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pythonpython-3.xgeneratoryield

Valid, but weird use of yield inside list

I was playing with generators a bit, when I suddenly wrote this piece of code.

Although I have written it, I don't know why it is valid, or why it works. 

>>> x = (lambda : [(yield 1), (yield 2)])()
>>> next(x)
1
>>> next(x)
2
>>> next(x)
StopIteration
Solution

  • You created a generator function, with the lambda syntax. yield is just another expression, so can be used inside a lambda.

    You basically did this:

    def foo():
    return [(yield 1), (yield 2)]

x = foo()

    but all in a single expression.

    Both yield expressions return None, so the StopIteration result value is set to [None, None]:

    >>> x = (lambda : [(yield 1), (yield 2)])()
>>> next(x), next(x)
(1, 2)
>>> try:
...     next(x)
... except StopIteration as si:
...     print(si.value)
...
[None, None]

    Instead of using next(), you could use generator.send() to make the yield values return something other than None:

    >>> x = (lambda : [(yield 1), (yield 2)])()
>>> next(x)  # advance to first yield
1
>>> x.send(42)   # set return value for first yield, continue to second
2
>>> try:
...     x.send(81)   # set return value for second yield
... except StopIteration as si:
...     print(si.value)
...
[42, 81]

    It works, because it is legal Python syntax. It's not very useful however.