How to determine whether it is an integer in javascript

Javascript method to determine whether it is an integer: 1. Use the remainder operator to determine; 2. Use Math.round, Math.ceil, Math.floor to determine; 3. Judge through parseInt; 4. Use bit operations judge.

The operating environment of this tutorial: Windows 7 system, JavaScript version 1.8.5, DELL G3 computer.

Javascript method to determine whether it is an integer:

Method 1. Use the remainder operator to determine

Any integer will Divisible by 1, the remainder is 0. Use this rule to determine whether it is an integer.

function isInteger(obj) { return obj%1 === 0 } isInteger(3) // true isInteger(3.3) // false　 isInteger('') // true isInteger('3') // true isInteger(true) // true isInteger([]) // true

Returns true for empty strings, string type numbers, Boolean true, and empty arrays. If you are interested in the internal conversion details of these types, please refer to: Weird False Values in JavaScript

Therefore, you need to first determine whether the object is a number, such as adding a typeof

function isInteger(obj) { return typeof obj === 'number' && obj%1 === 0 } isInteger('') // false isInteger('3') // false isInteger(true) // false isInteger([]) // false

Method 2: Use Math.round, Math.ceil, and Math.floor to determine whether the

integer is still equal to itself after rounding. Use this feature to determine whether it is an integer, Math.floor example, as follows

function isInteger(obj) { return Math.floor(obj) === obj } isInteger(3) // true isInteger(3.3) // false isInteger('') // false isInteger('3') // false isInteger(true) // false isInteger([]) // false

Method 3. Determine through parseInt

function isInteger(obj) { return parseInt(obj, 10) === obj } isInteger(3) // true isInteger(3.3) // false isInteger('') // false isInteger('3') // false isInteger(true) // false isInteger([]) // false、 //很不错，但也有一个缺点 isInteger(1000000000000000000000) // false

The reason is that parseInt forces the integer to be parsed before parsing it. The first parameter is parsed into a string. This method of converting numbers to integers is not a good choice.

Method 4. Determine through bit operations

function isInteger(obj) { return (obj | 0) === obj } isInteger(3) // true isInteger(3.3) // false isInteger('') // false isInteger('3') // false isInteger(true) // false isInteger([]) // false //这个函数很不错，效率还很高。但有个缺陷，上文提到过，位运算只能处理32位以内的数字，对于超过32位的无能为力 isInteger(Math.pow(2, 32)) // 32位以上的数字返回false了

Method 5. ES6 provides Number.isInteger

Number.isInteger(3) // true Number.isInteger(3.1) // false Number.isInteger('') // false Number.isInteger('3') // false Number.isInteger(true) // false Number.isInteger([]) // false

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