4 ms·

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?

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?