[Javascript?] help with functions: isOdd in terms of isEven

  • Java
  • Thread starter mindauggas
  • Start date
  • #1
mindauggas
127
0
Problem statement:

Define isOdd function, and then define isEven function in terms of isOdd function.

My attempt is ridiculous:

Code:
var isOdd = function (x){
	if (x % 2 ===0){
		return isOdd;
	}else{
		return "is even";
	}
};
 
 var isEven = function (){
	return;
 };
 
Last edited by a moderator:

Answers and Replies

  • #2
14,193
8,180


in java:

public boolean isOdd(int x) { return ( (x%2) == 1 ); } // returns boolean true if odd

public boolean isEven(int x) { return ( (x%2) == 0 ); } // returns boolean true if even

or alternatively:

public boolean isEven(int x) { return !isOdd(x); }

basically don't return strings for "is" style functions better to return clearly defined boolean values. With strings you have to contend with remembering whether to use upper / lowercase and a host of other little problems each of which can cause a programming error in your code.
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
7


Problem statement:

Define isOdd function, and then define isEven function in terms of isOdd function.

My attempt is ridiculous:

Code:
var isOdd = function (x){
	if (x % 2 ===0){
		return isOdd;
	}else{
		return "is even";
	}
};
 
 var isEven = function (){
	return;
 };

What language are you programming in? Javascript?

Usually x % 2 gives the remainder when dividing x by 2, so if x % 2 ==0, then is x odd or even?

When you write something like
Code:
var isOdd = function (x){
	//do stuff
        return isOdd;
	}
};

You are defining a function that returns itself, which doesn't make much sense. Instead, try returning a boolean value:

Code:
var isOdd = function (x){
	//if x is odd
        return true;
        //otherwise
        return false;
	}
};
 
  • #4
mindauggas
127
0


Sorry, it's Javascript (my justification for not mentioning the programming language is that I'm new in programming and I presumed that it is easy to identify the language just by glancing ant the syntax - i learned that i was wrong ... ).
 
  • #5
jtbell
Mentor
15,962
4,724


Another important reason to include the name of the language in the title of your post is to tell people what's inside, so they don't waste their time opening a post about a language they don't know anything about. It also attracts people who do know the language, and might be in a position to help you.
 
Last edited:
  • #6
mindauggas
127
0


What language are you programming in? Javascript?

Usually x % 2 gives the remainder when dividing x by 2, so if x % 2 ==0, then is x odd or even?

When you write something like
Code:
var isOdd = function (x){
	//do stuff
        return isOdd;
	}
};

You are defining a function that returns itself, which doesn't make much sense. Instead, try returning a boolean value:

Code:
var isOdd = function (x){
	//if x is odd
        return true;
        //otherwise
        return false;
	}
};

Yes, JavaScript. I would use the litteral booleans, if I had only one function to work with. However, I don't understand how to define the isEven function in term of isOdd (I need to use the logical operator "!" (not) somehow, but it is confusing. And from this confusion many kinds of non-workable ideas arise ...
 
  • #7
36,698
8,687


Yes, JavaScript. I would use the litteral booleans, if I had only one function to work with. However, I don't understand how to define the isEven function in term of isOdd (I need to use the logical operator "!" (not) somehow, but it is confusing. And from this confusion many kinds of non-workable ideas arise ...

In the definition of isEven() you can test the number x by calling isOdd(x). If isOdd(x) returns false, x must be even.
 
  • #8
mindauggas
127
0


In the definition of isEven() you can test the number x by calling isOdd(x). If isOdd(x) returns false, x must be even.

Here is the original problem statement:

PHP:
Remember the functions isOdd and isEven from Exercise 3.4?

We'd like you to code them here again! But this time, the aim is to define one function in terms of the other using the ! symbol.

Define isOdd, and then define isEven in terms of isOdd.

To define the isOdd function, you will need to use:

a) The modulo % symbol.
b) If / else statement in the body of the function

I have to have both the isEven and isOdd funtions.

Code:
var isOdd = function (x){
	if (!x % 2 ===0){
		return "x is odd";
	}
};
 
var isEven = function (x){
	isOdd(x);
	return !isOdd(x);
};

I know this is WRONG, just don't know how to do it :)
 
  • #9
DavidSnider
Gold Member
510
146


Think about what the type being returned from isOdd is.

Does using ! on that type make any sense?

Also
Code:
 if(!x % 2 ===0)

is incorrect. That says "IF (NOT X) MOD 2 ===0". Try to rewrite this using the != operator.
 
Last edited:
  • #10
36,698
8,687


I have to have both the isEven and isOdd funtions.

Code:
var isOdd = function (x){
	if (!x % 2 ===0){
		return "x is odd";
	}
};
David already pointed out what was wrong with your if expression. The other thing is that you DON'T want to return a string. A function can return a boolean constant (i.e., true or false).
Code:
var isEven = function (x){
	isOdd(x);
	return !isOdd(x);
};

I know this is WRONG, just don't know how to do it :)
Try this:
Code:
var isEven = function (x){
	[STRIKE]isOdd(x);[/STRIKE]
	return !isOdd(x);
};
 
  • #11
mindauggas
127
0


Code:
var isOdd = function (x){
[I][B]	if ((x % 2)!=== 0){[/B][/I]
		return true;
	}
};
 
var isEven = function (x){
	return !isOdd(x);
};

Tried it like that, but it seems to be sintacticaly incorrect. I got an error: "missing operand; found =" on line 2
 
  • #12
gabbagabbahey
Homework Helper
Gold Member
5,002
7


Code:
	if ((x % 2)!=== 0){[/B][/I]

Tried it like that, but it seems to be sintacticaly incorrect. I got an error: "missing operand; found =" on line 2

When you write it like that, order of operations is to first compute (x%2), then try to compare it to 0 using the comparison operator "!===", but "!===" is not a valid comparison operator in javascript. There is no javascript comparison operator for "is not exactly equal to", so instead, you could test whether (x%2) is exactly equal to 0, and then take the logical negative of the comparison:

Code:
var isOdd = function (x){
	if(!((x % 2)===0)){
		return true;
	}
};

Additionally, you should also specify that the function should return false when (x%2)===2, which could be done like

Code:
var isOdd = function (x){
	if(!((x % 2)===0)){
		return true;
        }else{
	        return false;
        }
};

or like

Code:
var isOdd = function (x){
	if(!((x % 2)===0)){
		return true;
        }
        return false;
};

or, without using the logical not operator "!",
Code:
var isOdd = function (x){
	if((x % 2)===0){
		return false;
        }else{
	        return true;
        }
};
 
  • #14
mindauggas
127
0
Solved, thanks to all :)
 

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