为什么这两个函数不能重载呢？

Question

为什么我不能重载这两个函数？

#include <iostream>
using namespace std;

class Base
{
    public:
    void run(int a);
    void run(const int a);
};
int main() {
    // your code goes here
    return 0;
}

输出：

prog.cpp:8:7: error: 'void Base::run(int)' cannot be overloaded
  void run(const int a);
       ^
prog.cpp:7:7: error: with 'void Base::run(int)'
  void run(int a);

Answer 1

按照标准：

13.1 Overloadable declarations
....

(3.4) — Parameter declarations that differ only in the presence or absence of const and/or volatile are equivalent. That is, the const and volatile type-specifiers for each parameter type are ignored when determining which function is being declared, defined, or called. [ Example:
   typedef const int cInt;
   int f (int);
   int f (const int); // redeclaration of f(int)
   int f (int) { /* ... */ } // definition of f(int)
   int f (cInt) { /* ... */ } // error: redefinition of f(int)
—end example ]

Only the const and volatile type-specifiers at the outermost level of the parameter type specification are ignored in this fashion; const and volatile type-specifiers buried within a parameter type specification are significant and can be used to distinguish overloaded function declarations. In particular, for any type T, “pointer to T,” “pointer to const T,” and “pointer to volatile T” are considered distinct parameter types, as are “reference to T,” “reference to const T,” and “reference to volatile T.”

为什么这两个函数不能重载呢？

Why these two functions cannot be overloaded?

c++

overloading