Discussion on lobsters: https://lobste.rs/s/oxjvv0/0_1_0_2

It’s not that Roc only supports base-10 arithmetic. It also supports the typical base-2 floating-point numbers, because in many situations the performance benefits are absolutely worth the cost of precision loss. What sets Roc apart is its choice of default; when you write decimal literals like 0.1 or 0.2 in Roc, by default they’re represented by a 128-bit fixed-point base-10 number that never loses precision, making it reasonable to use for calculations involving money.

In Roc, floats are opt-in rather than opt-out.

  • Andy@programming.devOPM
    2·
    5 months ago

    Even more:

    If a function takes a number—whether it’s an integer, a floating-point base-2 number, or a Dec—you can always write 5 as the number you’re passing in. (If it’s an integer, you’ll get a compiler error if you try to write 5.5, but 5.5 will be accepted for either floats or decimal numbers.)

    Because of this, it’s actually very rare in practice that you’ll write 0.1 + 0.2 in a .roc file and have it use the default numeric type of Dec. Almost always, the type in question will end up being determined by type inference—based on how you ended up using the result of that operation.

    For example, if you have a function that says it takes a Dec, and you pass in (0.1 + 0.2), the compiler will do Dec addition and that function will end up receiving 0.3 as its argument. However, if you have a function that says it takes F64 (a 64-bit base-2 floating-point number), and you write (0.1 + 0.2) as its argument, the compiler will infer that those two numbers should be added as floats, and you’ll end up passing in the not-quite-0.3 number we’ve been talking about all along.