Fix let? unwrap early return for option types #8091
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@HuijungYoon could you please run |
zth
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Dec 18, 2025
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Hey @HuijungYoon , thank you for taking this on!
Returning a variable is intentional (it means we don't need to recreate the value at runtime). So that should not be changed. The problem to fix is what this variable is called.
So, the fix here should most likely be to propagate the name of the variable used in the let binding:
let? Some(myVar) = someOptFn()
// We need to propagate the `myVar` name to where `x` is now used hard coded
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Thank you. Now you need to do these things:
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HuijungYoon
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When using `let? Some(myVar) = ...`, the variable name was hardcoded as "x" instead of using the actual pattern variable name "myVar". This fix extracts the variable name from the pattern and uses it in the early return cases, ensuring proper variable propagation and avoiding unnecessary runtime allocations. Fixes rescript-lang#8085 Signed-Off-By: [Huijung Yoon] <[markup3604@gmail.com]>
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@HuijungYoon Thank you! |
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When using
let? Some(x) = Nonein a function returningoption<T>, the early return was incorrectly returning a variable instead of None.This fixes the issue by explicitly returning None/Some values instead of using pattern-matched variables, ensuring correct type propagation and JS interop compatibility.
Fixes #8085