Cannot bind a plain reference to a literal?

Question

Example:

auto &h = 42;        // error：we can't bind a plain reference to a literal.
const auto &j = 42;  // right

I don't understand why compiler can't know &h is `const int& I mean ,"auto" is have two step:1.know what type of rvalue. 2.make sure lvalue become the type if the step is right,why we must add "const" when rvalue is literal?

Possible duplicate of [Why do const references extend the lifetime of rvalues?](https://stackoverflow.com/questions/39718268/why-do-const-references-extend-the-lifetime-of-rvalues) — , Sep 28 '17 at 06:05
You should know that it isn't `const int&`, the compiler isn't confused. It's an `int` pure rvalue. The fact a const reference binds to it, doesn't mean it's of a const reference type. — StoryTeller - Unslander Monica, Sep 28 '17 at 06:05
@StoryTeller _You should know_ sounds a lot like _C++ is an expert-friendly language_ indeed. :-D — skypjack, Sep 28 '17 at 06:19
@skypjack - We must accept it for what it is, and love it no less :P — StoryTeller - Unslander Monica, Sep 28 '17 at 06:20
@StoryTeller "The fact a const reference binds to it, doesn't mean it's of a const reference type",why not? I still couldn't understand@_@ — wuduojia, Sep 28 '17 at 08:13
@StoryTeller Oh, I see. That's why.you mean is the type on either side of the equals sign is unlikeliness? — wuduojia, Sep 28 '17 at 09:34

score 1 · Accepted Answer · answered Sep 28 '17 at 08:29

1

auto never deduces a const when the input was non-const. Since 42 has type int, then auto deduces to int and your code is equivalent to:

int& h = 42;

which is an error because a non-const lvalue reference cannot bind to a prvalue.

answered Sep 28 '17 at 08:29

M.M

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Cannot bind a plain reference to a literal?

1 Answers1