Why is `const T&` not sure to be const?

问题

template<typename T>
void f(T a, const T& b)
{
    ++a; // ok
    ++b; // also ok!
}

template<typename T>
void g(T n)
{
    f<T>(n, n);
}

int main()
{
    int n{};
    g<int&>(n);
}

Please note: b is of const T& and ++b is ok!

Why is const T& not sure to be const?

回答1:

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

so you have specified that T is a int&. When we apply a reference to an lvalue reference, the two references collapse to a single one, so int& & becomes just int&. Then we get to the rule from [dcl.ref]/1, which states that if you apply const to a reference it is discarded, so int& const just becomes int& (note that you can't actually declare int& const, it has to come from a typedef or template). That means for

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

---

Had you called g as

g<int>(n);
// or just
g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

回答2:

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& ) 

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

来源：https://stackoverflow.com/questions/54480599/why-is-const-t-not-sure-to-be-const