exercise3-6.md

Exercise 3-6

Write a version of itoa that accepts three arguments instead of two. The third argument is a minimum field width; the converted number must be padded with blanks on the left if necessary to make it wide enough.

Process

This one took not even half an hour, just two lines of code. while(i < w) s[i++] = 0x20;
I guess the question is to make sure you have an understanding of the concept. Which I believe I do now.

I was a little confused about the word blank. I initially thought it meant a 0 character but after checking my solution I realised it meant a blank space

Code

#include <stdio.h>
#include <string.h>
#define size 30

void itoa(int, char[], int);
void reverse(char s[]);

int main(void)
{
    char s[size];
    for (int w = 0; w < 20; w = w + 5)
    {
        itoa(50, s, w);
        printf("padding of %2d:%s\n", w, s);    
    }
}

/* itoa: convert n to characters in s with minimum field width */
void itoa(int n, char s[], int w)
{
    int i, sign;

    if ((sign = n) < 0) /* record sign */ 
        n = -n;         /* make n positive */
        
    i = 0;
    do {        /* generate digits in reverse order */
        s[i++] = (unsigned) n % 10 + '0';  /* get next digit */
    } while ((n = (unsigned) n / 10) > 0);    /* delete it */
    
    if (sign < 0)
        s[i++] = '-';
    
    while(i < w)
        s[i++] = 0x20;    
    
    s[i] = '\0';
    reverse(s);
}

/* reverse: reverse string s in place */
void reverse(char s[])
{
    int c ,i ,j;
    
    for (i = 0, j = strlen(s) -1; i < j; i++, j--) {
        c = s[i];
        s[i] = s[j];
        s[j] = c;
    }
}

Output

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