Reputation: 24944
I want to print a line similar as following:
====================================
And I need to control the count of the char, and able to specify which char to print.
I don't want to use loop.
Is it possible to do this with a single printf()
statement?
@Update
I ask this because I use printf in this way sometimes:
printf("%10s\n", "abc");
So, if printf could do this, then it's possible to do what I ask, I am just not sure ... now I know it can't ...
Ok, I wrote a simple util function to do this:
#include <stdio.h>
void printRepeatChar(char c, int count) {
char cs[count+1];
int i;
for(i=0; i<count; i++)
cs[i] = c;
cs[count] = '\0';
printf("%s\n", cs);
}
int main(int argc, char * argv[]) {
printRepeatChar('-', 6*4);
}
maybe use memset()
from string.h
instead of the direct loop makes the code shorter, just as in the answers.
And, thank you all for help.
Upvotes: 0
Views: 2674
Reputation: 2995
#include <stdio.h>
#include <string.h>
void PrintStuff( char to_print, int length ) {
// adjust buffer size as desired
char buffer[256];
// -1 for null terminator
if( length > sizeof(buffer)-1 ) length = sizeof(buffer)-1;
// fill buffer with desired character
memset( buffer, to_print, length );
// add null terminator
buffer[length] = 0;
// print to output
puts( buffer );
}
int main() {
PrintStuff( '=', 11 );
return 0;
}
And to answer the subquestion: no, printf cannot repeat a character as a formatting rule. It can only repeat spaces or 0's when padding.
Upvotes: 3
Reputation: 19874
#include <stdio.h>
#include <string.h>
int main(void) {
char c='=';
char a[20];
memset(a,c,(sizeof(a)-1));
a[19] = '\0';
printf("%s\n",a);
return 0;
}
Dynamic memory allocation and character scanning can be added to this .
Upvotes: 2