How to repeat a char using printf?
CPrintfC Problem Overview
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.
C Solutions
Solution 1 - C
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.
Solution 2 - C
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;
}
Solution 3 - C
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);
Solution 4 - C
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
Solution 5 - C
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.
Solution 6 - C
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.
Solution 7 - C
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'));
Solution 8 - C
#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)
}
Solution 9 - C
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
#####
Solution 10 - C
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 />
Solution 11 - C
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'.
Solution 12 - C
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);