Thread: Changing a variable

Hi. I'm trying to write a simple program where you need a key to open a door, and it requires you to get the key before opening the door. I'm getting some errors. This is what I have so far:

Code:

#include <stdio.h>
#define KEY 0

int a;

int main() {

    char c;
    int key;
    key = 0;

    printf("You stand before a locked door. There is a key nearby.\n\n");
    printf("a. Take key.\nb. Open door.\n");
    
    c=getch();
    
    if(c == 'a')  {
        printf("\nYou take the key.\n"); 
    }
    key = 1; 
    
        
    if(c == 'b') && KEY = 0 {
        printf("\nYou do not have the key.\n"); 
        
    if(c == 'b') && KEY = 1 {
        printf("\nThe door unlocks with ease.\n"); 
        
    }
}

Sorry for asking such a basic question, but if anybody can help me with this code, I'll appreciate it.

Thanks so much. but I don't know how to have it display "You do not have the key." I thought it would if you pressed b before getting the key.


EDIT: oh I found that by putting "key = 1;" inside the first If brace it will work.

What I don't understand is if this is setting the variable so that the key acts like a key: you need it to open the door. I don't know how to program that.

For the edit, of course it works, because you are saying that if the user pressed a, you print a message and set the variable key to 1. If you put it outside, key is set to 1 regardless if the user enters a or not.
And I don't understand the second question.

Firstly, I would get rid of the variable a, since it's global and never used.
Then I would write it like this to emphasize readability.

Code:

#include <stdio.h>

int main()
{
    int c; 
    int key; 
    key = 0; 
    
    printf("\n\nYou stand before a locked door. There is a key nearby.\n\n");
    printf("a. Take key.\nb. Open door.\n\n");
    
    c = getch(); 
    
    if (c == 'a')
    {
        printf("You take the key. What would you like to do now?\n\n");
        key = 1; 
        printf("a. Open the door!\n");
        printf("b. Examine the key.\n");
        c = getch(); 
        if (c == 'a' && key == 1) printf("You open the door..."); 
        if (c == 'b' && key == 1) printf("It's a nice enough key, shiny and gold.");
    }
        
    if (c == 'b' && key == 0) 
    {
        printf("You don't have the key.");
    }
}

But this is not enough.
Consider these lines...

Code:

        if (c == 'a' && key == 1) printf("You open the door..."); 
        if (c == 'b' && key == 1) printf("It's a nice enough key, shiny and gold.");

Key is always set to 1! So there is no point in checking it here... unless, of course, you don't intend to always set it to 1.

You've got a good beginning program here. I congratulate you on getting this far.

What you want is the same question asked over and over again until you open the door. You need a loop, which is the point of variables; making loops act differently different times through.

Lemme psuedocode something for you:

Code:

do 
{
  show description.
  if (the key is not picked up) describe key in room.
  show choices "a) get key, b) open door"
  get input.
  if (input == 'a') 
  {
    pick up key (set key = 1)
    show pickup key message "It's a nice enough key, shiny and gold."
  }
  else if (input == 'b') 
  {
    if (key is not picked up) 
    {
      show key not picked up message
    }
    else
    {
      open door (you'll need a door variable, I'm thinking)
      show door open message
    }
  }
  else
  {
    show "press 'a' or 'b'" message
  }
} while (the door is not open)
show exit message.

It's not as major a rewrite as it seems, but it is a major rethink. This way when you get out of the room you're out of the loop and you can move on with the next step without nesting the next step inside the previous.

Here's a Cymon's Games example: http://cymonsgames.retroremakes.com/letterguess/

FANTASTIC. I actually loaded it in my compiler and there was a small issue with the fact that getch() isn't in stdio, so I switched it to getchar() and it ran fine. Could use a little bit of formatting (a \n after open door for instance), but good job! You're on your way to being a coder.

Originally Posted by Meldreth

don't forget to return a value from main.

That said, with respect to the 1999 (but not 1990) edition of the C standard, main() implicitly returns 0 if control reaches its end without encountering a return statement.

Originally Posted by matthew82

That must be why Meldreth added conio.h.

Yes. However, note that <conio.h> is non-standard. In the first place, you would not need getchar() or getch() if you run your program from the command prompt, or use an IDE that pauses the program for you.

Right. Changing getch for getchar it responded to the 'enter' character that I had to type after each line by telling me to just type 'a' or 'b'. I probably should have just included conio too, tho that library is technically non-standard.

EDIT: Dang it laserlight. You slipped yours in a moment before mine.