C++98/03 reference collapsing and cv qualifiers

Question

The code below compiles (gcc 4.7.2 or icc 13) and produces \"1 2\" output. Which means that const qualifier is dropped, i. e., f ha

Answer 1

GCC 4.7.2 does not compile this when the flag -std=c++98 is specified. In fact, in C++98 (as well as in C++03) references to references do not collapse.

An attempt to instantiate f, where T = int&, produces the following function signature (here I intentionally switch the position of the argument type T and the const specifier, which is allowed because const T& is the same as T const&):

void f(int& const& t) // ERROR: reference to reference is illegal

The above is not legal in C++98, nor in C++03. Consistently, this is the error you get from GCC 4.7.2:

Compilation finished with errors:
source.cpp: In function 'int main()':
source.cpp:15:14: error: no matching function for call to 'f(int&)'
source.cpp:15:14: note: candidate is:
source.cpp:5:6: note: template void f(const T&)
source.cpp:5:6: note:   template argument deduction/substitution failed:
source.cpp: In substitution of 'template void f(const T&) [with T = int&]':
source.cpp:15:14:   required from here
source.cpp:5:6: error: forming reference to reference type 'int&'

Nevertheless, if you use the -std=c++11 flag, then the compiler performs reference collapsing when instantiating the template: an lvalue reference to an lvalue reference becomes an lvalue reference:

void f(int& const& t) == void f(int& t)

Here the const qualifier gets dropped, because it applies to the reference, and not to the referenced object. Since references cannot be reassigned, they are const by nature, which is why the const is considered superfluous and removed. See this Q&A on SO for an explanation.

That yields an lvalue reference to an lvalue reference, which resolves into a simple lvalue reference. Hence, the signature on the right side is instantiated.

The above is a viable candidate to resolve the call for f(a) and, therefore, it compiles without errors.