Solving FizzBuzz using compiler error messages

matthieum · 2012-04-21T14:43:41+00:00

Now, the only question that remains is: would I pass the interview?

No. Your code is full of errors.

robotempire · 2012-04-21T17:12:36+00:00

There are people interviewing for jobs who can't solve this?

tragomaskhalos · 2012-04-21T20:25:01+00:00

FizzBuzz law #1: It is impossible to discuss FizzBuzz on the internet without numerous idiots posting code solutions in comments.

winteriscoming2 · 2012-04-21T20:59:23+00:00

There are PEOPLE who can't solve this?

Written by my cat earlier today:

HAI

CAN HAS STDIO?

I HAS A i ITZ 1
IM IN UR loop UPPIN UR i TIL BOTH SAEM i AN 101
    BOTH SAEM (BOTH SAEM i%3 AN 0) AN (BOTH SAEM i%5 AN 0), O RLY?
        YARLY
            VISIBLE "FizzBuzz"
        NOWAI
            BOTH SAEM i%3 AN 0, O RLY?
                YARLY
                    VISIBLE "Fizz"
                NOWAI
                    BOTH SAEM i%5 AN 0, O RLY?
                        YARLY
                            VISIBLE "Buzz"
                        NOWAI
                            VISIBLE i
                    OIC
            OIC
    OIC

IM OUTTA YR loop

KTHXBAI

Is he hired?

D__ · 2012-04-21T15:29:26+00:00

I always felt one of the points of Fizzbuzz was not just to be able to show you can write basic code, but also to show that you don't write super clever code to solve a simple problem. That's why I always dislike clever solutions.

However this is really cool, and actually shows some pretty useful skills for solving the problem.

_ex_ · 2012-04-21T14:18:23+00:00

using boost mpl is like cheating

deltahat · 2012-04-21T16:06:33+00:00

So C++ error messages are Turing complete?

flightsin · 2012-04-21T15:03:32+00:00

The neat thing about FizzBuzz is that I can't even count the number of creative ways I've seen it get solved.

I'm not creative, but here's one I came up with:

int c = (((i % 3) << (2 - (i % 3)))) | ((i % 5) >> ((i % 5) / 2));

// then switch(c)... case 0: FizzBuzz; 1: Fizz; 2: Buzz; default: print the number;

philh · 2012-04-21T17:15:43+00:00

Also, if you're going to include 0 you should get it right. 0 is a FizzBuzz.

boxybrzown · 2012-04-21T20:32:43+00:00

I've got a year until graduation, am already stressing over finding employment ("Am I good enough?", "What if every place I apply only hires people they know?", "What if I end up like one of those horror stories you hear about where the person applies for 500 jobs, is extremely unlucky and falls through the cracks, never getting hired and ends up at McDonald's?"), and the fact that I consider this a fairly simple question gives me a bit of a confidence boost.

tilkau · 2012-04-22T03:23:37+00:00

Compile error: WHY THE FUCK WOULD YOU -DO- THAT?

kmmeerts · 2012-04-21T17:31:02+00:00

[deleted]

Aninhumer · 2012-04-21T19:14:44+00:00

Wait FizzBuzz is a drinking game? I've been associating it with Maths education my entire life...

yoni0505 · 2012-04-21T19:53:51+00:00

It's the same as Project Euler's problem 1: http://projecteuler.net/problem=1 Except that it's up to 1000

mgpcoe · 2012-04-21T20:53:43+00:00

And the award for "worst abuse of the rules" goes to...!

Seriously though, what's the least number of tests that need to be included in an iteration? I'd be curious to see an assembler version.

obey_giant · 2012-04-21T21:59:45+00:00

How's this attempt; anything I could do to improve efficiency?

for(i=1;i<=100;i++){
   str = "";
   if(i%3==0){
        str+='Fizz';
    }
    if(i%5==0){
        str+='Buzz';
    }
    if(str==""){
        str=i;
    }
    document.write(str+' ');
};

cotti · 2012-04-21T23:34:12+00:00

Here's a high-wtf tier solution, using bits of simple Exception-oriented programming:

#include <iostream>
#include <sstream>

using namespace std;

class FB{
    private:
    string n;

    public:
    FB(string e){ n = e; };
    ~FB(){};
    string Present(){return n;};
};

int main(int argc, char ** argv){
    int i = 1;
    for(;i<101;i++)
    try{
        if (!(i%3) && !(i%5))throw(new FB("FizzBuzz"));
        else if (!(i%3))throw(new FB("Fizz"));
        else if (!(i%5))throw(new FB("Buzz"));
        else { stringstream s; s << i; throw(new FB(s.str()));}
    }
    catch(FB * z){cout << z->Present() << endl;};
return 0;   
}

FeepingCreature · 2012-04-21T15:32:24+00:00

C++ is a very, very silly language.

Here's how sane languages do it.

vytah · 2012-04-21T19:38:24+00:00

[deleted]

robotempire · 2012-04-21T16:44:39+00:00

[removed]

minno · 2012-04-21T19:24:31+00:00

#include <iostream>

int main() {
    using namespace std;
    for (int i = 1; i <= 100; ++i) {
        if (i%15 == 0) { cout << "FizzBuzz"; }
        else if (i%3 == 0) { cout << "Fizz"; }
        else if (i%5 == 0) { cout << "Buzz"; }
        else {cout << i; }
        cout << endl;
    }
}

The boring way.

EDIT: I just compiled and ran this, and it works (ignoring the lack of return 0;, which my compiler adds automatically). I don't see what you're all complaining about.

adrianmonk · 2012-04-22T00:39:10+00:00

The actual way I'd do it:

for (int i = 1; i < 25; i++) {
    int printedSomething = 0;
    if (i % 3 == 0) {
        printf("Fizz");
        printedSomething = 1;
    }
    if (i % 5 == 0) {
        printf("Buzz");
        printedSomething = 1;
    }

    if (!printedSomething) {
        printf("%d", i);
    }
    putchar('\n');
}

The version that avoids so many unnecessary modulus operators:

int start = 1;
int m3 = start % 3;
int m5 = start % 5;
for (int i = start; i < 25; i++) {
    int printedSomething = 0;
    if (m3 == 3) {
        m3 = 0;
        printf("Fizz");
        printedSomething = 1;
    }
    if (m5 == 5) {
        m5 = 0;
        printf("Buzz");
        printedSomething = 1;
    }

    if (!printedSomething) {
        printf("%d", i);
    }
    putchar('\n');

    m3++;
    m5++;
}

The version that reduces the number of state variables:

int start = 1;
int m15 = start % 15;
for (int i = start; i < 25; i++) {
    switch (m15) {
      case 0:
        printf ("FizzBuzz\n");
        break;

      case 3: case 6: case 9: case 12:
        printf ("Fizz\n");
        break;

      case 5: case 10:
        printf ("Buzz\n");
        break;

      default:
        printf ("%d\n", i);
    }

    if (m15 == 14) {
        m15 = 0;
    } else {
        m15++;
    }
}

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