[Python-Dev] yield * (Re: Missing operator.call) (original) (raw)

Guido van Rossum guido at python.org
Sat Feb 7 19:25:54 CET 2009

We already have yield expressions and they mean something else...

On Sat, Feb 7, 2009 at 1:04 AM, Greg Ewing <greg.ewing at canterbury.ac.nz> wrote:

Guido van Rossum wrote:

It would be way too confusing to have "a different form of call" with totally different semantics that nevertheless used the same terminology as is used for regular calls. I expect you're right, so I won't argue for calling it "call" any more. I'd still like to find a good name for it, though. The other important thing is that my proposed construct should be usable as an expression, and its value should be whatever is returned by the called generator when it exits. E.g. if we continue spelling it "yield *" for the moment, then def f(): v = yield *g() print v def g(): yield 42 return "spam" for x in f(): pass should end up printing "spam". Would you entertain the idea of a "yield *" expression with those semantics? -- Greg

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-- --Guido van Rossum (home page: http://www.python.org/~guido/)

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