This position precludes the following use of the equality operator for scalar values:
a := 1 b := 1 a == b would be false under the approach below since a and b are not the same set of bits. I think most people would find this a little surprising. On Thu, 2016-06-30 at 09:24 -0700, Chad wrote: > It's a view in another array. > Why should they be equal? Unless the second slice is constructed by > subslicing the other as such `[:]`, the slices *are* different. > > If I were to access one of the slice backing array and mutate it, I > wouldn't expect the slices to be equal anymore. > > The argument that it would be surprising is perhaps flawed because it stems > from a misconception. > A slice is a dynamic view. For two slices to be equal it makes sense to be > looking at the same thing at all time. > > Taken mathematically, if we describe an indirection as a mathematical > function, a slice is such a function. Two functions are equals if they are > equals everywhere they are defined. A slice is equal to another slice if > the backign arrays always have the same values, whatever mutation occurs on > any of them. > > Comparing the snapshot view of a slice would then different. The behaviour > seems consistent to me. -- 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+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
