It appears you are right, thanks for the answer. On Thursday, October 13, 2016 at 9:25:07 PM UTC-7, Chris Manghane wrote: > > In the Go Language specification under operators ( > https://golang.org/ref/spec#Operators), there are a couple examples that > demonstrate this exact situation: > > var u2 = 1<<s != 1.0 // illegal: 1 has type float64, cannot shift > var v float32 = 1<<s // illegal: 1 has type float32, cannot shift > > The corresponding rule behind this is: > > If the left operand of a non-constant shift expression is an untyped > constant, it is first converted to the type it would assume if the shift > expression were replaced by its left operand alone. > So in your baz() example, because 1 is an untyped constant, the shift > expression acts as if 1 is the only operand and converts it to a float64 > type; therefore, 1 has type float64 and you cannot shift. To get around > this, just use an explicitly typed constant. > > On Thu, Oct 13, 2016 at 7:37 PM, Carl Mastrangelo <carl.mas...@gmail.com > <javascript:>> wrote: > >> https://play.golang.org/p/iZTogUaWWl >> >> In the program above, foo and bar compile but baz does not. It fails >> with the message: "invalid operation: 1 << b (shift of type float64)". >> This seems to be wrong on the surface, since the order of operations should >> imply the shift takes precedence. In the bar function, using a temporary >> variable without specifying the type shows that the compiler does know what >> to do. What am I missing? >> >> -- >> You received this message because you are subscribed to the Google Groups >> "golang-nuts" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to golang-nuts...@googlegroups.com <javascript:>. >> For more options, visit https://groups.google.com/d/optout. >> > >
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