Why don't const l-value references to r-values result in dangling references?

Question

Why is the following not an error?

const auto& foo = std::string("foo");

In my mental model of C++ I think of references as glorified non-null pointers that the language wraps in syntactic sugar for me. However the code below would be an error but the above is not.

const auto* foo = &(std::string("foo"));

In the reference case why is the string not immediately destructed after the r-value expression is evaluated?

https://stackoverflow.com/questions/39718268/why-do-const-references-extend-the-lifetime-of-rvalues — Mat, Nov 21 '21 at 17:05
Why do you call it a "const l-value reference"? Isn't that normally just called an "r-value reference"? — Patrick Roberts, Nov 21 '21 at 17:14
just meant that `foo` is a reference and it is l-value since it has a name. — jwezorek, Nov 21 '21 at 17:15

score 1 · Accepted Answer · answered Nov 21 '21 at 17:10

Because this is the language rule Lifetime of a temporary

Whenever a reference is bound to a temporary or to a subobject thereof, the lifetime of the temporary is extended to match the lifetime of the reference.

There is no such rule for const pointers.

Why don't const l-value references to r-values result in dangling references?

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