如何返回到我的菜单C++

Question

这是我的程序，亚姆很抱歉，因为我只是一个noobie所以我想问我怎么能返回到菜单，因为亚姆使用switch语句，我想问他后，他/她完成转换。请参阅程序中的评论。

#include <iostream> 
#include <stdlib.h> 
#include <conio.h> 

using namespace std; 

int main() 
{ 
int option; 
float C,F,result; 
char rpt; 

cout<<"Temperature Converter" 
    <<"\n1.Celsius to Fahrenheit" 
    <<"\n2.Fahrenheit to Celsius" 
    <<"\n3.Exit" 
    <<"\n\nEnter a number from 1 to 3: "; 
cin>>option; 

switch (option) 
{ 
    case 1: 
    system ("CLS"); 
    cout<<"Celsius to Fahrenheit!"<<endl; 
    cout<<"\nEnter Celsius: "; 
    cin>>C; 

    result=C*9/5+32; 

    cout<<C<<" Celsius is equivalent to "<<result<<" Fahrenheit."<<endl<<endl; 


    cout<<"Would you like to convert again? Type Y to try again OR N to exit. : "; 
    cin>>rpt; 

    if (rpt='Y') 

    //What Code should i put here? 


    break; 

Answer 1

你应该使用循环。像这样：

bool ended = false; 
while (!ended) 
{ 
    //some code 
} 

如果答案是N变化结束为true。

或者，正如有些人喜欢：

bool ended; 
do 
{ 
    //some code 
} while(!ended); 

Answer 2

把代码放到一个do while循环

int main() 
{ 
    do { 
    ..... 
    } while(rpt);// instead of if 

} 

Answer 3

你应该做一个功能，即显示菜单和返回所选择的选项 例：

 int showMenu() { 
     //Show menu 
     //Read option 
    //Return option 
     }  

Answer 4

询问用户是否要重复的代码属于后切换块。你也应该使用==运算符进行比较。因此，像

do { 

    // Get menu choice 
    ... 

    switch (option) 
    { 
     case 1: 
      ... 
      break; 

     case 2: 
      ... 
      break; 

     ... 
    } 

    cout<<"Would you like to convert again? Type Y to try again OR N to exit. : "; 
    cin>>rpt; 
} while (rpt == 'Y'); 

Answer 5

// put the label Start below the declaration // 

if (rpt == 'Y') 

system ("CLS"); 
goto Start; 

else 

exit (0);