Retrieve type alias alternate name using reflection
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Is it possible in any form or fashion to retrieve the name of a type alias? I realize that this is likely impossible because a type alias IS the same type, and indeed something such as the following definitely prints the original type.
type MyString string
type AliasString = MyString
var a AliasString = "hello"
fmt.Println(reflect.ValueOf(a).Type().Name())
// prints MyString
Is there ANY way for this to be able to print AliasString? Or is there just absolutely no way to distinguish them as an alias?
go
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
add a comment |
Is it possible in any form or fashion to retrieve the name of a type alias? I realize that this is likely impossible because a type alias IS the same type, and indeed something such as the following definitely prints the original type.
type MyString string
type AliasString = MyString
var a AliasString = "hello"
fmt.Println(reflect.ValueOf(a).Type().Name())
// prints MyString
Is there ANY way for this to be able to print AliasString? Or is there just absolutely no way to distinguish them as an alias?
go
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
add a comment |
Is it possible in any form or fashion to retrieve the name of a type alias? I realize that this is likely impossible because a type alias IS the same type, and indeed something such as the following definitely prints the original type.
type MyString string
type AliasString = MyString
var a AliasString = "hello"
fmt.Println(reflect.ValueOf(a).Type().Name())
// prints MyString
Is there ANY way for this to be able to print AliasString? Or is there just absolutely no way to distinguish them as an alias?
go
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
Is it possible in any form or fashion to retrieve the name of a type alias? I realize that this is likely impossible because a type alias IS the same type, and indeed something such as the following definitely prints the original type.
type MyString string
type AliasString = MyString
var a AliasString = "hello"
fmt.Println(reflect.ValueOf(a).Type().Name())
// prints MyString
Is there ANY way for this to be able to print AliasString? Or is there just absolutely no way to distinguish them as an alias?
go
go
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
asked Nov 16 '18 at 23:08
RayfenWindspearRayfenWindspear
3,8331530
3,8331530
add a comment |
add a comment |
1 Answer
1
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votes
Per the proposal, type aliases are mostly invisible at runtime. The exception is in the name of embedded fields.
With the types in the question, there's no way to distinguish the alias from the original type.
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Per the proposal, type aliases are mostly invisible at runtime. The exception is in the name of embedded fields.
With the types in the question, there's no way to distinguish the alias from the original type.
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
add a comment |
Per the proposal, type aliases are mostly invisible at runtime. The exception is in the name of embedded fields.
With the types in the question, there's no way to distinguish the alias from the original type.
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
add a comment |
Per the proposal, type aliases are mostly invisible at runtime. The exception is in the name of embedded fields.
With the types in the question, there's no way to distinguish the alias from the original type.
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
Per the proposal, type aliases are mostly invisible at runtime. The exception is in the name of embedded fields.
With the types in the question, there's no way to distinguish the alias from the original type.
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
answered Nov 16 '18 at 23:41
Cerise LimónCerise Limón
55.7k57395
55.7k57395
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
add a comment |
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
I figured as much, as I did read through the whole proposal and specification of aliases. Figured it couldn't hurt to ask though in the hopes that there is some sort of accessable metadata attached to them.
– RayfenWindspear
Nov 16 '18 at 23:44
add a comment |
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