How to determine whether it is an integer in javascript
Mar 30, 2021 pm 04:48 PM
Method: 1. Use the remainder operator to determine; 2. Use the "Math.round", "Math.ceil", and "Math.floor" methods to determine; 3. Use the parseInt function to determine ; 4. Judge through bit operations; 5. Use "Number.isInteger" to judge.
The operating environment of this tutorial: windows7 system, javascript version 1.8.5, Dell G3 computer.
Method 1. Use the remainder operator to determine
Any integer will be divisible by 1, that is, 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([]) // falseMethod 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. The Math.floor example is 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[Recommended learning: js basic tutorial]
Method 3. Through parseInt Judgment
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、
//很不錯，但也有一個缺點
isInteger(1000000000000000000000) // falseThe reason is that parseInt forces the first parameter to be parsed into a string before parsing the integer. 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
//這個函數(shù)很不錯，效率還很高。但有個缺陷，上文提到過，位運算只能處理32位以內的數(shù)字，對于超過32位的無能為力
isInteger(Math.pow(2, 32)) // 32位以上的數(shù)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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