How to repeat a char using printf?

Question

I'd like to do something like printf("?", count, char) to repeat a character count times.

What is the right format-string to accomplish this?

EDIT: Yes, it is obvious that I could call printf() in a loop, but that is just what I wanted to avoid.

Answer 1

You can use the following technique:

printf("%.*s", 5, "=================");

This will print "=====" It works for me on Visual Studio, no reason it shouldn't work on all C compilers.

Answer 2

Short answer - yes, long answer: not how you want it.

You can use the %* form of printf, which accepts a variable width. And, if you use '0' as your value to print, combined with the right-aligned text that's zero padded on the left..

printf("%0*d\n", 20, 0);

produces:

00000000000000000000

With my tongue firmly planted in my cheek, I offer up this little horror-show snippet of code.

Some times you just gotta do things badly to remember why you try so hard the rest of the time.

#include <stdio.h>

int width = 20;
char buf[4096];

void subst(char *s, char from, char to) {
    while (*s == from)
    *s++ = to;
}

int main() {
    sprintf(buf, "%0*d", width, 0);
    subst(buf, '0', '-');
    printf("%s\n", buf);
    return 0;
}

Answer 3

If you limit yourself to repeating either a 0 or a space you can do:

For spaces:

printf("%*s", count, "");

For zeros:

printf("%0*d", count, 0);

Answer 4

In c++ you could use std::string to get repeated character

printf("%s",std::string(count,char).c_str());

For example:

printf("%s",std::string(5,'a').c_str());

output:

aaaaa

Answer 5

There is no such thing. You'll have to either write a loop using printf or puts, or write a function that copies the string count times into a new string.

Answer 6

printf doesn't do that -- and printf is overkill for printing a single character.

char c = '*';
int count = 42;
for (i = 0; i < count; i ++) {
    putchar(c);
}

Don't worry about this being inefficient; putchar() buffers its output, so it won't perform a physical output operation for each character unless it needs to.

Answer 7

If you have a compiler that supports the alloca() function, then this is possible solution (quite ugly though):

printf("%s", (char*)memset(memset(alloca(10), '\0', 10), 'x', 9));

It basically allocates 10 bytes on the stack which are filled with '\0' and then the first 9 bytes are filled with 'x'.

If you have a C99 compiler, then this might be a neater solution:

for (int i = 0;  i < 10;  i++, printf("%c", 'x'));

Answer 8

#include <stdio.h>
#include <string.h>

void repeat_char(unsigned int cnt, char ch) {
    char buffer[cnt + 1];
    /*assuming you want to repeat the c character 30 times*/
    memset(buffer,ch,cnd); buffer[cnt]='\0';
    printf("%s",buffer)
}

Answer 9

you can make a function that do this job and use it

#include <stdio.h>

void repeat (char input , int count )
{
    for (int i=0; i != count; i++ )
    {
        printf("%c", input);
    }
}

int main()
{
    repeat ('#', 5);
    return 0;
}

This will output

#####

Answer 10

char buffer[41];

memset(buffer, '-', 40);    // initialize all with the '-' character<br /><br />
buffer[40] = 0;             // put a NULL at the end<br />

printf("%s\n", buffer);     // show 40 dashes<br />

Answer 11

printf("%.*s\n",n,(char *) memset(buffer,c,n));

n <= sizeof(buffer) [ maybe also n < 2^16]

However the optimizer may change it to puts(buffer) and then the lack of EoS will .....

And the assumption is that memset is an assembler instruction (but still a loop be it on chip).

Strictly seen there is no solution given you precondition 'No loop'.

Answer 12

For best performance you could use memset then send that string directly to printf:

    char buffer[40];
    memset(buffer, '-', 39); // Set 39 characters
    buffer[39] = 0; // Tag the end of the string
    printf(buffer); // Print it

For instance, this method could be used to produce a rectangle of a given width and height:

#include <stdio.h>
 
int main() {
    int width = 40; // External width
    int height = 7; // External height

    char buffer[width + 2];
    memset(buffer, '-', width); // Set the dashes
    buffer[width] = '\n';  // Add carriage return
    buffer[width + 1] = 0; // Tag the end of string

    printf(buffer);

    char buffer2[width + 2];
    memset(buffer2 + 1, ' ', width - 2);
    buffer2[0] = '|';
    buffer2[width - 1] = '|';
    buffer2[width] = '\n';
    buffer2[width + 1] = 0;

    int l = height - 2;
    while (l-- > 0)
        printf(buffer2);

    printf(buffer);
    
    return 0;
}

Output:

----------------------------------------
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
----------------------------------------

Answer 13

i think doing some like this.

void printchar(char c, int n){
     int i;
     for(i=0;i<n;i++)
         print("%c",c);
}

printchar("*",10);