How to put an actual variable n in printf("%nd",var) instead of any integer in C? - printf

Question

I'm using %nd in a loop where 'var' increments. Now i want 'n' to be the no. of digits of 'var' so that it aligns correctly

Answer 1

You can specify a variable as the width parameter in printfusing the * width specifier:
printf("%*d\n", digs, var);
This will use the value of the digs variable as the format width.
A simple, complete example:
#include <stdio.h>
int main()
{
for (int i = 1; i <= 10; ++i)
printf("%*d\n", i, i); // i is used both as the width specifier and the actual value printed!
}

Answer 2

I would do it by generating the desired format-string using sprintf() at run-time, like this:
#include <stdio.h>
int main(int, char **)
{
char formatBuf[100];
for (int i=0; i<10; i++)
{
// Note: %% specifies one literal %
// %d specifies the width-number to put in the format string
// ... and the final d is a literal 'd' to include in the format string
sprintf(formatBuf, "num=%%%dd", i);
printf("formatBuf=[%s]\n", formatBuf);
printf(formatBuf, i);
printf("\n");
}
return 0;
}
... which gives this output as a demonstration:
formatBuf=[num=%0d]
num=0
formatBuf=[num=%1d]
num=1
formatBuf=[num=%2d]
num= 2
formatBuf=[num=%3d]
num= 3
formatBuf=[num=%4d]
num= 4
formatBuf=[num=%5d]
num= 5
formatBuf=[num=%6d]
num= 6
formatBuf=[num=%7d]
num= 7
formatBuf=[num=%8d]
num= 8
formatBuf=[num=%9d]
num= 9