c++ - C++ 标准中 14.8.2 第 3 段和第 4 段的含义是什么？

Question

我很难理解这条规则，特别是下面加粗的句子(我的重点):

考虑下面代码片段中的注释 #2:函数类型是 f(int) 是什么意思？ ，但是 t是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 ]

Answer 1

考虑:

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

f<int>格式正确，但是 f<const int>不是，因为它试图分配给 const变量。

参见:Use of 'const' for function parameters