Chapter 2 - Types, Operators and Expressions 2

Answer to Exercise 2-2, page 42
Solutions by "Flippant Squirrel" :-) and Craig Schroeder

Exercise 2-2 discusses a for loop from the text. Here it is:

  for(i=0; i<lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
    s[i] = c;

 

Write a loop equivalent to the for loop above without using && or || .

#include <stdio.h>

#define MAX_STRING_LENGTH 100

int main(void)
{
 /*
 for (i = 0; i < lim-1 && (c=getchar()) != '\n' && c != EOF; ++i)
  s[i] = c;
 */

 int i = 0,
  lim = MAX_STRING_LENGTH,
  c;
 char s[MAX_STRING_LENGTH];

 while (i < (lim - 1))
 {
  c = getchar();

  if (c == EOF)
   break;
  else if (c == '\n')
   break;

  s[i++] = c;
 }

 s[i] = '\0';   /* terminate the string */

 return 0;
}

Here's a Category 1 solution from Craig Schroeder, which is not so much exegetic as - um - cute. :-)

#include <stdio.h>

#define lim 80

int main()
{
        int i, c;
        char s[lim];

        /* There is a sequence point after the first operand of ?: */

        for(i=0; i<lim-1 ? (c=getchar()) != '\n' ? c != EOF : 0 : 0 ; ++i)
                s[i] = c;
       
        return s[i] ^= s[i]; /* null terminate and return. */
}