What is difference between these function signatures?
void func(const std::shared_ptr<TestStruct>& ts) {...}
void func(std::shared_ptr<TestStruct> const& ts) {...}
void func(std::shared_ptr<TestStruct> &const ts) {...}
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Валентин Никин
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3well the last one wont compile. The first two are exactly the same. – NathanOliver Dec 01 '21 at 19:27
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All of these variants successfully compilled. VS 2019, C++14 – Валентин Никин Dec 01 '21 at 19:30
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Unrelated, but taking a `std::shared_ptr` by reference seems a bit weird. If that function (or something within it) needs to become a new owner of the data, it would make more sense to accept it by value. Alternatively, if you don't care about the shared-ownership offered by `std::shared_ptr` and really just care about the underlying `TestStruct`, you may as well just take the `TestStruct` directly by reference. – Brian61354270 Dec 01 '21 at 19:30
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1@ВалентинНикин Turn up your warning options. I get *warning C4227: anachronism used: qualifiers on reference are ignored* for the last one. – NathanOliver Dec 01 '21 at 19:32
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@NathanOliver ok, so is the third variant meaningless? – Валентин Никин Dec 01 '21 at 19:42
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Yes. references by their very nature are already `const` because the can't be changed after they are initialized. A constant reference is redundant and not legal grammar, unless it is formed in a template context – NathanOliver Dec 01 '21 at 19:45
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I see, thank you – Валентин Никин Dec 01 '21 at 20:02