基于 int 创建多个编号变量

发布于 2024-08-21 15:51:53 字数 660 浏览 11 评论 0原文

如何使用数组的计数创建多个 NSDictionary 变量？

这基本上是我想到的，但我不确定如何使用 Objective-C 语法来实现它。 doesntContainAnother 是一个 NSArray。我希望字典的名称使用 loopInt 的当前值。

int *loopInt = 0;
while (doesntContainAnother.count <= loopInt) {

    NSMutableDictionary *[NSString stringWithFormat:@"loopDictionary%i", loopInt] = [[[NSMutableDictionary alloc] init] autorelease];
    [NSString stringWithFormat:@"loopDictionary%i", loopInt] = [NSDictionary dictionaryWithObject:[array1 objectAtIndex:loopInt] 
                                                 forKey:[array2 objectAtIndex:loopInt]];
    loopInt = loopInt + 1;
}

原文

How would I create a number of NSDictionary variables using an array's count?

This is basically what I came up with, but I'm not sure how to make this work with Objective-C syntax. doesntContainAnother is an NSArray. I want the names of the dictionaries to use the current value of loopInt.

int *loopInt = 0;
while (doesntContainAnother.count <= loopInt) {

    NSMutableDictionary *[NSString stringWithFormat:@"loopDictionary%i", loopInt] = [[[NSMutableDictionary alloc] init] autorelease];
    [NSString stringWithFormat:@"loopDictionary%i", loopInt] = [NSDictionary dictionaryWithObject:[array1 objectAtIndex:loopInt] 
                                                 forKey:[array2 objectAtIndex:loopInt]];
    loopInt = loopInt + 1;
}

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荒岛晴空 2024-08-28 15:51:53

创建一个可变数组并循环，直到达到原始数组的计数，创建一个字典并在每次迭代时将其添加到可变数组中。

您的代码应该如下所示。

NSMutableArray *dictionaries = [[NSMutableArray alloc] init];
for (int i = 0; i < doesntContainAnother.count; i++) {
    [dictionaries addObject:[NSMutableDictionary dictionaryWithObject:[array1 objectAtIndex:i] forKey:[array2 objectAtIndex:i]]];
}

创建名称末尾带有数字的变量的方法是一种反模式，在 Objective-C 中甚至是不可能的。它相当于一个数组，但更笨重。

Create a mutable array and loop until you reach the original array's count, creating a dictionary and adding it to the mutable array on each iteration.

Your code should look like this.

NSMutableArray *dictionaries = [[NSMutableArray alloc] init];
for (int i = 0; i < doesntContainAnother.count; i++) {
    [dictionaries addObject:[NSMutableDictionary dictionaryWithObject:[array1 objectAtIndex:i] forKey:[array2 objectAtIndex:i]]];
}

The approach of creating variables with numbers at the end of their names is an antipattern and not even possible in Objective-C. It's equivalent to an array, but clunkier.

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栖迟 2024-08-28 15:51:53

您需要创建一个可变数组，然后将对象放入该数组中。您不能像您所做的那样创建与字符串内容同名的变量。例如：

NSMutableArray *arr = [[NSMutableArray alloc] initWithCapacity:[doesntContainAnother count]];
int i = 0;    // Note: type is int, not int*
for (i = 0; i < [doesntCountainAnother count]; i++) {
    [arr addObject:[NSMutableDictionary dictionary]];
}

// Later...
NSMutableDictionary *d1 = [arr objectAtIndex:3];

或者如果您想按名称将它们从列表中取出：

NSMutableDictionary *dict = [[NSMutableDictionary alloc] initWithCapacity:[doesntCountainAnother count]];
int i = 0;
for (i = 0; i < [doesntContainAnother count]; i++) {
    [dict setObject:[NSMutableDictionary dictionary] forKey:[NSString stringWithFormat:@"loopDictionary%d", i]];
}

// Later...
NSMutableDictionary *d1 = [dict objectForKey:@"loopDictionary3"];

但第一种方法可能是最简单的。

You need to create a mutable array, and then put the objects into the array. You can't create a variable with the same name as the contents of a string, as you have done. For example:

NSMutableArray *arr = [[NSMutableArray alloc] initWithCapacity:[doesntContainAnother count]];
int i = 0;    // Note: type is int, not int*
for (i = 0; i < [doesntCountainAnother count]; i++) {
    [arr addObject:[NSMutableDictionary dictionary]];
}

// Later...
NSMutableDictionary *d1 = [arr objectAtIndex:3];

Or if you want to pull them out of the list by name:

NSMutableDictionary *dict = [[NSMutableDictionary alloc] initWithCapacity:[doesntCountainAnother count]];
int i = 0;
for (i = 0; i < [doesntContainAnother count]; i++) {
    [dict setObject:[NSMutableDictionary dictionary] forKey:[NSString stringWithFormat:@"loopDictionary%d", i]];
}

// Later...
NSMutableDictionary *d1 = [dict objectForKey:@"loopDictionary3"];

But the first way is likely the easiest.

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