Add BuildNumberSpec #3098
Add BuildNumberSpec #3098
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| { | ||
| return (equal_with<T>(lhs, rhs) || ...); | ||
| } | ||
| } |
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I don't get the need for all this complexity. From what I understand (but I might be wrong), you want to ensure that the variant hold objects of the same type, which you can check with lhs.index() == rhs.index(). With this, you don't need all the equal overloads at all.
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No, I want to ensure they contains the same index and that the contained objects compare equal.
I cannot do it with operator== because that would mean specializing in namespace std.
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From the definitions I read here, it looks like objects are considered equal if they have the same type. Did I miss something?
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Ha I see what you mean. The code written does a proper variant comparison, but from the types included here (none have members), it is equivalent to only checking for type equality.
This is not what I wrote because this does not describe the proper intent. It may lead to subtle bugs should one modify/add a predicate.
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| auto operator==(const BuildNumberPredicate& lhs, const BuildNumberPredicate& rhs) -> bool | ||
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| return (lhs.m_build_number == rhs.m_build_number) && equal(lhs.m_operator, rhs.m_operator); |
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Actually you could even perform the variant comparison here and get rid of the auto equal(const std::variant<T...>& lhs, const std::variant<T...>& rhs) function.
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Again, I cannot specialized operator==(std::equal_to<int>, std::equal_to<int>) without UB.
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Is there a case where operator==(std::equal_to<int>, std::equal_to<int>) (or any overload) should return false? Otherwise, comparing the type is enough.
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