CODEWITHSUNDEEP
javaif-statementfor-loop
aluckii
aluckii

Reputation: 39

For loop how to print to console without leaving on first iteration java

I have written a for loop which counts each value in an array and if that array does NOT contain a given number (numberIn) that the user has entered, then a relavent error will appear.

However I noticed that I get a dead code warning on i++ as it only iterates once. Is there a way for the loop to iterate fully?

for (int i = 0; i < array.length; i++) {
    if (!(numberIn == array[i]))
        System.out.println("WRONG!!!!!");
        break;
}

EDIT: sorry everyone for the massive confusion!, what i meant to say was that i want an error to appear if the value is NOT in the array, sorry again! :(

Upvotes: 0

Views: 411

Answers (11)

Kyle Williamson
Kyle Williamson

Reputation: 2184

When you call for a break, it will leave the loop you are currently in. Simply take out the break; allowing it to run through all values.

EDIT: Put the break inside the brackets. I don't think he wants it to run forever, just until he finds the value.

Upvotes: 0

JLRishe
JLRishe

Reputation: 101662

As I understand it, you are trying to test for the case when numberIn is not in the array. In that case, the simplest approach is to loop through the array, keeping track of whether the value was found, and then check whether it was found after that:

boolean found = false;
for (int i = 0; i < array.length; i++) {
    if (numberIn == array[i]) {
        found = true;
        break;
    }
}  

if (!found) {
    System.out.println("WRONG!!!!!");
}

You can also do this a bit more cleanly if you use a method:

boolean contains(array, value) {
    for (int i = 0; i < array.length; i++) {
        if (value == array[i]) {
            return true;
        }
    }  
    return false;
}


if (!contains(array, numberIn)) {
    System.out.println("WRONG!!!!!");
}

Upvotes: 1

Davis Broda
Davis Broda

Reputation: 4125

you need curly brackets {} around the if statement, otherwise it will always exit after one iteration. Without {} an if statement only applies to the statement immediately below it. In this case the System.out.println("WRONG!!!!!"); statement. the break statement is not part of the if statement by default. The curly brackets indicate that everything inside of them is only to be executed in the event the condition is true, rather than just the next single statement.

EDIT: also, you had it writing the output in the event that it didn't detect a match. It should println() ony in the event that there is a match. remove the ! from the if statement, and it should work.

Fixed version

for (int i = 0; i < array.length; i++) {
        if (numberIn == array[i]){
             System.out.println("WRONG!!!!!");
             break;
         }
}

Upvotes: 0

Gregory Basior
Gregory Basior

Reputation: 290

I agree with the other posts that the break should be withing brackets in the if statement, but according to what you are saying that you the loop is supposed to do, the logic does not appear correct either. You have said that you only want the error message to appear if the value is found in the loop so if I am understanding correctly what the loop intends to do, it should read:

for (int i = 0; i < array.length; i++) {
        if ( numberIn == array[i] ) 
        {
             System.out.println("WRONG!!!!!");
             break;
        }
}

Upvotes: 0

Kevin F
Kevin F

Reputation: 169

public static void main (String[] args) throws java.lang.Exception
    {
        int array[] = {1,2,3};
        int numberIn = 3;
        for (int i = 0; i < array.length; i++) {
            if (!(numberIn == array[i])){
                 System.out.println("WRONG!!!!!");
                 break;
            }
        } 
    }

You need the brackets on if sentence :D

Upvotes: 0

Edd
Edd

Reputation: 3822

I think is a problem with your braces:

Without the if-statement body having braces, it's actually parsed as the following, which probably isn't what you intended:

for (int i = 0; i < array.length; i++) {
    if (!(numberIn == array[i])) {
        System.out.println("WRONG!");
    }
    break;
}

Hopefully this makes it clearer to see why your loop will always break after one iteration. Adding the braces, and putting the break back where you want it should resolve your issue (I've also slightly re-written your if condition):

for (int i = 0; i < array.length; i++) {
    if (numberIn != array[i]) {
        System.out.println("CORRECT!");
        break;
    }
}

Upvotes: 0

Ovy.Istrate
Ovy.Istrate

Reputation: 504

Your break hits because it's not in the if statement, you should put the break inside the if statement it will work

for (int i = 0; i < array.length; i++) {
            if (!(numberIn == array[i]))
               {
                 System.out.println("WRONG!!!!!");
                 break;
               }
        }

Upvotes: 0

Bob Billy
Bob Billy

Reputation: 285

Your mistake is forgetting to put brackets in your "if" statement. Your code looks like this:

 if (!(numberIn == array[i]))
                   System.out.println("WRONG!!!!!");
   break;

Therefore no matter if the if statement is run, the break will be run.

Put the break inside the if statement and it will work.

Upvotes: 0

Astra Bear
Astra Bear

Reputation: 2738

You need more braces eg:

for (int i = 0; i < array.length; i++) {
            if (!(numberIn == array[i])){
                 System.out.println("WRONG!!!!!");
                 break;

 }
        }

Upvotes: 0

alephtwo
alephtwo

Reputation: 322

You need to include the break in your if block. Right now, it's always executing.

See: 

for (int i = 0; i < array.length; i++) {
  if (!(numberIn == array[i])) {
    System.out.println("WRONG!!!!!");
    break;
  }
}

In general, the use of break is considered bad practice. Perhaps consider converting this to a while loop if you would like to stop after the first error is detected.

Upvotes: 0

TheHorse
TheHorse

Reputation: 234

You break from your loop regardless whether or not the if statement evaluates to true.

Fix:

for (int i = 0; i < array.length; i++) {
    if (!(numberIn == array[i])){
        System.out.println("WRONG!!!!!");
        break;
    }
}

Upvotes: 0

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