If type T2
is based on type T1
, other than sharing the same data fields, is there any relationship between T1
and T2
?
package main
import "fmt"
type T1 struct { s string }
func (v *T1) F1() string { return v.s }
type T2 T1
func (v *T2) F2() string { return v.s }
func main() {
var t1 = T1{ "xyz" }
var t2 = T2{ "pdq" }
s0 := t2.F1() // error - expected ok
s1 := ((*T1)(&t2)).F1() // ok - expected
s2 := ((*T2)(&t1)).F2() // ok - not expected
fmt.Println( s0, s1, s2 )
}
My understanding here is lacking
was hoping that
T2
would inheritT1
's methods, but such is not the case.was expecting
T2
could be coerced intoT1
, since it was derived fromT1
was surprised that
T1
could be coerced intoT2
, but so it is.it seems that the relationship between
T1
andT2
is completely symmetrical - I cannot find anything that breaks the symmetry despite the fact one is actually derived from the other - or is this an illusion?
(NOTE: I am not criticizing or judging - I entirely respect decisions made - just verifying I understand what is there that for me is counter-intuitive - I'm sure I'm not the only one!)
Answer
Go does not support object-oriented type inheritance.
Is Go an object-oriented language?
Why is there no type inheritance?
A method is bound to a single specific type.
A method declaration binds an
identifier to a method. The method is
said to be bound to the base type and
is visible only within selectors for
that type.
You can convert between types T1
and T2
.
A value
x
can be converted to type
T
[when]x
's type andT
have
identical underlying types.
For example,
package main
import (
"fmt"
)
type T1 struct{ i int }
func (t T1) String() string { return "T1" }
type T2 T1
func (t T2) String() string { return "T2" }
func main() {
t1 := T1{1}
t2 := T2{2}
fmt.Println(t1, t2)
c1 := T1(t2)
c2 := T2(t1)
fmt.Println(c1, c2)
t1 = T1(c2)
t2 = T2(c1)
fmt.Println(t1, t2)
}
Output:
T1 T2
T1 T2
T1 T2
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