# Define Lagrangian using struct
**URL:** https://discourse.julialang.org/t/define-lagrangian-using-struct/95193
**Category:** New to Julia
**Tags:** question
**Created:** [February 25, 2023, 4:51pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193 "2023-02-25T16:51:57Z")
**Posts on this page:** 6
**Page:** 1
### Author:  [@khaledharizb](https://discourse.julialang.org/u/khaledharizb)
#### Post date: [February 25, 2023, 4:51pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/1 "2023-02-25T16:51:57Z")
I’m new in Julia and I want to define a dynamical Lagrangian which is a difference between the kinetic energy and the potential energy L(x,v)=K(v)-U(x). I use this code in order to do trick for example if we but `a=Lagrangian` and `a.K` give me the kinetic energy.
```julia
mutable struct Lagrangian{T<:Float64}
x::T
v::T
function K{T}(v::T) where T<:Float64
return 0.5*v^2
end
function U{T}(x::T) where T<:Float64
return 0.5*x^2
end
end
```
UndefVarError: K not defined
Stacktrace:
[1] top-level scope
@ In[3]:1
---
### Author:  [@jmair](https://discourse.julialang.org/u/jmair)
#### Post date: [February 25, 2023, 6:02pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/2 "2023-02-25T18:02:46Z")
In Julia, structs are only really for data, and all functions are defined separately (maybe an exception for a constructor). So for your example:
```julia
mutable struct Lagrangian{T}
x::T
v::T
end
kintetic(L::Lagrangian) = L.v^2/2
potential(L::Lagrangian) = L.x^2/2
```
There is an option to override some functions like `getproperty` to do what you want but I would recommend the above method instead.
---
### Author:  [@khaledharizb](https://discourse.julialang.org/u/khaledharizb)
#### Post date: [February 25, 2023, 7:00pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/3 "2023-02-25T19:00:59Z")
Thank you @jmair for your reply. That was clear, but how can I evaluate `kintetic(L::Lagrangian) = L.v^2/2` let say for `v=1.5`
---
### Author:  [@khaledharizb](https://discourse.julialang.org/u/khaledharizb)
#### Post date: [February 25, 2023, 7:08pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/4 "2023-02-25T19:08:54Z")
I see in this example that it could use a function with structs like
```julia
struct Point{T<:Real}
x::T
y::T
Point{T}(x,y) where {T<:Real} = x^2+y^2
end
```
and then use `Point{Int}(1,1)` for the evaluation which gives `2`.
---
### Author:  [@jmair](https://discourse.julialang.org/u/jmair)
#### Post date: [February 25, 2023, 7:22pm UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/5 "2023-02-25T19:22:07Z")
Just call like a regular function:
```julia
julia> L = Lagrangian(0.0, 2.0);
julia> kinetic(L)==2.0
true
```
> [@khaledharizb](#):
>
> ```julia
> struct Point{T<:Real}
> x::T
> y::T
> Point{T}(x,y) where {T<:Real} = x^2+y^2
> end
>
> ```
This is what I would call a “functor” but again, I would avoid this unless there is a good reason.
```julia
mutable struct Point{T}
x::T
y::T
end
function (point::Point)()
return point.x^2+point.y^2
end
```
Although I am not sure what reason you have for the struct at all, these are all probably better done with normal plain functions.
---
### Author:  [@khaledharizb](https://discourse.julialang.org/u/khaledharizb)
#### Post date: [February 27, 2023, 8:45am UTC](https://discourse.julialang.org/t/define-lagrangian-using-struct/95193/6 "2023-02-27T08:45:32Z")
Great @jmair thank you!