C programming. The FizzBuzz program [closed]

Answer 1

This one avoids some code repetition but requires a temporary variable char t

void FizzBuzz( ) {
    char t = 0;
    for (unsigned char i = 1; i <= 100; ++i, t = 2) {
        (i % 3) ? --t : printf("Fizz");
        (i % 5) ? --t : printf("Buzz");
        if (!t) printf("%d", i);
        printf("\n");   
    }
}

Answer 2

#include <stdio.h>

char const * template[] = {
  "%i",
  "Buzz",
  "Fizz",
  "FizzBuzz"
};
const int __donotuseme3[] = { 2, 0, 0 };
const int __donotuseme5[] = { 1, 0, 0, 0, 0 };
#define TEMPLATE(x) (template[__donotuseme3[(x) % 3] | __donotuseme5[(x) % 5]])

int
main(void) {
  int i;
  for (i = 1; i <= 100; i++) {
    printf(TEMPLATE(i), i);
    putchar('\n');
  }
  return 0;
}

Answer 3

Obfuscated form of Mr Lister's answer

main(int i){while(i++<100){printf("number= %d %s%s",i,i%3?"":"Fizz",i%5?"":"Buzz");}}

Answer 4

You can do it using a String:

String s="";
if(num%3==0)
   s+="fizz";
if(num%5==0)
   s+="buzz";
if(s.length()==0)
   s+=num+"";

Answer 5

I'd go with a helper function :-)

#include <stdio.h>

int fbindex(int n) {
    int i = 0;
    if (n % 3 == 0) i += 1;
    if (n % 5 == 0) i += 2;
    return i;
}

int main(void) {                                 
    const char *fb[] = {"%d\n", "Fizz\n", "Buzz\n", "FizzBuzz\n"};
    for (int i = 1; i <= 100; i++) printf(fb[fbindex(i)], i);                
}