5 ms·

Why can't &3 work? Rob says 3 does not have a type and that's a problem. Would it be possible to change the Go compiler such that 3 has a type? (I'm guessing no

by uh_uh 6y ago

Why can't &3 work? Rob says 3 does not have a type and that's a problem. Would it be possible to change the Go compiler such that 3 has a type? (I'm guessing no, at least not easily, otherwise he'd be suggesting it, but I'm curious about the reason)

fred256 6y ago
From the language specification: > Numeric constants represent exact values of arbitrary precision and do not overflow.
- friseurtermin 6y ago
  Ok I think I'm not understanding this correctly then. Why does this return an error? package main import ( "fmt" ) const tst = 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 func main() { fmt.Println("%v", tst) } Error: ./prog.go:16:17: constant 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 overflows int See: https://play.golang.org/p/47l5qAsXD5r https://play.golang.org/p/47l5qAsXD5r
  - howeyc 6y ago
    https://golang.org/ref/spec#Constants https://golang.org/ref/spec#Constants > An untyped constant has a default type which is the type to which the constant is implicitly converted in contexts where a typed value is required. The default type for a number (integer) is int. If you were to add a period in that long string of zeros, the default for floating-point is float64.
  - HALtheWise 6y ago
    If you look closely, that error happens at the print line statement, not at the constant declaration. The constant is perfectly legal to exist as written, and you can even do constant operations like mathematics on it, you just can't legally assign it to a variable (implicitly in this case as part of the function call) if it's value is too large for the type of that variable.
    - friseurtermin 6y ago
      I see, that makes sense. Thanks!
masklinn 6y ago
> Would it be possible to change the Go compiler such that 3 has a type? (I'm guessing no, at least not easily, otherwise he'd be suggesting it, but I'm curious about the reason) Why not? In fact it already kind-of does: Go has "default types" for most untyped constants. When you write i := 3 absent an explicit type, Go will fall back to "int".
Laremere 6y ago
I asked that in the thread, response here: https://github.com/golang/go/issues/45624#issuecomment-822590091 https://github.com/golang/go/issues/45624#issuecomment-82259... With my reply (and further exploration) here: https://github.com/golang/go/issues/45624#issuecomment-822632731 https://github.com/golang/go/issues/45624#issuecomment-82263... tl;dr: Go uses casts to coerce values to the correct types, which means you couldn't get pointers to number literals for non-default number types.
majewsky 6y ago
If the syntax were &3, you could use it to obtain a *int, but not a *int32, *uint64 or such.

by uh_uh 6y ago

fred256 6y ago

From the language specification: > Numeric constants represent exact values of arbitrary precision and do not overflow.

friseurtermin 6y ago
Ok I think I'm not understanding this correctly then. Why does this return an error? package main import ( "fmt" ) const tst = 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 func main() { fmt.Println("%v", tst) } Error: ./prog.go:16:17: constant 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 overflows int See: https://play.golang.org/p/47l5qAsXD5r https://play.golang.org/p/47l5qAsXD5r
- howeyc 6y ago
  https://golang.org/ref/spec#Constants https://golang.org/ref/spec#Constants > An untyped constant has a default type which is the type to which the constant is implicitly converted in contexts where a typed value is required. The default type for a number (integer) is int. If you were to add a period in that long string of zeros, the default for floating-point is float64.
- HALtheWise 6y ago
  If you look closely, that error happens at the print line statement, not at the constant declaration. The constant is perfectly legal to exist as written, and you can even do constant operations like mathematics on it, you just can't legally assign it to a variable (implicitly in this case as part of the function call) if it's value is too large for the type of that variable.
  - friseurtermin 6y ago
    I see, that makes sense. Thanks!

masklinn 6y ago

> Would it be possible to change the Go compiler such that 3 has a type? (I'm guessing no, at least not easily, otherwise he'd be suggesting it, but I'm curious about the reason) Why not? In fact it already kind-of does: Go has "default types" for most untyped constants. When you write i := 3 absent an explicit type, Go will fall back to "int".

Laremere 6y ago

I asked that in the thread, response here: https://github.com/golang/go/issues/45624#issuecomment-822590091 https://github.com/golang/go/issues/45624#issuecomment-82259... With my reply (and further exploration) here: https://github.com/golang/go/issues/45624#issuecomment-822632731 https://github.com/golang/go/issues/45624#issuecomment-82263... tl;dr: Go uses casts to coerce values to the correct types, which means you couldn't get pointers to number literals for non-default number types.

majewsky 6y ago

If the syntax were &3, you could use it to obtain a *int, but not a *int32, *uint64 or such.