, getting a different output and going, "Que?" Your for loop should be either:

更新:我明白你在说什么,CRD。但是,对于任何可能正在阅读此内容的新程序员来说,只需要进行修正,获得不同的输出并继续“Que?”你的for循环应该是:

for (int i = 0; i < howManyTimes; i++)
            value = someBlock(value);

...or...

(i = 1; i <= howManyTimes; i++)

...to get the answer 243.

......得到答案243。

And, yes, this is exactly what I was initially trying to do with this code. At least that's what I thought was supposed to be happening. Turns out the author's intent wasn't to triple a number, store that value, triple the stored value, store that...etc., but rather just to print x * 3 for numbers 1-5 (3, 6, 9, 12, 15).

而且,是的,这正是我最初尝试使用此代码的原因。至少那是我认为应该发生的事情。原来,作者的意图不是将数字加倍,存储该值,存储值增加三倍,存储......等等,而只是为数字1-5打印x * 3(3,6,9, 12,15)。

Here is the finished product. I just typedef'd a block that takes an int and returns an int, called Tripler. I also changed the name of the argument from "someBlock" to "triple" to more clearly indicate the intended use of the block. I think those are the only changes to the code.

这是成品。我只是输入一个带有int的块并返回一个名为Tripler的int。我还将参数的名称从“someBlock”更改为“triple”以更清楚地指示块的预期用途。我认为这些是对代码的唯一更改。

/********************  interface  ********************/


#import <Foundation/Foundation.h>

typedef int (^Tripler)(int);

@interface AnObject : NSObject

-(void)iterateFromOneTo:(int)number withBlock:(Tripler)triple;

@end

/********************  implementation  ********************/


#import "AnObject.h"

@implementation AnObject

-(void)iterateFromOneTo:(int)number withBlock:(Tripler)triple {
    for (int i = 1; i <= number; i++) {
        NSLog(@"%d", triple(i));
    }
}

@end

/********************  main.m  ********************/


#import "AnObject.h"
#import <Foundation/Foundation.h>

int main(int argc, const char * argv[])
{
    @autoreleasepool
    {
        AnObject *obj = [[AnObject alloc] init];

        [obj iterateFromOneTo:5 withBlock:^(int number) {
            return number * 3;
        }];
    }
    return 0;
}

As you can probably imagine, the resulting output is:

您可以想象,结果输出是:

2012-05-05 17:10:13.418 Untitled 2[71735:707] 3
2012-05-05 17:10:13.445 Untitled 2[71735:707] 6
2012-05-05 17:10:13.446 Untitled 2[71735:707] 9
2012-05-05 17:10:13.446 Untitled 2[71735:707] 12
2012-05-05 17:10:13.447 Untitled 2[71735:707] 15

I was making it a lot more complicated than it needed to be. Sorry for explaining it so poorly in the OP. Thanks for your help! /thread? :)

我让它变得比它需要的复杂得多。很抱歉在OP中解释得太差了。谢谢你的帮助! /线? :)

3 个解决方案

#1

Just call the block like a regular C function.

只需像普通的C函数一样调用块。

-(int)repeat:(int)howManyTimes withBlock:(triple)someBlock {
    for (int i = 0; i <= howManyTimes; i++) {
        int blockReturnValue = someBlock(i);
        // do something with blockReturnValue
    }
}

Update after your "further edit"

“进一步编辑”后更新

No, the block you passed in as an argument is run five times, each pass through the for loop.

不,您作为参数传入的块运行五次,每次都通过for循环。

• The first time, it invokes the block with 1 as the argument, and gets back 3. It stores that in blockReturnValue, then goes on to the next iteration of the loop.

第一次,它调用带有1作为参数的块,然后返回3.它将它存储在blockReturnValue中,然后继续循环的下一次迭代。

• The second time, it invokes the block with 2 as the argument, and gets back 6. It stores that in blockReturnValue, completely overwriting the value we stored there in the previous pass.

第二次,它调用带有2作为参数的块,然后返回6.它将它存储在blockReturnValue中,完全覆盖我们在前一遍中存储的值。

• The third time, it inbokes the block with 3 as the argument, and gets back 9. Again, it overwrites the value in blockReturnValue.

第三次,它以3作为参数来阻止块,然后返回9.再次,它覆盖了blockReturnValue中的值。

• The fourth time, we store 12 in blockReturnValue.

第四次,我们在blockReturnValue中存储了12个。

• The fifth time, we store 15 in blockReturnValue.

第五次,我们在blockReturnValue中存储了15个。

Then we exit the for loop, and return 15. So yes, you're correct that you've made a pointless method to multiply by 3. But you're doing it in a way that also does a bunch of useless calculations.

然后我们退出for循环,并返回15.所以是的,你是正确的,你已经做了一个无意义的方法乘以3.但你这样做的方式也做了一堆无用的计算。

#2

From reading your question I understood, or maybe misunderstood, that your intention was to produce the result of applying your block n times; e.g. if you applied a tripling function twice you'd get the original value multiplied by nine.

从阅读你的问题我理解,或者可能被误解,你的意图是产生你的块的n次结果;例如如果你应用了两次三倍函数,你会得到原始值乘以9。

Just in case it helps, here is the code to do that:

万一它有帮助,这里有代码:

@interface AnObject

typedef int (^monadic)(int); // an function which takes an int and return an int

- (int) repeat:(int)howManyTimes for:(int)value withBlock:(monadic)someBlock;

@end

@implementation AnObject

- (int) repeat:(int)howManyTimes for:(int)value withBlock:(monadic)someBlock
{
   for (int i = 0; i < howManyTimes; i++)
      value = someBlock(value);

   return value;
}

@end

Now call this with:

现在打电话给:

AnObject *myObject = [AnObject new];

int z = [myObject repeat:5  
                     for:1 
               withBlock: ^(int number)
                          {
                             return number * 3;
                          }
        ];

and z will have the value 243.

和z将具有值243。

#3

@import Foundation;

typedef int(^MyBlockType)(int);

@implementation NSObject(extra)
+ (int)invokeBlock:(MyBlockType)block withArgument:(int)arg
{
    return block(arg);
}
@end;

int main() {
    @autoreleasepool {
        NSLog(@"executeBlock(3) returns %d",
              [NSObject invokeBlock:^(int param) { return param * param; } withArgument:3]);
    } return 0;
}