使用 NSString 制作变量名-移动开发问题

Make variable name out of NSString(使用 NSString 制作变量名)

本文介绍了使用 NSString 制作变量名的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个 for 循环，其中必须创建 9 个六边形(hexagon1 到 hexagon9)...但是我不能使用 hexString 作为 Sprite 的名称，因为它是一个 NSString，对吗?那么我该怎么做呢?

I have got a for loop where 9 hexagons (hexagon1 through hexagon9) have to be created... But I cannot use hexString as the name of the Sprite because it is a NSString, right ? So how would I make it right ?

hexString [<- 我希望 for 循环生成hexagon1"，然后是hexagon2"等等，而不是 NSString] = [self createHexagon:ccp(xVal,yVal) :i];


int hexCount = [[[itemPositions valueForKey:myString]valueForKey:@"hexposition"] count];

    for (int i=1;i<=hexCount;i++){
        NSString *hexString = [NSString stringWithFormat:@"hexagon%d",i];
        NSNumber *generatedXVal = [[[[itemPositions valueForKey: myString ]valueForKey:@"hexposition"] valueForKey: hexString]valueForKey: @"xVal"];
        int xVal = [generatedXVal integerValue];
        NSNumber *generatedYVal = [[[[itemPositions valueForKey: myString ]valueForKey:@"hexposition"] valueForKey: hexString ]valueForKey: @"yVal"];
        int yVal = [generatedYVal integerValue];

        hexString = [self createHexagon:ccp(xVal,yVal) : i];
        NSLog(@"%@", hexString);
    }


推荐答案
使用 NSMutableDictionary.
for (int i=1;i<=hexCount;i++){
    NSString *hexString = [NSString stringWithFormat:@"hexagon%d",i];
    CCSprite *sprite = [self doSomethingToGetSprite];

    [mutableDictionary setObject:sprite forKey:hexString];
}        

稍后您可以使用以下方法遍历字典中的所有精灵:
Later you can iterate over all the sprites in the dictionary using:
for (NSString *key in mutableDictionary) {
    CCSprite *sprite = [mutableDictionary objectForKey:key];
    [self doStuffWithSprite:sprite];
}

顺便说一句，你为什么要覆盖你在这里分配的 hexString:
By the way, why are you overwriting hexString that you assign here:
NSString *hexString = [NSString stringWithFormat:@"hexagon%d",i];

这里有一个:
hexString = [self createHexagon:ccp(xVal,yVal) : i];

并且该方法调用是一个明显的语法错误，其中悬挂 : i 部分.
And that method call is an obvious syntax error with the dangling : i part there.

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