The return type of std::forward depends on the template type, and can not be recovered from the args. Attempting to do so will result in a compile failure. For example, make_secure_unique<std::string>(std::string{}); does not compile on current master, but does with this pull.
Another example would be make_secure_unique<std::pair<std::string, std::unique_ptr<int>>>(std::string{}, std::make_unique<int>(21));
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For details see: https://corecheck.dev/bitcoin/bitcoin/pulls/31464.
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| Type | Reviewers |
|---|---|
| ACK | hodlinator, hebasto, TheCharlatan |
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ACK fa397177acfa1006ea6feee0b215c53e51f284de
Accidentally tested make_secure_unique<std::string>(std::string{}) with std::forward<T> in the implementation (misremembering the change in this PR). It worked, so I came up with a better test case which requires std::forward<Args>:
0make_secure_unique<std::pair<std::string, std::unique_ptr<int>>>(std::string{}, std::make_unique<int>(21));
T != Argsstd::unique_ptr in case it shakes out issues with l-values and our std::forward usage.My only nit would be changing the PR summary to an example closer to that.
ACK fa397177acfa1006ea6feee0b215c53e51f284de.
