通过指针枚举NSString字符 [英] Enumerate NSString characters via pointer

问题描述

如何通过拉出每个unichar来枚举NSString？我可以使用characterAtIndex，但是比做一个递增unichar *慢。我没有看到任何在苹果的文档，不需要将字符串复制到第二个缓冲区。

How can I enumerate NSString by pulling each unichar out of it? I can use characterAtIndex but that is slower than doing it by an incrementing unichar*. I didn't see anything in Apple's documentation that didn't require copying the string into a second buffer.

这样的东西是理想的：

for (unichar c in string) { ... }

或

unichar* ptr = (unichar*)string;

推荐答案

您可以加快 characterAtIndex：，首先将其转换为IMP格式：

You can speed up -characterAtIndex: by converting it to it's IMP form first:

NSString *str = @"This is a test";

NSUInteger len = [str length]; // only calling [str length] once speeds up the process as well
SEL sel = @selector(characterAtIndex:);

// using typeof to save my fingers from typing more
unichar (*charAtIdx)(id, SEL, NSUInteger) = (typeof(charAtIdx)) [str methodForSelector:sel];

for (int i = 0; i < len; i++) {
    unichar c = charAtIdx(str, sel, i);
    // do something with C
    NSLog(@"%C", c);
}  

EDIT：看起来 CFString / a>包含以下方法：

It appears that the CFString Reference contains the following method:

const UniChar *CFStringGetCharactersPtr(CFStringRef theString);

这意味着您可以执行以下操作：

This means you can do the following:

const unichar *chars = CFStringGetCharactersPtr((__bridge CFStringRef) theString);

while (*chars)
{
    // do something with *chars
    chars++;
}

如果你不想分配内存来处理缓冲区，要走的路。

If you don't want to allocate memory for coping the buffer, this is the way to go.

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