startswith function - incorrect arguements - not throwing error

Question

Can someone tell me why this doesn't throw an error? It prints True when the user types in http:// and false when they type in https://. I can't understand why it would work at all.

URL = input("Enter an URL address: ")
URL.startswith("http://" or "https://")

Possible duplicate of [Strange use of "and" / "or" operator](https://stackoverflow.com/questions/47007680/strange-use-of-and-or-operator) — NoDataDumpNoContribution, Mar 06 '19 at 15:50
We can say "starts with a or b" but what we mean by that is "starts with a or starts with b". Python needs the conditions rather explicit. — NoDataDumpNoContribution, Mar 06 '19 at 15:52

DjaouadNM · Accepted Answer · 2019-03-06T16:02:23.153

4

"http://" or "https://" is a boolean expression which evaluates to "http://", because that's what an or statement is (because "http://" is the first True-ish value encountered in the or statement), you need to do this instead:

URL.startswith("http://") or URL.startswith("https://")

Also, as @ShadowRanger suggested, you could make this shorter and faster by passing a tuple of accepted starting strings to the startswith method, it will then return True if any of the strings in the tuple matched with the start of the string:

URL.startswith(("http://", "https://"))

edited Mar 06 '19 at 16:02

answered Mar 06 '19 at 15:50

DjaouadNM

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There's a better way actually; [`startswith` also accepts a `tuple` of prefixes to check for](https://docs.python.org/3/library/stdtypes.html#str.startswith), and returns `True` if any of them match. So you could write `URL.startswith(("http://", "https://"))` and it would behave identically to the `or` separated version you wrote (just shorter to type, and faster to run). – ShadowRanger Mar 06 '19 at 16:00
@ShadowRanger Brilliant, I just added your note to my answer. – DjaouadNM Mar 06 '19 at 16:03

startswith function - incorrect arguements - not throwing error

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