How can I print multiple character with one printf? - c

Question

I want to print multiple character using printf. My approach up to now is this-
#include <stdio.h>
int main()
{
printf("%*c\n", 10, '#');
return 0;
}
But this only prints 9 spaces before the #.
I want to print it like this-
##########
I am unable to figure out how to do this. Please help me?

Answer 1

You can not use printf like that to print repetitive characters in Ansi C. I suggest you to use a loop like this -
#include <stdio.h>
int main()
{
int i;
for(i = 0; i < 10; i++) putchar('#');
return 0;
}
Or if you have absolutely no desire to use loops, you can do something like this-
#include <stdio.h>
int main()
{
char out[100];
memset(out, '#', 10);
out[10] = 0;
printf("%s", out);
return 0;
}
By the way, using printf like this also works-
#include <stdio.h>
int main()
{
printf("%.*s", 10, "############################################");
return 0;
}

Answer 2

I think the best approach, if you have an upper limit to the number of characters to be output is:
printf("%.*s", number_of_asterisks_to_be_printed,
"**********************************************************************");
I think this will also be the most efficient, portable way to do that.

Answer 3

this will print ten # characters, followed by a newline
char tenPounds[] = "##########";
printf( "%s\n", tenPounds);

Answer 4

I am working on a similar problem in "The C Programming Language" book (exercises 1-13 and 1-14). My own program, to start simply, is to count the occurrences of the digits 0 to 9 in a given input, and print a horizontal histogram made of '=' bars according to each count.
To do this, I created the following program;
main() {
int c, ix, k;
int nDigit[10];
//Instantiate zero values for nDigits array
for(ix = 0; ix < 10; ix++) {
nDigit[ix] = 0;
}
//Pull in input, counting digit occurrences and
//incrementing the corresponding value in the nDigit array
while ((c = getchar()) != EOF) {
if (c >= '0' && c <= '9') {
++nDigit[c-'0'];
}
}
//For each digit value in the array, print that many
//'=' symbols, then a new line when completed
for (ix = 0; ix < 10; ix++) {
k = 0;
while (k <= nDigit[ix]) {
putchar('=');
k++;
}
printf("\n");
}
}
Note that this is a work in progress. A proper histogram should include axes labels, and most importantly this program does not account for digits of zero count. If the input includes five 1's but no 0's, there is no visual way to show that we have no zeros. Still, the mechanism for printing multiple symbols works.