CODEWITHSUNDEEP
java
MBS
MBS

Reputation: 11

could you hep me to explain this part of code please?

its about break with label, java programming. This program searches for the number 1 in an array.

   public static void main(String[] args) {
        int[][] arrayOfInts =
        {{  32,     87,     3,      589},
        {   12,     1076,   2000,   8},
        {   622,    127,    77,     955}};

        int searchfor = 1;

        int i = 0;
        int j = 0;
        boolean foundIt = false;

        search:
        for (; i < arrayOfInts.length; i++) {
            for (j = 0; j < arrayOfInts[i].length; j++){
                if (arrayOfInts [i][j] == searchfor) {
                    foundIt = true;
                    break search;
                }
            }
        }
        if (foundIt) {
            System.out.println("Found "+ searchfor + " at " + i +", "+ j);
        } else {
            System.out.println(searchfor + " not in the array");
        }
    }
}

the part of code that i didnt understand is

search:
        for (; i < arrayOfInts.length; i++) {
            for (j = 0; j < arrayOfInts[i].length; j++){
                if (arrayOfInts [i][j] == searchfor) {
                    foundIt = true;
                    break search;
                }
            }
        }

and why there is ";" in front of "i"??

Upvotes: 1

Views: 75

Answers (3)

geffchang
geffchang

Reputation: 3340

This is the format of the for loop. The initialization, termindation, and increment part can be blank.

for (initialization; termination; increment) {
    statement(s)
}

In your code example, i was initialized before the for loop.

It will be similar to this:

  public static void main(String[] args) {
    int[][] arrayOfInts = {{1,2}, {3, 4}};
    int searchfor = 8;
    boolean foundIt = false;
    for (int i = 0; i < arrayOfInts.length; i++) {
      for (int j = 0; j < arrayOfInts[i].length; j++){
        if (arrayOfInts [i][j] == searchfor) {
          foundIt = true;
          break;
        }
      }
    }
  }

See: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/for.html

Upvotes: 0

QuickSilver
QuickSilver

Reputation: 4045

For loop to (<intialize counter>;<check loop condition>;<modify counter>) In you first for loop there is nothing to initialize hence ;

or (; i < arrayOfInts.length; i++) { // Selects i'th row of your array
            for (j = 0; j < arrayOfInts[i].length; j++){ // Scans array by j'th column of the i'th row
                if (arrayOfInts [i][j] == searchfor) { // matches the i'th row and j'th column value of the array 
                    foundIt = true;
                    break search; // comes out of the column for loop
                }
            }
        }

Upvotes: 2

Deepak Tatyaji Ahire
Deepak Tatyaji Ahire

Reputation: 5309

It's just an another way of writing for loop.

As i is initialised outside before the for loop, there is no initialisation required in for loop.

Code 1:

for(int i=0;i<n;i++){}

Code 2:

int i=0;
for(;i<n;i++){}

The Code 1 and Code 2 mean the same thing.

Upvotes: 2

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