# Get non-missing type in the case of parametric type
**URL:** https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109
**Category:** General Usage
**Tags:** parametric-types
**Created:** [September 24, 2019, 1:05pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109 "2019-09-24T13:05:44Z")
**Posts on this page:** 5
**Page:** 1
### Author:  [@tlienart](https://discourse.julialang.org/u/tlienart)
#### Post date: [September 24, 2019, 1:05pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109/1 "2019-09-24T13:05:44Z")
So I was looking at how to get `T` from a `Union{Missing,T}` and came up with
```julia
nonmissing(TT::Type{Union{Missing,T}}) where T = T
nonmissing(TT::Type{T}) where T = T
```
(PS: incidentally I found that @bkamins had answered something similar on stackoverflow [Is there a type subtraction operation in Julia? - Stack Overflow](https://stackoverflow.com/questions/52118361/is-there-a-type-subtraction-operation-in-julia) so it looks like this is not a bad idea)
This works fine for non-parametric types including abstract types:
```julia
julia> nonmissing(Union{Missing,Real}) == Real
```
However when there’s a parametric type, it returns the union
```julia
julia> struct Foo{N} end
julia> nonmissing(Union{Missing,Foo}) == Union{Missing,Foo}
```
Is there a way to recuperate `Foo` in this case? One way I found around it that works is to write a macro but I feel that’s overkill here…
```julia
macro nonmissing(ex)
ex isa Symbol && return ex
types = ex.args[2:end]
types[1] == :Missing && return types[2]
return types[1]
end
```
thanks
---
### Author:  [@tlienart](https://discourse.julialang.org/u/tlienart)
#### Post date: [September 24, 2019, 1:40pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109/2 "2019-09-24T13:40:11Z")
Hmm I guess this does the job:
```julia
nomiss(::Type{T}) where T = T isa Union ? ifelse(T.a == Missing, T.b, T.a) : T
```
but I’ll take anything that’s more elegant than that or the macro 🙂
---
### Author:  [@bkamins](https://discourse.julialang.org/u/bkamins)
#### Post date: [September 24, 2019, 1:43pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109/3 "2019-09-24T13:43:51Z")
Well now (SO answer is old 🙂 and things move forward) you have `nonmissingtype` defined as:
```
nonmissingtype(::Type{S}) where {S} = Core.Compiler.typesubtract(S, Missing)
```
and it does the job as expected (please correct me if I missed something).
---
### Author:  [@tlienart](https://discourse.julialang.org/u/tlienart)
#### Post date: [September 24, 2019, 1:47pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109/4 "2019-09-24T13:47:15Z")
Nice, thanks, unfortunately that won’t work on older versions of Julia right? it doesn’t on 1.0. as far as I can tell. (for `Union{Missing,Foo}` it will return `Union{Missing,Foo}`)
I guess I could do a version check and either use that or the `nomiss` thing.
Thanks either way!
---
### Author:  [@nalimilan](https://discourse.julialang.org/u/nalimilan)
#### Post date: [September 24, 2019, 3:04pm UTC](https://discourse.julialang.org/t/get-non-missing-type-in-the-case-of-parametric-type/29109/5 "2019-09-24T15:04:14Z")
You can use Missings.jl, which exports `nonmissingtype` on older Julia versions.