[已解决]用函数模板交换字符串问题

YJSzhou

为什么用函数模板，交换函数中只是简单传字符串交换，输出结果为
交换前：s1=布丁，s2=补丁
交换后：s1=补丁，s2=布丁

template<class T>
void swap(T a, T b)
{
        T temp = a;
        a = b;
        b = temp;
}

int main()
{
        std::string s1 = "布丁";
        std::string s2 = "补丁";
        std::cout << "交换前：s1=" << s1 << ",s2=" << s2 << "\n";
        swap(s1, s2);
        std::cout << "交换后：s1=" << s1 << ",s2=" << s2 << "\n";

        return 0;
}

最佳答案

月排行榜 / 总排行榜

人造人

2022-8-7 10:24:48

#include <iostream>

#if 0
template <class T>
void swap(T &a, T &b) {
    T temp = a;
    a = b;
    b = temp;
}

int main() {
    std::string s1 = "布丁";
    std::string s2 = "补丁";
    std::cout << "交换前：s1=" << s1 << ",s2=" << s2 << "\n";
    swap(s1, s2);
    std::cout << "交换后：s1=" << s1 << ",s2=" << s2 << "\n";
    return 0;
}
#else
template <class T>
void swap(T *a, T *b) {
    T temp = *a;
    *a = *b;
    *b = temp;
}

int main() {
    std::string s1 = "布丁";
    std::string s2 = "补丁";
    std::cout << "交换前：s1=" << s1 << ",s2=" << s2 << "\n";
    swap(&s1, &s2);
    std::cout << "交换后：s1=" << s1 << ",s2=" << s2 << "\n";
    return 0;
}
#endif

跳转到最佳答案楼层

人造人

#include <iostream>

#if 0
template <class T>
void swap(T &a, T &b) {
    T temp = a;
    a = b;
    b = temp;
}

int main() {
    std::string s1 = "布丁";
    std::string s2 = "补丁";
    std::cout << "交换前：s1=" << s1 << ",s2=" << s2 << "\n";
    swap(s1, s2);
    std::cout << "交换后：s1=" << s1 << ",s2=" << s2 << "\n";
    return 0;
}
#else
template <class T>
void swap(T *a, T *b) {
    T temp = *a;
    *a = *b;
    *b = temp;
}

int main() {
    std::string s1 = "布丁";
    std::string s2 = "补丁";
    std::cout << "交换前：s1=" << s1 << ",s2=" << s2 << "\n";
    swap(&s1, &s2);
    std::cout << "交换后：s1=" << s1 << ",s2=" << s2 << "\n";
    return 0;
}
#endif

ba21

1. 没明白你想说什么。
2. 你的代码输出结果应为：

交换前：s1=布丁,s2=补丁
交换后：s1=布丁,s2=补丁

3. void swap(T a, T b) 是传值，函数内的修改对函数外没影响。
该代码要实现交换的正确方式是传址(传引用) void swap(T &a, T &b)

aaron0919

当变量传入函数时，变为形参，接下来的事就与s1,s2无关，
加上&取地址符后，就是对原值操作
或者加上inline ，即在main内执行