Question

我已经实现了两个数字的交换功能，这很有效。但是，我现在尝试交换两个字符串，但我收到的顺序相同。有人会知道我哪里出错了，或者我怎么办才能让名字改变位置？这是以下代码的示例：

#include "stdafx.h"
#include <cstring>
#include <iostream>
#include <string>

using namespace std;

void swapages (int &age1, int &age2);           
void swapname(char *person1, char *person2);

int _tmain(int argc, _TCHAR*argv[])
{
    char person1[] = "Alex";
    char person2[] = "Toby";
    int age1 = 22;
    int age2 = 27;


    cout << endl << "Person1 is called " << person1;
    cout << " and is " << age1 << " years old." << endl;
    cout << "Person2 is called " << person2;
    cout << " and is " << age2 << " years old." << endl;

    swapname(person1,person2);
    swapages(age1,age2);
    cout << endl << "Swap names..." << endl;
    cout << endl << "Person1 is now called " << person1;
    cout << " and is " << age1 << " years old." << endl;
    cout << "Person2 is now called " << person2;
    cout << " and is " << age2 << " years old." << endl;

    system("pause");
    return 0;
}

void swapages(int &age1, int &age2)
{
    int tmp = age2;
    age2 = age1;
    age1 = tmp;
}

void swapname(char *person1, char *person2)
{
    char* temp = person2;
    person2 = person1;
    person1 = temp;
}

Answer 1

您已将其标记为C ++，并且您已经包含<string>标头，那么为什么不使用std:string而不是所有这些指针和数组？

void swapname(string &person1, string &person2)
{
    string temp(person2);
    person2 = person1;
    person1 = temp;
}

int _tmain(int argc, _TCHAR*argv[])
{
    string person1 = "Alex";
    string person2 = "Toby";

    swapname(person1, person2);
}

Answer 2

问题在于，当您需要交换字符串时，您正在尝试交换作为函数局部变量的指针。所以你需要将一个字符串复制到另一个字符串中。此外，琴弦可以有不同的长度，因此不可能完全交换。如果函数的参数是相同大小的数组（对数组的引用），则可以毫无问题地完成。例如

void swap_name( char ( &lhs )[5], char ( &rhs )[5] )
{
    char tmp[5];

    std::strcpy( tmp, lhs );
    std::strcpy( lhs, rhs );
    std::strcpy( rhs, tmp );
}

Answer 3

您需要进行微小的更改才能让它按照您希望的方式工作。

void swapname(char **person1, char **person2);
.
.
char *person1 = "Alex";
char *person2 = "Toby";
.
.
swapname(&person1, &person2);
.
.

和

void swapname(char **person1, char **person2)
{
    char* temp = *person2;

    *person2 = *person1;  
    *person1 = temp;    
}

Answer 4

在你的代码中，person1和person2被定义为2个char数组而不是char指针变量，你不能交换它们，如果你将2个数组传递给swapname函数，它接受2个指针作为参数，它甚至不应该编译。

C ++：通过ref交换字符

4 个答案: