103 lines
2.4 KiB
Markdown
103 lines
2.4 KiB
Markdown
|
---
|
||
|
|
title: undocumented idris2 features
|
||
|
|
date: 2022-11-12
|
||
|
|
tags: [computer, idris]
|
||
|
|
header-includes: |
|
||
|
|
<style>
|
||
|
|
.sidebyside {
|
||
|
|
display: grid;
|
||
|
|
grid-template-columns: 1fr 1fr;
|
||
|
|
}
|
||
|
|
.sidebyside :is(.input, .goal) {
|
||
|
|
box-sizing: border-box;
|
||
|
|
width: 95%;
|
||
|
|
height: 95%;
|
||
|
|
}
|
||
|
|
.sidebyside .input { grid-area: 0 / 1; }
|
||
|
|
.sidebyside .goal { grid-area: 1 / 2; }
|
||
|
|
</style>
|
||
|
|
...
|
||
|
|
|
||
|
|
if these are already in the documentation somewhere, i didn't find it.
|
||
|
|
|
||
|
|
## quantities on case, let, with
|
||
|
|
|
||
|
|
occasionally, idris can't infer that these expressions should be non-ω. usually
|
||
|
|
when there are still holes in the definition, it seems. so you can specify the
|
||
|
|
quantity you want directly after the keyword, for [example][viewLsb]:
|
||
|
|
|
||
|
|
```idris
|
||
|
|
export
|
||
|
|
viewLsb : (n : Nat) -> ViewLsb n
|
||
|
|
viewLsb n =
|
||
|
|
-- ↓ here
|
||
|
|
let 0 nz : NonZero 2 = %search in
|
||
|
|
rewrite DivisionTheorem n 2 nz nz in
|
||
|
|
rewrite multCommutative (divNatNZ n 2 nz) 2 in
|
||
|
|
viewLsb' (modNatNZ n 2 nz) (divNatNZ n 2 nz) (boundModNatNZ n 2 nz)
|
||
|
|
```
|
||
|
|
|
||
|
|
[viewLsb]: https://git.rhiannon.website/rhi/quox/src/branch/ope/lib/Quox/NatExtra.idr#L110-L116
|
||
|
|
|
||
|
|
|
||
|
|
## syntactic with-abstraction
|
||
|
|
|
||
|
|
using `with` can be costly, since it has to evaluate the expression being
|
||
|
|
abstracted, as well as the types of the goal and bound variables, to find
|
||
|
|
occurrences of it. maybe you know they are all already syntactically equal.
|
||
|
|
in that case, you can say `… with %syntactic (expr)`:
|
||
|
|
|
||
|
|
|
||
|
|
:::sidebyside
|
||
|
|
``` {.idris .input}
|
||
|
|
blah : (n : Nat) -> 2 * n = n + (n + 0)
|
||
|
|
blah n with (2 * n)
|
||
|
|
blah n | w = ?blah_rhs
|
||
|
|
```
|
||
|
|
|
||
|
|
``` {.idris .goal}
|
||
|
|
w : Nat
|
||
|
|
n : Nat
|
||
|
|
------------------------------
|
||
|
|
blah_rhs : w = w
|
||
|
|
```
|
||
|
|
|
||
|
|
``` {.idris .input}
|
||
|
|
blah2 : (n : Nat) -> 2 * n = n + (n + 0)
|
||
|
|
blah2 n with %syntactic (2 * n)
|
||
|
|
blah2 n | w = ?blah2_rhs
|
||
|
|
```
|
||
|
|
|
||
|
|
``` {.idris .goal}
|
||
|
|
w : Nat
|
||
|
|
n : Nat
|
||
|
|
------------------------------
|
||
|
|
blah2_rhs : w = plus n (plus n 0)
|
||
|
|
```
|
||
|
|
:::
|
||
|
|
|
||
|
|
in `blah2`, only the exact syntactic occurrence of `2 * n` is replaced, and
|
||
|
|
the `n + (n + 0)` is left alone.
|
||
|
|
|
||
|
|
|
||
|
|
## equality proof in with
|
||
|
|
|
||
|
|
a `with`-abstraction can _also_ have a proof of equality between the pattern
|
||
|
|
and the original expression, like the old inspect pattern.
|
||
|
|
|
||
|
|
:::sidebyside
|
||
|
|
``` {.idris .input}
|
||
|
|
blah : (n : Nat) -> 2 * n = n + (n + 0)
|
||
|
|
blah n with (2 * n) proof eq
|
||
|
|
blah n | w = ?blah_rhs
|
||
|
|
```
|
||
|
|
|
||
|
|
``` {.idris .goal}
|
||
|
|
w : Nat
|
||
|
|
n : Nat
|
||
|
|
eq : plus n (plus n 0) = w
|
||
|
|
------------------------------
|
||
|
|
blah_rhs : w = w
|
||
|
|
```
|
||
|
|
:::
|