[go-nuts] Go predicts the end of the universe
Go predicts the end of the time, i.e. the end of the universe 😂
https://play.golang.org/p/THHvbdo_IS
package main
import (
"fmt"
"math"
"time"
)
func main() {
endOfTheTime := time.Unix(math.MaxInt64,81394542089).UTC()
fmt.Println(endOfTheTime.String())
}
Hal
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[go-nuts] [generics] Why is constraint optional?
"If a constraint is specified for any type parameter, every type parameter
must have a constraint. If some type parameters need a constraint and some
do not, those that do not should have a constraint of interface{}."
"interface{}" equals to "Any Type" in the context of generics. So it seems
that we don't have to support optional constraint, for the following
reasons:
* the syntax defining generic function is verbose on purpose (type
keyword), not only for clarification, but also a remind of the cost and
complexity behind generics, so it is not a bad thing to be explicit about
the the default constraint interface{}
* normal parameter list does not support default type, to be consistent,
type parameter list should not either
* multiple generic types without constraints can be written as "type T1,
T2, T3 interface{}", not too much boilerplate anyway
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[go-nuts] Re: [generics] Why is constraint optional?
Yes, an alternative syntax I can come up with (underscore as a placeholder):
func Foo(type T1 _, T2 Bar)
On Thursday, 18 June 2020 03:28:39 UTC+1, Andrey Tcherepanov wrote:
>
> Wouldn't it be nice to have just
>
> func Foo(type T1, type T2 Bar)
>
> (type as keyword splitting it into 2 type declarations)
>
> On Wednesday, June 17, 2020 at 5:50:47 AM UTC-6, Brian Candler wrote:
>>
>> Consider a generic where you want T1 unconstrained but T2 constrained.
>> If you write
>>
>> func Foo(type T1, T2 Bar)
>>
>> then Bar constrains both. Hence the need for
>>
>> func Foo(type T1 interface{}, T2 Bar)
>>
>
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