Thread: Having trouble with an if statement. please help!!

Hi

Im working on an if statement with the following specifications:

If the methods argument is between 1 and 4 it should set the value of its
"setting" instance variable to the value of the int argument. The method
should then return a string showing the current setting – "setting1",
"setting2", "setting3" or "setting4" depending on whether the "setting" instance variable
has been set to 1, 2, 3 or 4. If the argument is not in the range it should return the string "out of range"

the code I have so far is:

 public String whatSetting(int aNumber);
    {
         if ((aNumber > 0) && (aNumber < 5))
        {
            this.setting = aNumber;
            if (this.setting() == 1))

I am now stuck on how to get it to return the different strings depending on the value of "setting" and return the "out of range" string if the argument is out of range.

thanks in advance to anyone who can help

Just return the appropriate string based upon the condition you already wrote:

 
public String whatSetting(int aNumber)
{
         if ((aNumber > 0) && (aNumber < 5))
        {
            this.setting = aNumber;
           return "setting" + Integer.toString(aNumber);
         }else{
           return "out of range";
         }
}

(Technically you don't need the else, but I left it there for clarity)

Hi helloworld922, thankyou very much for your help so far.

I am unable to use arrays at this point. ive been working on this code some more but am still having problems! My new code is:

public String whatSetting(int aNumber)
    {
        if ((aNumber < 1) || (aNumber > 4))
        {
            return "out of range";
        }
        else
        if ((aNumber > 0) && (aNumber < 5))
             {
                 this.setting = aNumber;
                 if (this.setting == 1)
                 {
                     return "off";
                 }
                 if (this.setting == 2)
                 {
                     return "low";
                 }
             }
        }

I am getting the error "missing return statement" I know why this is but i just cant see a way to write the code to avoid the error!

How about this!

public String whatSetting(int aNumber)
    {
        if ((aNumber > 0) && (aNumber < 5))
             {
                 this.setting = aNumber;
                 if (this.setting == 1)
                 {
                     return "off";
                 }
                 if (this.setting == 2)
                 {
                     return "low";
                 }
                 If (this.setting == 3)
                 {
                      return "medium";
                 }
                 if (this.setting == 4) 
                 {
                      return "high";
                 }
             }
             else {
                return "out of range";
             }
        }

Then a return is always executed.

Hi JavaDaveUK,

Thankyou for your help.

I had already tried that earlier and it still said "missing return statement". I just tried re-writing it again as to your post and it still returns the same error.

are you sure you have the else from the first if, not the if that results in the high setting.

i.e.

if ((aNumber > 0) && (aNumber < 5))
{
             ... set diff. settings based on 1 to 4
             ..etc
}
else {
    return "out of range";
}

Hi just to let you know i work nights so im at work now. il be back working on this tomorrow so il look into it then. thankyou very much for all your help so, I will be sure to click the thankyou user when the thread is finished.

Hi, I have checked the code and i have it down exactcly as you said but it still returns the same error. any ideas???

sorry my mistake, remove the else at the end, so the return is guaranteed.

ie.

if ((aNumber > 0) && (aNumber < 5))
{
             ... set diff. settings based on 1 to 4
             ..etc
}
return "out of range";

As stated on another post arrays would be the best way of doing this and having the out of range as the first if and then returning the correct element in the array.
But in your case, this will suffice.

Here you go - you don't need to IF crap

	public String whatSetting(int aNumber)
	{
		this.setting = aNumber;
		switch(aNumber)
		{
		case 1: return "off";
		case 2: return "low";
		case 3: return "medium";
		case 4: return "high";
		default: return "out of range";
		}
	}

This will handle ANY int you pass. Also, if the settings change to include 0,5 or anything else it is easy to maintain.