Thread: Need Help

Code:

#include<conio.h>
#include<iostream>
using namespace std;


int opcio1();
int opcio2();



void main()
{

	char sel;
do{

        sel=getch();

    }while((sel<'1' || sel>'2')&&sel!=27);

switch(sel){

        case '1':

            opcio1();

            break;

 

        case '2':

            opcio2();

            break;

			case 27:

            return;

}


int opcio1()
{

	char sentence[100];
	
	cout<<"Ask what you should do in the market today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	 int theNumber;
		 theNumber = 1 + rand() % 3;		 
     switch (theNumber)  
	 {        
case 1:                        
     cout<<"You should definetly buy!"<<endl;                     
     break;         
case 2:
     cout<<"Sell!Sell!Sell!"<<endl;
     break;
case 3: 
     cout<<"Heck if I know. Go ask your brocker."<<endl;
     break;
default:
     break; 
}

     


int opcio2()
{
	char sentence[100];
cout<<"Do you feel lucky today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	 int theNumbers;
		 theNumbers = 1 + rand() % 4;
		 if (theNumbers = 1)
		 {
			 cout<<"No way!"<<endl;
		 }
		 if (theNumbers = 2)
		 {
			 cout<<"Most Likely, yes!"<<endl;
		 }
		 if (theNumbers = 3)
		 {
			 cout<<"Try again tomorrow!"<<endl;
		 }
		 if (theNumbers = 4)
		 {
			 cout<<"Yes. I belive so!"<<endl;
		 }

}
		 system("PAUSE");
     return 0;
}

Can somebody help me with this.

At the beginning I have menu so the user can pick what function to enter.
But I have errors with getch, and with the definition of functions. Can somebody help me

Thank you for your help.

Here I made some improvements, but still I have small error:
error C2146: syntax error : missing ';' before identifier 'cout'

Here is the new version

Code:

#include<conio.h>
#include<iostream>
using namespace std;

int opcio1();
int opcio2();


 
int main()
{
  int choice
cout << "==========================================" << endl;
cout << " Menu" << endl;
cout << "==========================================" << endl;
cout << "1.Fortune Teller" << endl;
cout << "2.Stock Market" << endl;
cout << "==========================================" << endl;
 
cout<<"For fortune teller press 1, for stock market press 2"<<endl;
cin>>choice;
      if (choice == 1) opcio1();      
      else if (choice == 2) opcio2();      
      else cout << "Use only number shown in the menu" << endl;  
      

 
system ("PAUSE");
return 0;    
}

int opcio1()//Stock Market
{
	char sentence[100];	
	cout<<"Ask what you should do in the market today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	int theNumber;
	theNumber = 1 + rand() % 3;

	switch (theNumber)  
	{        
		case 1:                        
			cout<<"You should definetly buy!"<<endl;                     
			break;

		case 2:
			cout<<"Sell!Sell!Sell!"<<endl;
			break;
		case 3:
			cout<<"Heck if I know. Go ask your brocker."<<endl;
			break;
		default:
			break; 
	}
	system ("PAUSE");
		return 0;
 }

int opcio2()//Fortune teller
{
	char sentence[100];
	cout<<"Do you feel lucky today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	int theNumbers;
	theNumbers = 1 + rand() % 4;
	// = is assign, == is compare, you can just as well use a switch.
	if (theNumbers == 1)
	{
		cout<<"No way!"<<endl;
	}
	else if (theNumbers == 2)
	{
		cout<<"Most Likely, yes!"<<endl;
	}
	else if (theNumbers ==3)
	{
		cout<<"Try again tomorrow!"<<endl;
	}
	else if (theNumbers == 4)
	{
		cout<<"Yes. I belive so!"<<endl;
	}
	system ("PAUSE");
		return 0;
}

And I have one more question, after the program is done, how can I make a menu which can ask the user
Do you like to exit OR
Do you like to go back in the beginning?

Thank You

Code:

int choice;

Note that error messages include line numbers, which would help greatly in finding such a thing. (In this case, the line number would have been for the line with the cout while the ; is missing on the line above, but since it tells you "before" that's easy to spot.

If you want things to happen over and over and over again, you will need to put the things that happen over and over and over again inside a loop structure. (You know how to put a menu on the screen and ask for input, you've got it in your program already.)

Thank you gays!!

Finally I dont have any errors in my script.

But I dont know why does not go back to main in opcio3()

I need to get back in main() and to run the script as many times as the user want to.

Any suggestions

Code:

#include<conio.h>
#include<iostream>
#include <cstdlib>
#include <stdlib.h>
using namespace std;

int opcio1();
int opcio2();
int opcio3();


 
int main()
{
  int choice;
cout << "==========================================" << endl;
cout << " Menu" << endl;
cout << "==========================================" << endl;
cout << "1.Fortune Teller" << endl;
cout << "2.Stock Market" << endl;
cout << "==========================================" << endl;
 
cout<<"For fortune teller press 1, for stock market press 2"<<endl;
cin>>choice;
      if (choice == 1) opcio1();      
      else if (choice == 2) opcio2();      
      else cout << "Use only number shown in the menu" << endl;  
      

  
}

int opcio1()//Stock Market
{
	char sentence[100];	
	cout<<"Ask what you should do in the market today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	int theNumber;
	theNumber = 1 + rand() % 3;
			
	switch (theNumber)  
	{        
		case 1:                        
			cout<<"You should definetly buy!"<<endl;                     
			break;

		case 2:
			cout<<"Sell!Sell!Sell!"<<endl;
			break;
		case 3:
			cout<<"Heck if I know. Go ask your brocker."<<endl;
			break;
		default:
			break; 
	}
			 
		system("cls");
		return opcio3();
 }

int opcio2()//Fortune teller
{
	char sentence[100];
	cout<<"Do you feel lucky today!"<<endl;
	cin.getline (sentence,100);
	cin.ignore();
	int theNumbers;
	theNumbers = 1 + rand() % 4;
	
	if (theNumbers == 1)
	{
		cout<<"No way!"<<endl;
	}
	else if (theNumbers == 2)
	{
		cout<<"Most Likely, yes!"<<endl;
	}
	else if (theNumbers ==3)
	{
		cout<<"Try again tomorrow!"<<endl;
	}
	else if (theNumbers == 4)
	{
		cout<<"Yes. I belive so!"<<endl;
	}

	system("cls");
		return opcio3();
}
int opcio3()
{
int choice1;

cout<<"If you want to exit press 1, for starting over press 2"<<endl;
cin>>choice1;
      	       
      if (choice1 == 1) exit(-1);
	  else if(choice1 == 2) main(); 
	  else cout << "Use only number shown in the menu" << endl;
	  	
}

Add a loop to main that contains all the code you want to repeat, then add a way to exit that loop and end the program.

I allready have a way to exit the program.
I need the script to repeat indefinetly until the user want to exit.

Do you have any idea how to create the loop.

Please, I need this scipt for tomorrow.

I would get rid of opcio3() and move it's logic up to main(). Then, make the 3rd choice in main() for the user to exit. Finally, wrap the logic you want repeating in main() inside a while(1) block, break if the choice is 3.

Todd

Be specific.

What is wrong with my script????

I've started learning C++ 2 months ago, so I'm not an expert.
Plus, my native language is not English.

Thank you. I know that you are trying to help me, but please be more specific.