为什么函数重载在C ++中会产生模糊错误？

Question

In the following code snippets, In function call f(1) , 1 is a literal of type int and in first function void f(double d) argument type is double and second function void f(short int i) argument type is short int .

Here 1 is an int type not a double type, then Why does compiler generated ambiguous error?

#include <iostream>
using namespace std;

void f(double d)  // First function
{
    cout<<d<<endl;
}

void f(short int i) // Second function
{
    cout<<i<<endl;
}

int main()
{
    f(1); // 1 is a literal of type int
    return 0;
}

Answer 1

Because, as your comment notes, 1 is a literal of type int .

To the compiler, an implicit conversion of int to short int is equally as valid as an implicit conversion of int to double ( cf . the C++ language standard, §13.3).

Thus, because the compiler can't decide between the double and short int overloads, it gives up and issues a diagnostic.

Note that the magnitude of the function parameter doesn't matter: just the type.

(It would be annoying if the compiler chose, at runtime, the short int overload if the calling argument was appropriate, and the double one in other instances.)