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.

```
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

jmair
February 25, 2023, 6:02pm
2
In Julia, structs are only really for data, and all functions are defined separately (maybe an exception for a constructor). So for your example:

```
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.

2 Likes

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`

I see in this example that it could use a function with structs like

```
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`

.

jmair
February 25, 2023, 7:22pm
5
Just call like a regular function:

```
julia> L = Lagrangian(0.0, 2.0);
julia> kinetic(L)==2.0
true
```

This is what I would call a “functor” but again, I would avoid this unless there is a good reason.

```
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.

1 Like