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Java
package test;
import java.lang.* ;



public class bubbleSort {
	
	static void printArr(int[] arr) {
		for(int k=0;k<arr.length;k++) {//print arr
			System.out.print(arr[k] +" ");
			
		}
		
		
	}
	
	 static void check(int[] arr) {
		
		for(int j=0;j<arr.length-1;j++) {
		
			if(arr[j]<arr[j+1] || arr[j]==arr[j+1]) {
				if(j+1==arr.length-1) {//if last two elements are sorted too,print arr
					printArr(arr);
					
				}
			}
			else {
				sort(arr);//if it is not sorted,send to sort method again.
			}
			
		}
			
		
	}
	
	 static void sort(int[] arr) {
		int temp;
		for(int i=0;i<arr.length-1;i++) {
			
			if(arr[i]>arr[i+1]) {//swap
				temp=arr[i];
				arr[i]=arr[i+1];
				arr[i+1]=temp;
			}
		}
		check(arr);
	}

	public static void main(String[] args) {
		
		
		int arr[]= {64, 34, 25, 12, 22, 11, 90,86} ;
		sort(arr);

	}

}


As you can see,it is a basic bubble sort algorithm.Sort method works correct,after each turn it sends array to check which compares array items if it is correctly sorted or not.if it is not,it sends back to sort method.
İf it is correct(j+1==arr.length-1),it sends array to printArr.Everything is fine up to here,but it prints it multiple times...

What I have tried:

Putting System.exit(0) after print loop solves the problem,but I wonder actual problem.
Posted
Updated 17-Mar-21 11:46am

Your print is inside the loop.
And since a bubble sort is intended to move the highest element to the end each time round the loop, and check calls sort, which calls check, which calls sort ... you get the idea.
Bubble sort need not, and should not be recursive: it will crash you app very quickly once the array size starts to get large as the stack is really pretty small ...
 
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Comments
CPallini 18-Mar-21 3:07am    
5.
Quote:
but I wonder actual problem.

I made some changes to your code, run it and you will see what is going on:
Java
package test;
import java.lang.* ;



public class bubbleSort {
	
	static void printArr(int[] arr) {
		for(int k=0;k<arr.length;k++) {//print arr
			System.out.print(arr[k] +" ");
			
		}
		
		
	}
	
	 static void check(int[] arr) {
		System.out.print("Start check");
		for(int j=0;j<arr.length-1;j++) {
		
			if(arr[j]<arr[j+1] || arr[j]==arr[j+1]) {
				if(j+1==arr.length-1) {//if last two elements are sorted too,print arr
					printArr(arr);
					
				}
			}
			else {
				sort(arr);//if it is not sorted,send to sort method again.
			}
			
		}
			
		
		System.out.print("Exit check");
	}
	
	 static void sort(int[] arr) {
		System.out.print("Start sort");
		int temp;
		for(int i=0;i<arr.length-1;i++) {
			
			if(arr[i]>arr[i+1]) {//swap
				temp=arr[i];
				arr[i]=arr[i+1];
				arr[i+1]=temp;
			}
		}
		check(arr);
		System.out.print("Exit sort");
	}

	public static void main(String[] args) {
		
		
		int arr[]= {64, 34, 25, 12, 22, 11, 90,86} ;
		sort(arr);

	}

}

Your code is recursive: sort calls check which calls sort.

Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

jdb - The Java Debugger[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.
 
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Comments
CPallini 18-Mar-21 3:07am    
5.
Patrice T 18-Mar-21 3:09am    
Thank you

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