3 ms·

I'm not sure why but I have a visceral, negative response to this. It might be the best solution, but it definitely /feels/ like the worst of all the worlds.

sk0g 5y ago
Is it? In most languages you do something like `time.Sleep(1 * 1e6)` instead at which point it could be a second, a few minutes, a day, who really knows? I'm just not seeing any major downsides of this, keep in mind `time.Second` isn't the only one of its kind, you have millisecond, minute, hour, etc etc.
syntaxfree 5y ago
Overloading the asterisk is always weird because multiplication is expected to be associative etc etc.
- ts4z 5y ago
  time.Second*1 gets the same value; it's not overloaded. Well, ok, so what actually happens is that time.Second is a time.Duration, and Duration*int yields Duration (and int*Duration yields Duration). But the value of time.Second is actually a Duration with value 1000000, IIRC -- it's microseconds. It's just the type that's special, and the int handling here is general over a lot of types. It really is nice in practice.
  - ts4z 5y ago
    (As noted elsewhere it's nanoseconds.)
- xyzzy_plugh 5y ago
  It's not overloaded. It's a unit. You can just type time.Sleep(time.Second) but this reads nicely time.Sleep(3 * time.Minute)
- rwcarlsen 5y ago
  It's not really overloading. it is associative: 2*time.Hour+1*time.second == time.Second+time.Hour*2
  - wtallis 5y ago
    I believe that's illustrating commutativity, not associativity.
    - thaumasiotes 5y ago
      The illustration is wrong, but the claim is correct; multiplication between units and scalars is just as associative as you'd expect. Multiplying one kilowatt by an hour gives you exactly the same result as multiplying 1 by a kilowatt-hour.
- Laremere 5y ago
  There is no operator overloading here. The type Duration is effectively an int64: https://pkg.go.dev/time#Duration https://pkg.go.dev/time#Duration
- vitus 5y ago
  I'm not sure I follow. 2 * 3 * time.Second is the same whether you group 2 * (3 * time.Second) or (2 * 3) * time.Second (namely, the implicit grouping under left-associativity). You wouldn't normally write time.Sleep(time.Second * time.Second) because your units wouldn't work out. (Apparently you can write that in Golang; it's just very sketchy and results in a 30-year sleep.)
- thaumasiotes 5y ago
  But from a mathematical point of view, the relationship between a unit and its coefficient is that you're multiplying them together. Why would it be weird to overload the multiplication operator to represent multiplication?
cwojno 5y ago
Having used it extensively, it's actually quite nice.

sk0g 5y ago

Is it? In most languages you do something like `time.Sleep(1 * 1e6)` instead at which point it could be a second, a few minutes, a day, who really knows? I'm just not seeing any major downsides of this, keep in mind `time.Second` isn't the only one of its kind, you have millisecond, minute, hour, etc etc.

syntaxfree 5y ago

Overloading the asterisk is always weird because multiplication is expected to be associative etc etc.

ts4z 5y ago
time.Second*1 gets the same value; it's not overloaded. Well, ok, so what actually happens is that time.Second is a time.Duration, and Duration*int yields Duration (and int*Duration yields Duration). But the value of time.Second is actually a Duration with value 1000000, IIRC -- it's microseconds. It's just the type that's special, and the int handling here is general over a lot of types. It really is nice in practice.
- ts4z 5y ago
  (As noted elsewhere it's nanoseconds.)
xyzzy_plugh 5y ago
It's not overloaded. It's a unit. You can just type time.Sleep(time.Second) but this reads nicely time.Sleep(3 * time.Minute)
rwcarlsen 5y ago
It's not really overloading. it is associative: 2*time.Hour+1*time.second == time.Second+time.Hour*2
- wtallis 5y ago
  I believe that's illustrating commutativity, not associativity.
  - thaumasiotes 5y ago
    The illustration is wrong, but the claim is correct; multiplication between units and scalars is just as associative as you'd expect. Multiplying one kilowatt by an hour gives you exactly the same result as multiplying 1 by a kilowatt-hour.
Laremere 5y ago
There is no operator overloading here. The type Duration is effectively an int64: https://pkg.go.dev/time#Duration https://pkg.go.dev/time#Duration
vitus 5y ago
I'm not sure I follow. 2 * 3 * time.Second is the same whether you group 2 * (3 * time.Second) or (2 * 3) * time.Second (namely, the implicit grouping under left-associativity). You wouldn't normally write time.Sleep(time.Second * time.Second) because your units wouldn't work out. (Apparently you can write that in Golang; it's just very sketchy and results in a 30-year sleep.)
thaumasiotes 5y ago
But from a mathematical point of view, the relationship between a unit and its coefficient is that you're multiplying them together. Why would it be weird to overload the multiplication operator to represent multiplication?

cwojno 5y ago

Having used it extensively, it's actually quite nice.