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AK: Don't implicitly convert Optional<T&> to Optional<T>
C++ will jovially select the implicit conversion operator, even if it's complete bogus, such as for unknown-size types or non-destructible types. Therefore, all such conversions (which incur a copy) must (unfortunately) be explicit so that non-copyable types continue to work. NOTE: We make an exception for trivially copyable types, since they are, well, trivially copyable. Co-authored-by: kleines Filmröllchen <filmroellchen@serenityos.org>
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committed by
Ali Mohammad Pur
parent
8468fb9ae5
commit
d7596a0a61
Notes:
github-actions[bot]
2024-12-04 00:59:23 +00:00
Author: https://github.com/yyny Commit: https://github.com/LadybirdBrowser/ladybird/commit/d7596a0a61e Pull-request: https://github.com/LadybirdBrowser/ladybird/pull/2567 Reviewed-by: https://github.com/alimpfard Reviewed-by: https://github.com/gmta ✅
@@ -39,7 +39,7 @@ public:
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return m_map.contains(name);
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}
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[[nodiscard]] Optional<ByteString> get(ByteString const& name) const
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[[nodiscard]] Optional<ByteString const&> get(ByteString const& name) const
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{
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return m_map.get(name);
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}
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