[Python-Dev] Re: Why doesn't peephole optimise away operations with fast locals?

[Patrick Reader]

On 10/10/2021 18:26, Brandt Bucher wrote:
> The first reason is that this operation *does* have a side-effect: if a fast 
> local is unbound, the load will raise a NameError!
>
> def f():
>      x  # This should always raise.
>      x = None  # This makes x a fast local.
Ah, ok, that certainly explains it - thanks!
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