const int变量和数字的不同类型推导

Question

template<typename T>
void func(T&);

int x=0;
func(x); // OK, int&

const int cx=0;
func(cx); // OK, const int&

func(0); // invalid initialization of non-const reference of type 'int&' from an rvalue of type 'int'

But why doesn't the 'const int&' type get deduced in the third case? What is the rationale behind this type deduction rule?

After all, we can perfectly bind a number to const lvalue reference.

Answer 1

It's because the type of 0 is int , not const int . (There are no cv-qualified prvalues of scalar type.) Since the deduction only takes into account the type, and not the value category, T must be deduced as int here just as it was in the case of x --- even though that results in an attempt to perform an invalid reference binding.

(A forwarding reference T&& has special rules that take into account the argument's value category when deducing T . So passing in x and passing in 0 would deduce T differently. But for an ordinary reference T& , only the argument's type matters.)

Answer 2

I find it easier to imagine whats going wrong here if you understand that a reference is just a pointer with syntax. So if your method was:

template<typename T>
void func(T*);

int x=0;
func(&x); // OK, you can take the address of x

const int cx=0;
func(&cx); // OK, you can take the address of cx

func(&0); // Taking the address of 0 doesn't make sense.