Given a struct that looks like
type foo struct {
i *int
}
if I want to set i to 1, I must
throwAway := 1
instance := foo { i: &throwAway }
Is there any way to do this in a single line without having to give my new i value it's own name (in this case throwaway)?
As pointed in the mailing list, you can just do this:
func intPtr(i int) *int {
return &i
}
and then
instance := foo { i: intPtr(1) }
if you have to do it often. intPtr gets inlined (see go build -gcflags '-m' output), so it should have next to no performance penalty.
No this is not possible to do in one line.