> (a := b)
Will `with` start using this syntax instead of `b as a`: `with open('foo') as a:` == `with a := open('foo'):`?
Update: found my answer on the pep. `with EXPR as VAR` actually calls `EXPR.__enter__()` so it's not the same.
Will `with` start using this syntax instead of `b as a`: `with open('foo') as a:` == `with a := open('foo'):`?
Update: found my answer on the pep. `with EXPR as VAR` actually calls `EXPR.__enter__()` so it's not the same.
Seems very unnecessary to have both options.