iOS Objective C NSString.length不返回期望值

Question

我遇到了最基本的客觀C方法問題。 基於以下建議： How can I truncate an NSString to a set length?

我想寫一個方法來返回截斷的NSString。但是，它不起作用。例如，當我發送「555」時，長度（如變量'test'所示）返回爲0.我通過在行int測試後面設置斷點並將鼠標懸停在變量fullString和test上來確定此位置。我是否需要將指針取消引用到fullString或其他某些東西？我是一個完整的新手在客觀C.非常感謝

-(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; 

} 

基於評論和研究，我得到它的工作。謝謝大家。如果有人有興趣：

-(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

檢查，如果你是路過@"555"這種方法，不"555"。 另外，更好的方法是

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