%s minimum field width in the presence of unicode characters

Question

So, here's my problem:

If someone wants to output visually aligned strings using printf, they'll obviously use %<n>s (where <n> is the minimum field width). And this works just fine, unless one of the strings contains unicode (UTF-8) characters.

Take this very basic example:

#include <stdio.h>

int main(void)
{
    char* s1 = "\u03b1\u03b2\u03b3";
    char* s2 = "abc";

    printf("'%6s'\n", s1);
    printf("'%6s'\n", s2);

    return 0;
}

which will produce the following output:

'αβγ'
'   abc'

This isn't all that surprising, because printf of course doesn't know that \u03b1 (which consists of two characters) only produces a single glyph on the output device (assuming UTF-8 is supported).

Now assume that i generate s1 and s2, but have no control over the format string used to output those variables. My current understanding is that nothing i could possibly do to s1 would fix this, because i'd have to somehow fool printf into thinking that s1 is shorter than it actually is. However, since i also control s2, my current solution is to add a non-printing character to s2 for each unicode character in s1, which would look something like this:

#include <stdio.h>

int main(void)
{
    char* s1 = "\u03b1\u03b2\u03b3";
    char* s2 = "abc\x06\x06\x06";

    printf("'%6s'\n", s1);
    printf("'%6s'\n", s2);

    return 0;
}

This will produce the desired output (even though the actual width no longer corresponds to the specified field width, but i'm willing to accept that):

'αβγ'
'abc'

For context:

The example above is only to illustrate the unicode-problem, my actual code involves printing numbers with SI-prefixes, only one of which (µ) is a unicode character. Therefore i would generate strings containing only up to one normal or unicode character (which is why i can accept the resulting offset in the field-width).

So, my questions are:

• Is there a better solution for this?
• Is \x06 (ACK) a sensible choice (i.e. a character without undesired side-effects)?
• Can you think of any problems with this approach?

Answer 1

Since the non ascii is restricted to µ, I believe there is a solution. I've taken value of µ to be \u00b5. Replace it with the correct value

I've coded a small function myPrint which takes input the string and the width n. You should be able to modify the code below to fit to your needs.

The function searches for all occurrences of µ and increments that much of width to the string

#include <stdio.h>

void myPrint(char* string, int n)
{
  char* valueOfNu = "\u00b5";
  for(int i=0;string[i]!='\0';i++)
  {
    if(string[i]==valueOfNu[0] && string[i+1]==valueOfNu[1])
      n++;
  }

  printf("%*s",n,string);
}


int main(void)
{
    char* s1 = "ab\u00b5";
    char* s2 = "abc";

    myPrint(s1,6);
    printf("\n");
    myPrint(s2,6);
    printf("\n");

    return 0;
}