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

Answer 1

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 2

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);

Answer 3

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 4

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 5

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 6

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 />