Write a C program that accepts up to six arguments at the command line prompt wh

Question

Write a C program that accepts up to six arguments at the command line prompt when you run the program. The program should print the first character of any odd- numbered arguments (numbers 0, 2, and 4), and the second character of any even-numbered arguments (numbers 1, 3, and 5). The characters printed should be separated by spaces. The program should inform the user of the correct program usage if fewer than two or more than six arguments are provided. Assume each argument contains at least two characters.

Sample Input/Output:

Explanation / Answer

#include <stdio.h>

int main(int argc, char *argv[]) {
   if(argc > 1 && argc < 7) {
       int i;
       for(i = 0; i < argc; ++i) {
           if(i % 2 == 0) {
               printf("%c ", argv[i][0]);
           } else {
               printf("%c ", argv[i][1]);
           }
       }
       printf(" ");
   } else {
       printf("Must have 2 to 6 arguments. ");
   }
   return 0;
}

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