类中的函数重载

Posted 2021-09-08 阿弥陀佛.a

#include <stdio.h>

class Test
{
    int i;
public:
    Test()
    {
        printf("Test::Test()\\n");
        this->i = 0;
    }
    
    Test(int i)
    {
        printf("Test::Test(int i)\\n");
        this->i = i;
    }
    
    Test(const Test& obj)
    {
        printf("Test(const Test& obj)\\n");
        this->i = obj.i;
    }
    
    static void func()
    {
        printf("void Test::func()\\n");
    }
    
    void func(int i)
    {
        printf("void Test::func(int i), i = %d\\n", i);
    }
    
    int getI()
    {
        return i;
    }
};

void func()
{
    printf("void func()\\n");
}

void func(int i)
{
    printf("void func(int i), i = %d\\n", i);
}

int main()
{
    func();
    func(1);
    
    Test t;        // Test::Test()
    Test t1(1);    // Test::Test(int i)
    Test t2(t1);   // Test(const Test& obj)
    
    func();        // void func()
    Test::func();  // void Test::func()
    
    func(2);       // void func(int i), i = 2;
    t1.func(2);    // void Test::func(int i), i = 2
    t1.func();     // void Test::func()
    
    return 0;
}

#include <stdio.h>
#include <string.h>

char* strcpy(char* buf, const char* str, unsigned int n)
{
    return strncpy(buf, str, n);
}

int main()
{
    const char* s = "D.T.Software";
    char buf[8] = {0};
    
    //strcpy(buf, s);
    strcpy(buf, s, sizeof(buf)-1);
    
    printf("%s\\n", buf);
    
    return 0;
}

意义在于扩展

小结