why the answer is not 53.but output 56 25+28=53

Question

// Example program
#include <iostream>
using namespace std;

int main()
{
    

    
    char gender,length,style,answer;
    int Price,Counter=0;
    
do{
    cout<<"Enter Your Gender(F/M):";
    cin>>gender ;
    
    cout<<"Enter Your Hair Length(S/M/L):" ;
    cin>>length ;
    
    cout<<"Enter Your Style(R/S):" ;
    cin>>style ;
    
    
switch (gender){
    
    
    case 'F':
      if (((length=='S')&&(style=='R')))
      {
      Price=25;
      }
      else if (((length=='S')&&(style=='S')))
      {
      Price=35;
      }
      else if (((length=='M')&&(style=='R')))
      {
      Price=30;
      }
      else if (((length=='M')&&(style=='S')))
      {
      Price=40;
      }
      else if (((length=='L')&&(style=='R')))
      {
      Price=35;
      }
      else
      {
      Price=45;
      }
      cout<<"Price:RM"<<Price<<endl;
      break;
      
    case 'M':
      
     if (((length=='S')&&(style=='R')))
      {
      Price=28;
      }
      else if (((length=='S')&&(style=='S')))
      {
      Price=33;
      }
      else if (((length=='M')&&(style=='R')))
      {
      Price=31;
      }
      else if (((length=='M')&&(style=='S')))
      {
      Price=36;
      }
      else if (((length=='L')&&(style=='R')))
      {
      Price=35;
      }
      else
      {
      Price=40;
      }
      cout<<"Price:RM"<<Price<<endl;
      break;
      default:
      cout<<"Not valid"<<endl;;
      
}

    cout<<"Add another customer?(Y/N):" ;
    cin>>answer ;
    Counter++ ;
    }while(answer =='Y') ;
   
    
    if(Counter==1)
    goto result1;
    
    else
     Price+=Price;
     goto result2;
    
    
    result1:
    cout<<"Total:RM"<<Price<<endl;
    return 0;
    
    result2:
    cout<<"Total:RM"<<Price<<endl;
    return 0;    
}

Answer 1

The actual output depends on the input you provided to the program. So it's hard to tell what happened without knowing what you exactly did.

Also, using the goto command is not recommended as it makes the code hard to read and follow.