Confusion about captures

[hot lion]

So assuming this following snippet:

var my_var: ?SomeType = ...;
...
if (my_var) |captured| {
  ...
}

Why is "captured" const here unless I do:

var my_var: ?SomeType = ...;
...
if (my_var) |*captured| {
  ...
}

I'm confused because I don't understand why I need to capture a pointer to "captured" in order to be able to modify it.

I'd understand if it was:

const my_var: ?SomeType = ...

Because here "my_var" is const, so I'd understand the need to capture a pointer

[simple phoenix]

*captured is capturing a pointer to the non-null part of my_var, so modifying it is modifying my_var itself. if you want captured to be a mutable copy youd have to make a temp var like |captured_const| { var captured = captured_const; …

captures are const by default the same reason parameters are, prevents some bugs and iirc helps optimization

[formal ravine]

additionally, with your solution, changes wont be visible outside of the if block, whereas if you capture by pointer changes will be.

[hot lion]

Woah... Alright that's actually a huge footgun and good to know

[simple phoenix]

i dont really see the footgun part :P pointers mutate the things they point to and copies are a completely new piece of memory. its just how languages work.

|captured| is creating a copy of the non-null part, |*captured| is getting a pointer to the non-null part. simple as

[hot lion]

Interesting.. I guess I was just confused as to why the capture wouldn't inherit the const-ness of the thing I'm capturing, but it seems like that's because said capture could potentially be a copy