iOS Objective C NSString.length不返回期望值

Question

我遇到了最基本的客观C方法问题。 基于以下建议： How can I truncate an NSString to a set length?

我想写一个方法来返回截断的NSString。但是，它不起作用。例如，当我发送“555”时，长度（如变量'test'所示）返回为0.我通过在行int测试后面设置断点并将鼠标悬停在变量fullString和test上来确定此位置。我是否需要将指针取消引用到fullString或其他某些东西？我是一个完整的新手在客观C.非常感谢

-(NSString*) getTruncatedString:(NSString *) fullString { 

    int test = fullString.length; 
    int test2 = MIN(0,test); 
    NSRange stringRangeTest = {0, MIN([@"Test" length], 20)}; 

    // define the range you're interested in 
    NSRange stringRange = {0, MIN([fullString length], 20)}; 

    // adjust the range to include dependent chars 
    stringRange = [fullString rangeOfComposedCharacterSequencesForRange:stringRange]; 

    // Now you can create the short string 
    NSString* shortString = [_sentToBrainDisplay.text substringWithRange:stringRange]; 

    return shortString; 

} 

基于评论和研究，我得到它的工作。谢谢大家。如果有人有兴趣：

-(NSString*) getTruncatedString:(NSString *) fullString { 

if (fullString == nil || fullString.length == 0) { 
    return fullString; 
} 
NSLog(@"String length: %d", fullString.length); 

// define the range you're interested in 
NSRange stringRange = {MAX(0, (int)[fullString length]-20),MIN(20, [fullString length])}; 


// adjust the range to include dependent chars 
stringRange = [fullString rangeOfComposedCharacterSequencesForRange:stringRange]; 

// Now you can create the short string 
NSString* shortString = [fullString substringWithRange:stringRange]; 

return shortString; 

}

Answer 1

检查，如果你是路过@"555"这种方法，不"555"。 另外，更好的方法是

NSLog(@"String length: %d", fullString.length)。