分配int指针值的简单方法？

Question

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)?

Answer 1

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.