初始化 NSString 时的 EXC_BAD_ACCESS

发布于 2024-12-11 09:15:35 字数 413 浏览 0 评论 0原文

我想用长度为 1 的字符初始化 NSString，但出现错误。这是我的代码：

for(int i=0;i<[word length];i++)
    {

        letterMap = [self stringToAscii:[word substringWithRange:NSMakeRange(i, 1) ]];
        NSInteger codec = (letterMap + shifter) % 26;
        unichar ch = [self asciiToString:codec];

        NSString * codedLetter = [NSString stringWithCharacters:ch length:[ch length]];

    }

原文

I want to initialise an NSString with a character with length of 1, but I get an error. Here's my code:

for(int i=0;i<[word length];i++)
    {

        letterMap = [self stringToAscii:[word substringWithRange:NSMakeRange(i, 1) ]];
        NSInteger codec = (letterMap + shifter) % 26;
        unichar ch = [self asciiToString:codec];

        NSString * codedLetter = [NSString stringWithCharacters:ch length:[ch length]];

    }

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活雷疯 2024-12-18 09:15:35

unichar 不是 ObjC 对象，它是单个 unicode 字符的原始 C 类型（或 typedef）。您需要像这样调用 NSString 方法（它可以在类似 C 的数组中接受多个字符）：（

NSString * codedLetter = [NSString stringWithCharacters:&ch length:1];

您的代码行有两个问题：首先，您传递的是 unicharunichar 的值code> 作为第一个参数，而不是指向它的指针，第二个参数是您尝试在不是 ObjC 对象的对象上调用 -length ，甚至是指针它会因它而爆炸。后者当它试图向某些人发送消息时随机内存地址。）

unichar isn't an ObjC object, it's a primitive C type (or typedef) for a single unicode character. You need to call the NSString method (which can take multiple characters in a C-like array) like this:

NSString * codedLetter = [NSString stringWithCharacters:&ch length:1];

(There are two problems in your line of code: first that you're passing the value of a unichar as the first argument, instead of a pointer to it, and the second that you're trying to call -length on something that's not an ObjC object, or even a pointer. It's blowing up due the latter when it tries to send a message to some random memory address.)

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