Variable sized padding in printf

Question

Is there a way to have a variable sized padding in printf?

I have an integer which says how large the padding is:

void foo(int paddingSize) {
    printf("%...MyText", paddingSize);
}

This should print out ### MyText where the paddingSize should decide the number of '#' symbols.

Answer 1

Yes, if you use * in your format string, it gets a number from the arguments:

printf ("%0*d\n", 3, 5);

will print "005".

Keep in mind you can only pad with spaces or zeros. If you want to pad with something else, you can use something like:

#include <stdio.h>
#include <string.h>
int main (void) {
    char *s = "MyText";
    unsigned int sz = 9;
    char *pad = "########################################";
    printf ("%.*s%s\n", (sz < strlen(s)) ? 0 : sz - strlen(s), pad, s);
}

This outputs ###MyText when sz is 9, or MyText when sz is 2 (no padding but no truncation). You may want to add a check for pad being too short.

Answer 2

You could write like this :

void foo(int paddingSize) {
       printf ("%*s",paddingSize,"MyText");
}

Answer 3

printf( "%.*s", paddingSize, string );

For example:

const char *string = "12345678901234567890";
printf( "5:%.*s\n", 5, string );
printf( "8:%.*s\n", 8, string );
printf( "25:%.*s\n", 25, string );

displays:

5:12345
8:12345678
25:12345678901234567890

Answer 4

Be careful though - some compilers at least (and this may be a general C thing) will not always do what you want:

char *s = "four";
printf("%*.s\n", 5, s); // Note the "."

This prints 5 spaces;

char *s = "four";
printf("%*s\n", 3, s);  // Note no "."

This prints all four characters "four"

Answer 5

better use std::cout

#include<iomanip>
#include<iostream>
using namespace std;
cout << setw(9)                       //set output width
     << setfill('#')                  // set fill character
     << setiosflags(ios_base::right)  //put padding to the left
     << "MyText";

should produce:

###MyText