C programming. The FizzBuzz program [closed] - c

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I had a quiz and I wrote this code:
Print Fizz if it is divisible by 3 and it prints Buzz if it is
divisible by 5. It prints FizzBuss if it is
divisible by both. Otherwise, it will print the numbers between 1 and 100.
But after I arrived home, I wondered if could have
writen it with less code. However, I could not come out
with a shorter code.
Can I do it with a shorter code? Thanks.
This is what I wrote and I think it works well. But can I have done it
with less code.
#include <stdio.h>
int main(void)
{
int i;
for(i=1; i<=100; i++)
{
if(((i%3)||(i%5))== 0)
printf("number= %d FizzBuzz\n", i);
else if((i%3)==0)
printf("number= %d Fizz\n", i);
else if((i%5)==0)
printf("number= %d Buzz\n", i);
else
printf("number= %d\n",i);
}
return 0;
}

Answer 1

You could also do:
#include <stdio.h>
int main(void)
{
int i;
for(i=1; i<=100; ++i)
{
if (i % 3 == 0)
printf("Fizz");
if (i % 5 == 0)
printf("Buzz");
if ((i % 3 != 0) && (i % 5 != 0))
printf("number=%d", i);
printf("\n");
}
return 0;
}
A few lines shorter, and a lot easier to read.

Answer 2

I'm not sure when you'd start calling it unreadable, but there's this.
#include <stdio.h>
int main(void)
{
int i = 1;
for (; i<=100; ++i) {
printf("number= %d %s%s\n", i, i%3?"":"Fizz", i%5?"":"Buzz");
}
return 0;
}

Answer 3

If a number is divisible by both 3 and 5, then it's divisible by 15, so:
for each number 1 to 100:
if number % 15 == 0:
print number, "fizzbuzz"
else if number % 5 == 0:
print number, "buzz"
else if number % 3 == 0:
print number, "fizz"
else:
print number
Other than that, you probably won't get it much shorter, at least in a conventional language like C (and I'm assuming you don't want the normal code-golf style modifications that make your code unreadable).
You could also get the whole thing into two lines if you packed the entire main function onto a single large line, but I would hope you wouldn't be after that sort of trickery either.
You can possibly get it faster (though you should check all performance claims for yourself) with something like:
static const char *xyzzy[] = {
"", "", "fizz", "", "buzz",
"fizz", "", "", "fizz", "buzz",
"", "fizz", "", "buzz", "fizzbuzz",
// Duplicate those last three lines to have seven copies (7x15=105).
};
for (int i = 1; i <= 100; i++)
printf ("%d %s\n", i, xyzzy[i-1]);
As an aside, that array of char pointers is likely to be less space-expensive than you think, thanks to constant amalgamation - in other words, it will be likely that there will only be one of each C string.
As I say, whether it's faster should be tested. In addition, your original specs only called for the shortest code so it may be irrelevant.

Answer 4

#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 5

I would say that modulo is expensive while comparisons are cheap so only perform the modulo once. That would yield something like this.
int i;
for( i = 0; i!=100; ++i ) {
bool bModThree = !(i % 3);
bool bModFive = !(i % 5);
if( bModThree || bModFive ) {
if( bModThree ) {
printf( "Fizz" );
}
if( bModFive ) {
printf( "Buzz" );
}
} else {
printf( "%d", i );
}
printf( "\n" );
}

Answer 6

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 7

i would write something like that
main(){
if (i % 3 == 0){
cout<<"Fizz";
}
if (i % 5 == 0){
cout<<"Buzz";
}
// So if both are true, it will print “FizzBuzz” and augment the two strings
}

Answer 8

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);
}

Answer 9

void main()
{
int i = 0;
char h[4];
while (++i <= 100)
{
sprintf(h, "%d", i);
printf("%s%s%s\n", i%3 ? "" : "fizz", i%5 ? "" : "buzz", (i%3 && i%5) ? h: "");
}
}

Answer 10

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 11

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");}}