What is the meaning of 14.8.2 paragraphs 3 and 4 in the C++ Standard?

Question

I'm struggling to understand this rule, specially the sentences in bold below (my emphasis):

Consider the comment #2 in the snippet below: what does it mean to say that the function type is f(int), but t is const?

§14.8.2/3:
After this substitution is performed, the function parameter type adjustments described in 8.3.5 are performed. [ Example: A parameter type of “void ()(const int, int[5])” becomes “void(*)(int,int*)”. —end example ] [ Note: A top-level qualifier in a function parameter declaration does not affect the function type but still affects the type of the function parameter variable within the function. —end note ] [ Example:
template <class T> void f(T t);
template <class X> void g(const X x);
template <class Z> void h(Z, Z*);
int main() {
    // #1: function type is f(int), t is non const
    f<int>(1);
    // #2: function type is f(int), t is const
    f<const int>(1);
    // #3: function type is g(int), x is const
    g<int>(1);
    // #4: function type is g(int), x is const
    g<const int>(1);
    // #5: function type is h(int, const int*)
    h<const int>(1,0);
}

—end example ]

§14.8.2/4:

[ Note: f<int>(1) and f<const int>(1) call distinct functions even though both of the functions called have the same function type. —end note ]

Related: [Inconsistent use of const qualifier between declaration and definition](http://stackoverflow.com/a/20659598/1708801) — Shafik Yaghmour, Jul 07 '14 at 18:40

score 7 · Accepted Answer · edited May 23 '17 at 12:08

7

Consider:

template <class T> void f(T t) { t = 5; }

f<int> is well-formed, but f<const int> is not, because it attempts to assign to a const variable.

See: Use of 'const' for function parameters

edited May 23 '17 at 12:08

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answered Jul 07 '14 at 18:37

ecatmur

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Not particularly; the first sentence of 14.8.2.1p3 is more relevant. – ecatmur Jul 07 '14 at 19:16

score 6 · Answer 2 · answered Jul 07 '14 at 18:43

If you had two functions

void f(int x);
void g(const int x);

then both functions would have the same function type. This type is denoted void(int), and a function pointer of type void (*)(int) would be able to point to either function.

This is what is meant when we say that the top-level const-qualifiers on the parameters to a function do not affect the type of the function.

However, this does not mean that the const is meaningless. In the definition of f, you would be able to modify x, but in the definition of g, you would not be able to modify x, since x has type const int.

@Mao毛泽东 No, not really. If you're curious what that bullet point means, post another question. — Brian Bi, Jul 07 '14 at 19:14

What is the meaning of 14.8.2 paragraphs 3 and 4 in the C++ Standard?

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