In Objective-C, how to print out N spaces? (using stringWithCharacters)

问题

The following is tried to print out N number of spaces (or 12 in the example):

NSLog(@"hello%@world", [NSString stringWithCharacters:" " length:12]);

const unichar arrayChars[] = {' '};
NSLog(@"hello%@world", [NSString stringWithCharacters:arrayChars length:12]);

const unichar oneChar = ' ';
NSLog(@"hello%@world", [NSString stringWithCharacters:&oneChar length:12]);

But they all print out weird things such as hello ÔÅÓñüÔÅ®Óñü®ÓüÅ®ÓñüÔ®ÓüÔÅ®world... I thought a "char array" is the same as a "string" and the same as a "pointer to a character"? The API spec says it is to be a "C array of Unicode characters" (by Unicode, is it UTF8? if it is, then it should be compatible with ASCII)... How to make it work and why those 3 ways won't work?

回答1:

You can use %*s to specify the width.

NSLog(@"Hello%*sWorld", 12, "");

Reference:

A field width, or precision, or both, may be indicated by an asterisk ( '*' ). In this case an argument of type int supplies the field width or precision. Applications shall ensure that arguments specifying field width, or precision, or both appear in that order before the argument, if any, to be converted.

回答2:

This will get you what you want:

NSLog(@"hello%@world", [@"" stringByPaddingToLength:12 withString:@" " startingAtIndex:0]);

回答3:

I think the issue you have is you are misinterpreting what +(NSString *)stringWithCharacters:length: is supposed to do. It's not supposed to repeat the characters, but instead copy them from the array into a string.

So in your case you only have a single ' ' in the array, meaning the other 11 characters will be taken from whatever follows arrayChars in memory.

If you want to print out a pattern of n spaces, the easiest way to do that would be to use -(NSString *)stringByPaddingToLength:withString:startingAtIndex:, i.e creating something like this.

NSString *formatString = @"Hello%@World";
NSString *paddingString = [[NSString string] stringByPaddingToLength: n withString: @" " startingAtIndex: 0];
NSLog(formatString, paddingString);

回答4:

This is probably the fastest method:

NSString *spacesWithLength(int nSpaces)
{
    char UTF8Arr[nSpaces + 1];

    memset(UTF8Arr, ' ', nSpaces * sizeof(*UTF8Arr));
    UTF8Arr[nSpaces] = '\0';

    return [NSString stringWithUTF8String:UTF8Arr];
}

The reason your current code isn't working is because +stringWithCharacters: expects an array with a length of characters of 12, while your array is only 1 character in length {' '}. So, to fix, you must create a buffer for your array (in this case, we use a char array, not a unichar, because we can easily memset a char array, but not a unichar array).

The method I provided above is probably the fastest that is possible with a dynamic length. If you are willing to use GCC extensions, and you have a fixed size array of spaces you need, you can do this:

NSString *spacesWithLength7()
{
    unichar characters[] = { [0 ... 7] = ' ' };
    return [NSString stringWithCharacters:characters length:7];
}

Unfortunately, that extension doesn't work with variables, so it must be a constant.

Through the magic of GCC extensions and preprocessor macros, I give you.... THE REPEATENATOR! Simply pass in a string (or a char), and it will do the rest! Buy now, costs you only $19.95, operators are standing by! (Based on the idea suggested by @JeremyL)

// step 1: determine if char is a char or string, or NSString.
// step 2: repeat that char or string
// step 3: return that as a NSString
#define repeat(inp, cnt) __rep_func__(@encode(typeof(inp)), inp, cnt) 

// arg list: (int siz, int / char *input, int n)
static inline NSString *__rep_func__(char *typ, ...)
{
    const char *str = NULL;
    int n;

    {
        va_list args;
        va_start(args, typ);

        if (typ[0] == 'i')
            str = (const char []) { va_arg(args, int), '\0' };
        else if (typ[0] == '@')
            str = [va_arg(args, id) UTF8String];
        else
            str = va_arg(args, const char *);

        n = va_arg(args, int);

        va_end(args);
    }

    int len = strlen(str);
    char outbuf[(len * n) + 1];

    // now copy the content
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < len; j++) {
            outbuf[(i * len) + j] = str[j];
        }
    }

    outbuf[(len * n)] = '\0';
    return [NSString stringWithUTF8String:outbuf];
}

回答5:

The stringWithCharaters:length: method makes an NSString (or an instance of a subclass of NSString) using the first length characters in the C array. It does not iterate over the given array of characters until it reaches the length.

The output you are seeing is the area of memory 12 Unicode characters long starting at the location of your passed 1 Unicode character array.

This should work.

NSLog(@"hello%@world", [NSString stringWithCharacters:"            " length:12]);

来源：https://stackoverflow.com/questions/10702678/in-objective-c-how-to-print-out-n-spaces-using-stringwithcharacters

标签

objective-c

ios

nsstring

core-foundation