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identical.go
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identical.go
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package typeutil
import (
"go/types"
)
// Identical reports whether x and y are identical types.
// Unlike types.Identical, receivers of Signature types are not ignored.
// Unlike types.Identical, interfaces are compared via pointer equality (except for the empty interface, which gets deduplicated).
// Unlike types.Identical, structs are compared via pointer equality.
func Identical(x, y types.Type) (ret bool) {
if !types.Identical(x, y) {
return false
}
switch x := x.(type) {
case *types.Struct:
y, ok := y.(*types.Struct)
if !ok {
// should be impossible
return true
}
return x == y
case *types.Interface:
// The issue with interfaces, typeutil.Map and types.Identical
//
// types.Identical, when comparing two interfaces, only looks at the set
// of all methods, not differentiating between implicit (embedded) and
// explicit methods.
//
// When we see the following two types, in source order
//
// type I1 interface { foo() }
// type I2 interface { I1 }
//
// then we will first correctly process I1 and its underlying type. When
// we get to I2, we will see that its underlying type is identical to
// that of I1 and not process it again. This, however, means that we will
// not record the fact that I2 embeds I1. If only I2 is reachable via the
// graph root, then I1 will not be considered used.
//
// We choose to be lazy and compare interfaces by their
// pointers. This will obviously miss identical interfaces,
// but this only has a runtime cost, it doesn't affect
// correctness.
y, ok := y.(*types.Interface)
if !ok {
// should be impossible
return true
}
if x.NumEmbeddeds() == 0 &&
y.NumEmbeddeds() == 0 &&
x.NumMethods() == 0 &&
y.NumMethods() == 0 {
// all truly empty interfaces are the same
return true
}
return x == y
case *types.Signature:
y, ok := y.(*types.Signature)
if !ok {
// should be impossible
return true
}
if x.Recv() == y.Recv() {
return true
}
if x.Recv() == nil || y.Recv() == nil {
return false
}
return Identical(x.Recv().Type(), y.Recv().Type())
default:
return true
}
}