Determine whether the natural numbers are odd or even
Topic request
Code implementation
<script>
// The user enters a number
var n = Number(prompt("Please enter a number"));
// Check if it is a natural number.
if (0 <= n && Math.floor(n) === n) {
// is a natural number
// Check whether it is odd
if (n % 2! =0) {
alert("Is odd.");
} else {
alert("Even"); }}else {
alert("Not a natural number.");
}
</script>
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Topic difficulty
Summary of several methods to judge integers
1. Use modular arithmetic
if(num%1= = =0) {
// num is an integer
}
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2. Use Math methods: Round, floor, ceil
if(Math.round(num) === num) {
// num is an integer
}
if(Math.floor(num) === num) {
// num is an integer
}
if(Math.ceil(num) === num) {
// num is an integer
}
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3. Use the Number method isInteger
if(Number.isInteger(num)) {
// num is an integer
}
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This method is a new one in ES6, and polyfill is shown below
Number.isInteger = Number.isInteger || function(value) {
return typeof value === 'number' &&
isFinite(value) &&
Math.floor(value) === value;
};
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