FizzBuzz Javascript Solution - Bodyless for loop

FizzBuzz.js

// Concise
var f='Fizz', b='Buzz', i=0, d3, d5;
for (i; ++i <= 100; d3 = !(i % 3), d5 = !(i % 5), console.log(d3 ? d5 ? f+b : f : d5 ? b : i));

// Multi-line, commented
var               /* Declare our variables outside the loop, a performance best-practice */
    f='Fizz',     /* Variable `f` so we don't repeat 'Fizz' twice - DRY */
    b='Buzz',     /* Variable `b` so we don't repeat 'Buzz' twice - DRY */
    i=0,          /* For-loop counter, start at 0 */
    d3,           /* setup a variable for checking divisibility by 3 */
    d5;           /* setup a variable for checking divisibility by 5 */
for (                 /* Bodyless for loop */
    i;                /* Iterator on i */
    ++i <= 100;       /* ++i pre-increments to 1, stops at 100 */
    d3 = !(i % 3),    /* Logically we're testing for i%3 === 0 to be TRUE, indicating
                       * i is divisible by 3. Instead we test for 0, which also
                       * represents FALSE in Javascript. Use the ! NOT operator
                       * to coerce 0 to boolean value 1- which is TRUE. */
    d5 = !(i % 5),    /* Testing that NOT (i % 5) is TRUE using same logic as above */
    console.log(      /* ouput the result of a ternary operation to the console */
        d3 ?          /* was i divisible by 3? */
            d5 ?      /* was i divisible by 5? *? */
                f+b : /* Both conditions d3 and d5 are divisible TRUE: return string 'FizzBuzz': else */
            f :       /* only !(i%3) is true, just return string 'Fizz' */
        d5 ?          /* BOTH 3 AND 5 weren't divisible, was 5 divisible by itself TRUE? */
        b :           /* yes, return 'Buzz' */
        i             /* Nope, just return i */
    )                 /* end of console.log */
);                    /* end of for-loop */