Ed, I'm not seeing this perceived problem either. if we have >>> d = delayed {'a': 1, 'b': 2} # I'm not sure how this is delayed exactly, but sure >>> k = delayed string.ascii_lowercase[0] >>> d[k] 1 I'm not sure how the delayedness of any of the subexpressions matter, since evaluating the parent expression will evaluate all the way down. --Josh On Fri, Feb 17, 2017 at 4:53 PM Ed Kellett <edk141@gmail.com> wrote:
On Fri, 17 Feb 2017 at 21:21 Joseph Jevnik <joejev@gmail.com> wrote:
About the "whatever is d[k]" in five minutes comment: If I created an explict closure like: `thunk = lambda: d[k]` and then mutated `d` before evaluating the closure you would have the same issue. I don't think it is that confusing. If you need to know what `d[k]` evaluates to right now then the order of evaluation is part of the correctness of your program and you need to sequence execution such that `d` is evaluated before creating that closure.
If you create an explicit closure, sure. With delayed expressions, you could explicitly delay d[k], too. If you have an existing d and k, potentially passed to you by somebody else's code, the delayedness of d and k should not inflict arbitrarily-delayed sequencing on your attempt to find out what d[k] is now. _______________________________________________ Python-ideas mailing list Python-ideas@python.org https://mail.python.org/mailman/listinfo/python-ideas Code of Conduct: http://python.org/psf/codeofconduct/