One common trick is to handle the %3 and %5 with down-counters that you reset to 3 or 5 when they reach zero. Or better, unroll the loop by 3 so you only need one down-counter.

(Also, no, `%5` isn't "horrifically" slow when it's a compile-time-constant modulus: you or the compiler can do it with a multiply and shift for the integer division, and then x%5 = x - (x/5 * 5). https://godbolt.org/g/3HwBrF. It still sucks though.)

I wrote an x86 assembly FizzBuzz for fun a while ago, intended to be an example of how to optimize (https://stackoverflow.com/a/37494090/224132). Some parts of it kind of suck, though. I made some parts efficient (like appending "buzz\n" to a buffer with a mov [rdx], r13 / add rdx, 5), but left in too many conditional branches and a function call instead of inlining everything when unrolling.

I'm glad so many people like my post that the OP linked. It was fun to write. :)

Well, there is this approach:

    #include <stdio.h>
    #include <stdint.h>
    
    #define FIZZ(x)         (x & 0x0924)
    #define BUZZ(x)         (x & 0x0210)
    #define FIZZBUZZ(x)     (x & 0x4000)
    #define NONE(x)         (x & 0x34cb)
    
    void fizzbuzz(size_t n_max)
    {
        uint32_t x = 1;
        for (size_t i = 1; i <= n_max; i++)
        {
            if (FIZZ(x))
              printf("fizz\n");
            if (BUZZ(x))
              printf("buzz\n");
            if (FIZZBUZZ(x))
              printf("fizzbuzz\n");
            if (NONE(x))
              printf("%d\n", i);
            x <<= 1;
            x |= (x >> 15);
        }
    }
    
    int main(int argc, char **argv)
    {
        fizzbuzz(100);
    }

It's less costly than integer division, and I'm not sure it's what you had in mind. The output here still duplicates the strings.

Good nerdsnipe.

I decided to reject your conditions and replace them with my own, for no apparent reason. Mine has some of the redundancy you wish to eliminate, but the loop body is completely branchless, aside from the call to printf, of course, which we're apparently ignoring.

    #include <stdio.h>
    
    #define NUM "\x06" "%zu\n\0"
    #define FIZZ "\x07" "Fizz\n\0"
    #define BUZZ "\x07" "Buzz\n\0"
    
    const char *x =
        NUM
        NUM
        FIZZ
        NUM
        BUZZ
        FIZZ
        NUM
        NUM
        FIZZ
        BUZZ
        NUM
        FIZZ
        NUM
        NUM
        "\xa6" "FizzBuzz\n";
    
    void fizzbuzz(size_t upTo) {
        for(size_t i = 1; i <= upTo; i++) {
            printf(x + 1, i);
            x += *x;
        }
    }
    
    int main(int argc, char **argv) {
        fizzbuzz(100);
    }

    struct state {
        int next;
        char fmt[6];
    };

Let's optimize the if statements away, too (no extra memory accesses or branches, either).

  #include <stdio.h>
  #include <stdint.h>
  
  #define F(n) (1 << (2*(n)-2))
  #define B(n) (1 << (2*(n)-1))
  
  int main() {
    uint32_t m = F(3) | F(6) | F(9) | F(12) | F(15) | B(5) | B(10) | B(15);
    int i;
    static char t[4][16]= { "%d\n", "Fizz\n", "Buzz\n", "FizzBuzz\n" };
    for (i = 1; i <= 100; i++) {
      printf(t[m&3], i);
      m = (m >> 2) | (m << 28);
    }
    return 0;
  }

Transforming that into an if-based version is left as an exercise for the reader. :)

This probably wouldn't be a very good optimization puzzle as the cost of printing would so dominate the arithmetic operations. :(

  #define FIZZ 1
  #define BUZZ 2
  static const uint8_t divs[] = 
 {FIZZ+BUZZ,0,0,FIZZ,0,BUZZ,FIZZ,0,0,FIZZ,BUZZ,0,FIZZ,0,0};
  static const char* strs[] = {
    [0] = "%u\n",
    [FIZZ] = "fizz\n",
    [BUZZ] = "buzz\n",
    [FIZZ+BUZZ] = "fizzbuzz\n"
  };

  void fizzbuzz(uint32_t upTo)
  {
    uint32_t i, state;

    for(i = state = 0; i < upTo; i++) {
      printf(strs[divs[state]], i);
      if(++state == sizeof(divs) / sizeof(*divs))
          state = 0;
  }

No divisibility tests at all

One level of arrays may be skipped at cost of extra .rodata size (store string pointers directly in divs). But in a modern cpu cost of the extra load is small

  #define FIZZ 1
  #define BUZZ 2


  struct {
    uint8_t flags;
    uint8_t next;
  } static const nfo[] = {
    {FIZZ+BUZZ,1},
    {0,2},
    {0,3},
    {FIZZ,4},
    {0,5},
    {BUZZ,6},
    {FIZZ,7},
    {0,8},
    {0,9},
    {FIZZ,10},
    {BUZZ,11},
    {0,12},
    {FIZZ,13},
    {0,14},
    {0,0}
  } __attribute__((align(YOUR_L1D_LINE_SIZE)));
  static const char* strs[] = {
    [0] = "%u\n",
    [FIZZ] = "fizz\n",
    [BUZZ] = "buzz\n",
    [FIZZ+BUZZ] = "fizzbuzz\n"
  };

  void fizzbuzz(uint32_t upTo)
  {
    uint32_t i, state;

    for(i = state = 0; i < upTo; i++) {
      printf(strs[nfo[state].flags], i);
      state = nfo[state].next;
    }
  }

  .data: 0
  .rodata: 30 + 4 * sizeof(void*) + strings
  .text: depending on arch, but not much

Hm. I don't quite see how to do it, but I think I know one of the tricks you have in mind. If you have the book Hacker's Delight handy, open it up to 10-17, "Test for Zero Remainder after Division by a Constant". With n<=100 you can get an intermediate result with a single 32-bit multiply and mask that will allow testing divisibility by 3 and 5 each with a single operation (and a bit of union trickery to access high bits without counting as an operation). I just don't see how to make the test 'not a multiple of 3 or 5' come out of that in one C operation yet. So maybe this isn't the right technique.

Assuming 32-bit unsigned integers (but similarly for other integer sizes), multiplying by 0x11111111 and shifting right 28 bits would give a function of period 15, at least for inputs between 1 and 0x1000000e inclusive. Faster than dividing by 15.