Question

关于C语言的一个程序问题

#include<stdio.h>
#include<ctype.h>

void main()
{
double number1=0.0;
double number2=0.0;
char operation=0;
char ch=0;

printf("Please input number1,number2 and operation.");
scanf("%lf,%c,%lf",&number1,&operation,&number2);

switch(operation)
{
case '+':
printf("=%.2lf",number1+number2);
printf("\nEnter 'y' or 'Y' and 'n' or 'N' for next calculation");
/*scanf("%c",&ch);*/
/*toupper(ch);
if('ch'=='Y')*/
/* printf("%c",ch);*/
/*goto case ;*/
/*else*/
break;

case '-':
printf("\n=%.2lf",number1-number2);
break;

case '*':
printf("\n%.2lf",(number1*number2));
break;

case '/':
if(number2!=0)
printf("\n%.2lf",number1/number2);
else
printf("\nError!");
break;

case '%':
if(number2!=0)
printf("\n%.1d",(int)number1%(int)number2);
else
printf("\nError!");
break;

default :
printf("\nTry it again!");
break;
}
}
我想只用goto语句实现，给用户两种选择，输入y或Y ，就执行下一个计算，输入n或N，结束程序，我在程序插入了goto语句，但出现问题。

Answer 1

给，已经编译运行通过：
#include<stdio.h>
#include<ctype.h>

void main()
{
double number1=0.0;
double number2=0.0;
char operation=0;
char ch=0;

START: //加上这个

printf("Please input number1,number2 and operation.");
scanf("%lf,%c,%lf",&number1,&operation,&number2);

switch(operation)
{
case '+':
printf("=%.2lf",number1+number2);
printf("\nEnter 'y' or 'Y' and 'n' or 'N' for next calculation");
/*scanf("%c",&ch);*/
/*toupper(ch);
if('ch'=='Y')*/
/* printf("%c",ch);*/
/*goto case ;*/
/*else*/
break;

case '-':
printf("\n=%.2lf",number1-number2);
break;

case '*':
printf("\n%.2lf",(number1*number2));
break;

case '/':
if(number2!=0)
printf("\n%.2lf",number1/number2);
else
printf("\nError!");
break;

case '%':
if(number2!=0)
printf("\n%.1d",(int)number1%(int)number2);
else
printf("\nError!");
break;

default :
printf("\nTry it again!");
break;
}

printf("\nDo you want to try again?"); //加上这个
fflush(stdin);//加上这个
ch=getchar();//加上这个
if(ch=='y'||ch=='Y') goto START; //加上这个
}