请问汉诺塔当n==1时为什么它又变成了a，b，c不是重新给他赋值了吗？
程序如下：
#include<stdio.h>

void hannuota(int n,char A,char B,char C)
{
    if(n==1)
    {
        printf("将编号为%d的盘子直接从%c柱子移动到%c柱子\n",n,A,C);
    }
   
    else
    {
        hannuota(n-1,A,C,B);
        
        printf("将编号为%d的盘子直接从%c柱子移动到%c柱子\n",n,A,C);
        
        hannuota(n-1,B,A,C);   
    }

}

int main()
{
    int n;
   
    char ch1='E';
   
    char ch2='F';
   
    char ch3='G';
   
    printf("请输入要移动盘子的个数:");
   
    scanf("%d",&n);
   
    hannuota(n,ch1,ch2,ch3);
   
    return 0;
}

但是……ABC里面的内容会不同耶~

但是执行到n==1的时候他会变成一样的，你可以复制调试一下，实在想不明白

递归就有点类似“死循环”  所以一定要有一个终止这个“循环”并退出的条件

而本题目里的
if(n==1)
    {
        printf("将编号为%d的盘子直接从%c柱子移动到%c柱子\n",n,A,C);
    }
就是递归终止的条件

对哦~当n==1的时候是一样的~上楼已经说了那个是递归终止条件~

请问汉诺塔当n==1时为什么它又变成了a，b，c不是重新给他赋值了吗？

这句话什么意思？哪个被重新赋值？

就是执行到hannuota(n-1,A,C,B);这一句时不是给形参赋值了吗，但当n-1为1时，形参的值就又变成了a,b,c

#include<stdio.h>        
void hanoi(int N,char A,char B,char C)        
{        
    if(N==1)/*将A座上剩下的第N个盘子移动到C座上*/   
        printf("move dish %d from %c to %c\n",N,A,C);/*打印移动步骤*/
    else   
    {   
        hanoi(N-1,A,C,B);/*借助C座将N-1个盘子从A座移到B座*/
        printf("move dish %d from %c to %c\n",N,A,C);/*打印移动步骤*/
        hanoi(N-1,B,A,C);/*借助A座将N-1个盘子从B座移到C座*/
    }   
}        
main()        
{        
    int n;   
    printf("Please input the nuber of dishes;");   
    scanf("%d",&n);/*输入要移动的盘子个数*/   
    printf("The steps to move %2d dishes are:\n",n);   
    hanoi(n,'A','B','C');/*调用递归函数*/   
}        


Please input the nuber of dishes;3
The steps to move  3 dishes are:
move dish 1 from A to C
move dish 2 from A to B
move dish 1 from C to B
move dish 3 from A to C
move dish 1 from B to A
move dish 2 from B to C
move dish 1 from A to C
Press any key to continue

Please input the nuber of dishes;4
The steps to move  4 dishes are:
move dish 1 from A to B
move dish 2 from A to C
move dish 1 from B to C
move dish 3 from A to B
move dish 1 from C to A
move dish 2 from C to B
move dish 1 from A to B
move dish 4 from A to C
move dish 1 from B to C
move dish 2 from B to A
move dish 1 from C to A
move dish 3 from B to C
move dish 1 from A to B
move dish 2 from A to C
move dish 1 from B to C
Press any key to continue