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I want to write a Python generator function that never actually yields anything. Basically it's a "do-nothing" drop-in that can be used by other code which expects to call a generator (but doesn't always need results from it). So far I have this:
def empty_generator(): # ... do some stuff, but don't yield anything if False: yield Now, this works OK, but I'm wondering if there's a more expressive way to say the same thing, that is, declare a function to be a generator even if it never yields any value. The trick I've employed above is to show Python a yield statement inside my function, even though it is unreachable.
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You can use return once in a generator; it stops iteration without yielding anything, and thus provides an explicit alternative to letting the function run out of scope. So use yield to turn the function into a generator, but precede it with return to terminate the generator before yielding anything.
You can assign this generator to a variable in order to use it. When you call special methods on the generator, such as next() , the code within the function is executed up to yield . When the Python yield statement is hit, the program suspends function execution and returns the yielded value to the caller.
It is fairly simple to create a generator in Python. It is as easy as defining a normal function, but with a yield statement instead of a return statement. If a function contains at least one yield statement (it may contain other yield or return statements), it becomes a generator function.
Another way is
def empty_generator(): return yield Not really "more expressive", but shorter. :)
Note that iter([]) or simply [] will do as well.
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