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How to avoid one C++ foot gun (pboyd.io)

submitted 6 months ago by [email protected] to c/[email protected]

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[–] [email protected] 4 points 6 months ago (1 children)

I got stuck on cheese counting:

inventory.count(cheeseName) == 1 ? "Yes" : "No"

1 cheese == some, 2 cheese == none :/

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[–] [email protected] 1 points 6 months ago

Yeah, that's confusing to read. A hashset can only contain at most one of each cheese kind, in c++20 you can write inventory.contains(cheeseName), before that you had to use .count or inventory.find(cheeseName) != inventory.end() or something like that

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