In Go, as in most languages, spaces do not *have to* delimit tokens. A token
ends when the leftmost longest matching token is found, and the rest is placed
back in the input buffer for the next token. So
&&& true
1) read & — this can be &, &=, &&, &^, or &^=, so read again
2) read & — we now have &&, can't go further so give && to parser
3) read & — this can be &, &=, &&, &^, or &^=, so read again
4) read space — no match; give the & we have saved to the parser, then
ignore the space
5) read "true" (which itself breaks down into steps like the above)
I know there are some languages (like Swift) which allow you to create your own
operators, and thus have certain rules for significant spaces; to my knowledge,
these are in the minority.
> On Oct 31, 2016, at 3:45 PM, mhhc...@gmail.com wrote:
>
> Hi,
>
> noticed that an instant ago,
>
> package main
> import (
> "fmt"
> )
>
> func main() {
> if true &&& true &&& false {
> fmt.Println("ok")
> }
> }
>
> Gives this output,
>
> tmp/sandbox074568120/main.go:8: cannot take the address of true
> tmp/sandbox074568120/main.go:8: invalid operation: true && &true (mismatched
> types bool and *bool)
> tmp/sandbox074568120/main.go:8: cannot take the address of false
>
> the triple ampersands is parsed as :
> true && &true
>
> No syntax error is reported :/ &&& is not intended, its a typo and should be
> reported so.
>
> In my simple example its still ok because another error pops in to prevent
> that.
>
> But in my real life example it passes, This code builds fine, maybe with side
> effects, i m not sure.
> func (i *InputResolver) setResult(value interface{}, err error) {
> if err!=nil &&& i.LastValue!=nil &&& i.LastErr==nil{
> return // skip the new set as it reports error in producing and that the
> last value did produce properly.
> }
> i.LastValue = value
> i.LastErr = err
> }
>
>
>
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