print number of spaces using printf in C

Question

I was wondering how can I do it ,to print certain number of spaces using printf in CI was thinking something like this,but also my code doesn't print after the first printf statement,my program compiles perfectly fine tho.I'm guessing I have to print N-1 spaces but I'm not quite sure how to do so.

Thanks.

#include <stdio.h>
#include <limits.h>
#include <math.h>

int f(int);

int main(void){
    int i, t, funval,tempL,tempH;
    int a;

    // Make sure to change low and high when testing your program
    int low=-3, high=11;
    for (t=low; t<=high;t++){
        printf("f(%2d)=%3d\n",t,f(t));                 
    }
    printf("\n");
    if(low <0){
        tempL = low;
        tempL *=-1;
        char nums[low+high+1];
        for(a=low; a <sizeof(nums)/sizeof(int);a+5){
            printf("%d",a);
        }
    }
    else{
        char nums[low+high];
        for(a=low; a <sizeof(nums)/sizeof(int);a+5){
            printf("%d",a);
        }
    }

    // Your code here...
    return 0;
}


int f(int t){
    // example 1
    return (t*t-4*t+5);

    // example 2
    // return (-t*t+4*t-1);

    // example 3
    // return (sin(t)*10);

    // example 4
    // if (t>0)
    //  return t*2;
    // else
    //  return t*8;
}

the output should be something like this:

   1       6       11      16      21      26     31
   |       |       |       |       |       |       |  

Answer 1

Printing n spaces

printf has a cool width specifier format that lets you pass an int to specify the width. If the number of spaces, n , is greater than zero:

printf("%*c", n, ' ');

should do the trick. It also occurs to me you could do this for n greater than or equal to zero with:

printf("%*s", n, "");

Printing 1, 6, 11, ... pattern

It's still not fully clear to me what you want, but to generate the exact pattern you described at the bottom of your post, you could do this:

for (i=1; i<=31; i+=5)
    printf("%3d   ", i);
printf("\n");
for (i=1; i<=31; i+=5)
    printf("  |   ");
printf("\n");

This outputs:

  1     6    11    16    21    26    31   
  |     |     |     |     |     |     |   

Answer 2

Had your objective been :

Start printing at a specified width using printf

You could achieve it like below :

printf("%*c\b",width,' ');

Add the above stuff before printing actual stuff, eg. before a for-loop.

Here the \\b positions the cursor one point before the current position thereby making the output appear to start at a particular width, width in this case.