方法一、此方法当a和b的数非常大时，可能会出现溢出的问题

#include <stdio.h>
int main()
{
    int a = 2;
    int b = 3;
    printf("交换前 a=%d, b=%d\n", a, b);
    
    a = a + b;  //a为两数之和，b此时不变还是3
    b = a - b;  //此时b的值为5-3=2
    a = a - b;  //此时a的值为5-2=3
    printf("交换后 a=%d, b=%d\n", a, b);
    return 0;
}

方法二、可解决方法一缺点

#include <stdio.h>
int main()
{
    int a = 2;
    int b = 3;
    printf("交换前a=%d,b=%d\n",a,b);
    
    a = a ^ b;  //010^011→001
    b = a ^ b;  //001^011→010
    a = a ^ b;  //001^010→011
    printf("交换后a=%d,b=%d\n",a,b);
    return 0;
}