I came across http://rindovincent.blogspot.com/p/javascript.html where there was a simple Javascript program to find whether the number is odd or even. I am pasting the same code with permission here.
var n = prompt(“Enter a number to find odd or even”, “Type your number here”);
n = parseInt(n);
if (isNaN(n))
{
alert(“Please Enter a Number”);
}
else if (n == 0)
{
alert(“The number is zero”);
}
else if (n%2)
{
alert(“The number is odd”);
}
else
{
alert(“The number is even”);
}
I’ve read that the == comparison is dangerous because it does type coercion to make comparisons so it’s better practice to use === which will return true if both the left/right side are truly identical.
Generally, it’s best practice to use === but in this example, value % 2 is guaranteed to return an integer or NaN, which allows us to absolutely know the type of variable we’re comparing on either side of num % 2 or 0, but not num itself.
The problem comes down to if you accidentally get a false or true as an input, which loosely converts to a 0 or 1 when using the % operator. This might cause some problems in your code if false returns true. You normally would use a equality (===) operator to make this comparison, but since the boolean gets converted to a number on the % operator, equality won’t work here as most have typed the function.
Given the unknown nature of the use for this function and given it’s name, I’d say the following would be safe to use:
function isEven(num) {
return isNaN(num) && num !== false && num !== true ? false : num % 2 == 0;
}
or for those who enjoy readable code:
function isEven(num) {
if(num !== false && num !== true && !isNaN(num)) {
return num % 2 == 0;
} else return false;
}
Why not just
function isEven(value) {
return (value%2 == 0);
}
And the PHP version only needs an extra dollar sign:
function isEven($int){
return ($int%2 == 0);
}
In these times, I can assume most webdevvers use a javascript library like jQuery or similar. These libraries have functions like isEven built in.
For those who don’t use such libraries: great snippet ;)
I came across http://rindovincent.blogspot.com/p/javascript.html where there was a simple Javascript program to find whether the number is odd or even. I am pasting the same code with permission here.
var n = prompt(“Enter a number to find odd or even”, “Type your number here”);
n = parseInt(n);
if (isNaN(n))
{
alert(“Please Enter a Number”);
}
else if (n == 0)
{
alert(“The number is zero”);
}
else if (n%2)
{
alert(“The number is odd”);
}
else
{
alert(“The number is even”);
}
function isEven(value)
{
return /^\d+$/.test(value.toString()) ? (value%2 === 0 ? true : false ) : false;
};
//so, my turn =)
function isEven(value){
return isFinite(+value) && !(+value % 2)
}
function isOdd(n){
return 1&n;
}
Do i win?
start
read n
reminder=n/2
if (reminder=0)
print”number is even”
else
print”number is odd”
stop
I’ve read that the == comparison is dangerous because it does type coercion to make comparisons so it’s better practice to use === which will return true if both the left/right side are truly identical.
Generally, it’s best practice to use
===but in this example,value % 2is guaranteed to return an integer orNaN, which allows us to absolutely know the type of variable we’re comparing on either side ofnum % 2or0, but notnumitself.The problem comes down to if you accidentally get a
falseortrueas an input, which loosely converts to a0or1when using the%operator. This might cause some problems in your code iffalsereturnstrue. You normally would use a equality (===) operator to make this comparison, but since the boolean gets converted to a number on the%operator, equality won’t work here as most have typed the function.Given the unknown nature of the use for this function and given it’s name, I’d say the following would be safe to use:
or for those who enjoy readable code:
include<stdio.h>
include<conio.h>
main()
{
int n;
clrscr();
printf(“Enter a Number:\n”);
scanf(“%d”,&n);
if(n%2==0)
{
printf(“Even Number”);
}
else
{
printf(“Odd Number”);
}
getch();
}
Odd And Even syntax
Odd Syntax is:
Having the Even function…
And what if the “value” isn’t a base 10 number?
function isEven(value) { value.toString(10)%2==0?true:false; }var a;
var b=”ODD”;
var c=”EVEN”;
a=parseInt(window.prompt(“Enter The Value:”));
document.write(“Your Value is: “,chk(a));
function chk(a)
{
if (a%2==0)
return c;
else
return b;
}