C++如何使用按值传递和按引用传递重载方法

Question

如何使以下重载工作

#include <iostream>

using namespace std;

int subtractFive (int a)
{
    a = a - 5;

    return a;
}

int subtractFive (int &a)
{
    a = a -5;

    return a -5;
}

int main()
{
    int A = 10;

    cout << "Answer: " << subtractFive(*A) << endl;
    cout << "A Value "<< A << endl;

    cout << "Answer: " << subtractFive(A) << endl;
    cout << "A Value "<< A << endl;

    return 0;
}

试过但没有编译

    #include <iostream>

using namespace std;

int subtractFive (int a)
{
    a = a - 5;

    return a;
}

void subtractFive (int* a)
{
    *a = *a -5;
}

int main()
{
    int A = 10;

    cout << "Answer: " << subtractFive(A) << endl;
    cout << "A Value "<< A << endl;

    subtractFive(A);
    cout << "A Value "<< A << endl;

    return 0;
}

Answer 1

您可以尝试指定一个以地址作为参数的重载:

int subtractFive (int *a)
{
    *a = *a -5;

    return *a -5;
}