Why the following code fails the test ?

Question

1.00/5 (2 votes)

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I find no error and the following code , the code fails the test.

I used JUnit for testing. Can someone help me ?

Java

public class BinString {
	
	public BinString() {}
	
	public String convert(String s) {
		return binarise(sum(s));
	}
	
	public int sum(String s){
		if(s=="") 
			return 0;
		if(s.length()==1) 
			return ((int) (s.charAt(0)));
		return ((int) (s.charAt(0)))+sum(s.substring(1));
	}
	
	public String binarise(int x) {
		if(x==0)
			return "";
		if(x%2==1) 
			return "1"+binarise(x/2);
		return "0"+binarise(x/2);
	}
}

What I have tried:

This is my test class

Java

import static org.junit.Assert.*;

import org.junit.Before;
import org.junit.Test;

public class BinStringTest {

	private BinString binString;
	
	@Before protected void setUp(){
		binString = new BinString();
	}
	
	@Test public void testSumFunction() {
		int expected =0;
		
		assertEquals(expected, binString.sum(""));
		expected = 100;
		assertEquals(expected, binString.sum("d"));
		expected = 256;
		assertEquals(expected, binString.sum("Add"));
		
		
	}
	
	@Test public void testBinariseFunction(){
		String expected = "101";
		assertEquals(expected, binString.binarise(5));
		expected = "11111100";
		assertEquals(expected, binString.binarise(252));
				
	}
	
	@Test public void testTotalConversion() {
		String expected = "10000001";
		assertEquals(expected, binString.convert("A"));
	}
	

}

Posted 21-May-16 5:43am

Laurentiu Bobora

Updated 21-May-16 6:58am

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Comments

Patrice T 21-May-16 11:52am

What is supposed to do your code ?
How does it fail ? Show actual result and expected result.

Laurentiu Bobora 21-May-16 11:54am

Class tested together a string of ASCII characters by collecting its values and returns a binary representation of
amount

Laurentiu Bobora 21-May-16 12:04pm

Do not pass the test, and displays a window with the message :
Method 'initializationError' not found. Opening the test class

Patrice T 21-May-16 12:06pm

Use Improve question to update your question.
So that everyone can pay attention to this information.

Sergey Alexandrovich Kryukov 21-May-16 12:29pm

This is perfectly normal. Failing the test is not the same as "error". This is the whole purpose of unit testing.
In other words, there is nothing you need help with. Well, except using the debugger and improving test development skills.
—SA

1 solution

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Patrice T · Accepted Answer · 2016-05-21T06:58:00

Solution 1

I think you will want to replace this code

Java

public String binarise(int x) {
    if(x==0)
        return "";
    if(x%2==1)
        return "1"+binarise(x/2);
    return "0"+binarise(x/2);
}

with

Java

public String binarise(int x) {
    if(x==0)
        return "";
    if(x%2==1)
        return binarise(x/2)+"1";
    return binarise(x/2)+"0";
}

as I suspect a bug there.

Posted 21-May-16 6:58am

Patrice T

Comments

Laurentiu Bobora 22-May-16 5:36am

I tried the code above and still not pass the test

Patrice T 22-May-16 5:41am

Do not state what the code don't do, state what the code do and why it is wrong with examples: parameters of test, expected result, real result.
Use Improve question to update your question.
So that everyone can pay attention to this information.