Return in if statement

Question

I've been coding a text adventure game in C#. I want to do a scene where you encounter a pokemon. I want to have 3 random pokemon to encounter, and I made a method that returns the name of the pokemon you find. Code is:

public string choosePokemon()  
{  
    Random random = new Random();  
    int pokemonChosen = random.Next();  
    if (pokemonChosen == 1)  
    {  
        string pokemon = "Pidgey";  
        return pokemon;  
    }  
    if (pokemonChosen == 2)   
    {  
        string pokemon = "Charmander";  
        return pokemon;  
    }  
    if (pokemonChosen == 3)  
    {  
        string pokemon = "Geodude";  
        return pokemon;  
    }  
    return "missingno";  
} 

Whenever I execute the code it says "You encountered a missingno!". I want it to return the name and break the method, returning to the method of the scene. I put a return "missingno"; at the end because Studio would give me an error that not all code paths return a value.

Solution 1

Lets start with the Random number generator. To be effective, you should only instantiate Random once. To do this, you might make a static instance of it for the class. If you don't, it's possible to get the same number back each time you call random.Next().

public class PokemonChooser
{
    private static Random random = new Random();

    public static string ChoosePokemon() { ... }
}

Then, we need to realize that random.Next() returns any non-negative integer value, not just the ones you want. So, we need to round the result to something that's useful for your case. Since you have 3 items from which you would like to randomly choose, we can round it using the modulus operator (%), which basically gives you the remainder of a division operation (/). For instance: (int)(5 / 2) == 2, and 5 % 2 == 1.

var pokemonChosen = random.Next() % 3;

Next, since this result is zero-based, we'll need to account for that in the if statements.

      if (pokemonChosen == 0)
      {
        string pokemon = "Pidgey";
        return pokemon;
      }

      if (pokemonChosen == 1)
      {
        string pokemon = "Charmander";
        return pokemon;
      }

      if (pokemonChosen == 2)
      {
        string pokemon = "Geodude";
        return pokemon;
      }

Since the Pokemon names are pretty obvious, we can get rid of the pokemon variable without losing any readability.

      if (pokemonChosen == 0)
      {
        return "Pidgey";
      }

      if (pokemonChosen == 1)
      {
        return "Charmander";
      }

      if (pokemonChosen == 2)
      {
        return "Geodude";
      }

We could probably make this clearer and shorter by using a case statement instead of multiple if statements.

      switch (pokemonChosen) {
        case (0): return "Pidgey";
        case (1): return "Charmander";
        case (2): return "Geodude";
      }

And since we simply return "missingno" if the we cant find the pokemon by number, we can use a default case:

      switch (pokemonChosen) {
        case (0): return "Pidgey";
        case (1): return "Charmander";
        case (2): return "Geodude";
        default: return "missingno";
      }

All in all, we end up with:

public class PokemonChooser
{
    private static Random random = new Random();

    public static string ChoosePokemon()
    {
      var pokemonChosen = random.Next() % 3;

      switch (pokemonChosen) {
        case (0): return "Pidgey";
        case (1): return "Charmander";
        case (2): return "Geodude";
        default: return "missingno";
      }
    }
}

And we use it like so:

var pokemon = PokemonChooser.ChoosePokemon();

However, you may want to choose a higher number than 3 in random.Next() % 3, otherwise, "missingno" will never be returned.

Solution 2

Random.Next() (with no parameters) will return a random non-negative integer. You probably meant to constrain it to a certain range, like this:

int pokemonChosen = random.Next(1, 4);

Also note, you can make your code a bit cleaner by just returning a constant inside your if-block like this:

if (pokemonChosen == 1)  
{  
    return "Pidgey";  
}  

Or even better, encapsulate your random options in an array. That way you can get rid of all your if-blocks entirely:

var options = new [] { "Pidgey", "Charmander", "Geodude" };
int pokemonChosen = random.Next(options.Length); // array indexes start at 0
return options[pokemonChosen];

Solution 3

The problem is that your random.Next() call is returning a number anywhere in the range of random - You need an alternate version of random.Next() that returns only 1, 2, or 3, or you need to perform that yourself. Try doing this instead: int pokemonChosen = random.Next(1, 4) (The lower bound is inclusive while the upper bound is exclusive)

Solution 4

A good way to do this, as I've found through trials with my own Pokemon based game, is to put the New Random() outside the function. Also, you could instantiate the string outside the function, as well, in case you have different Pokemon options for different parts of the game. This way, you wouldn't have to return anything, just call the function then use the string. Assuming you're doing a windows based with a label for story output:

Random random = new Random();
string pokemon;

    public void choosePokemon()  
    {   
        int pokemonChosen = random.Next(2);  
        if (pokemonChosen == 0) {
        pokemon = "Pidgey";
        }
        if (pokemonChosen == 1) {
        pokemon = "Charmander";
        }
        if (pokemonChosen == 2) {
        pokemon = "Geodude";
        }
    } 

    lblStory.Text += "You encountered a wild " + pokemon + "!";

I did something very similar for a game I made for a friend.

EDIT: Also, something else I did in my game was to shorten pokemon in variable names to pkmn. Shorter to type, and as seen through the main games, commonly accepted. Not that the players will read your code.

Solution 5

You need to specify the upper bound for random.Next():

int pokemonChosen = random.Next(4);

would return an integer in the range 0 - 3.

To have it return 1-4 you could use:

int pokemonChosen = random.Next(1, 5);

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