严格相等运算符将告诉您两个对象类型是否相等。然而,是否有一种方法来判断两个对象是否相等,就像Java中的哈希码值一样?

堆栈溢出问题JavaScript中有hashCode函数吗?类似于这个问题,但需要一个更学术的答案。上面的场景说明了为什么有必要有一个,我想知道是否有等效的解决方案。


当前回答

在React中,你可以使用' React -fast-compare'中的isEqual。这个答案可能不适用于纯JavaScript,但在使用React时可能有用。

console.log(isEqual({ hello: 'world' }, { hello: 'world' })) // returns true

React中最快的深度相等比较。非常快速的通用深度比较。非常适合React。备忘录和shouldComponentUpdate。

更多信息可以在这里找到:https://www.npmjs.com/package/react-fast-compare。

其他回答

这取决于你对平等的定义。因此,作为类的开发人员,要由您来定义它们的相等性。

有时会使用一种情况,如果两个实例指向内存中的相同位置,则认为它们是“相等的”,但这并不总是您想要的。例如,如果我有一个Person类,如果两个Person对象具有相同的Last Name、First Name和Social Security Number(即使它们指向内存中的不同位置),我可能会认为它们是“相等的”。

另一方面,我们不能简单地说两个对象是相等的,如果它们的每个成员的值都相同,因为,有时,你并不想这样。换句话说,对于每个类,由类开发人员定义组成对象“标识”的成员并开发适当的相等操作符(通过重载==操作符或Equals方法)。

Saying that two objects are equal if they have the same hash is one way out. However you then have to wonder how the hash is calculated for each instance. Going back to the Person example above, we could use this system if the hash was calculated by looking at the values of the First Name, Last Name, and Social Security Number fields. On top of that, we are then relying on the quality of the hashing method (that's a huge topic on its own, but suffice it to say that not all hashes are created equal, and bad hashing methods can lead to more collisions, which in this case would return false matches).

我建议不要使用散列或序列化(正如JSON解决方案所建议的那样)。如果需要测试两个对象是否相等,则需要定义相等的含义。这可能是两个对象中的所有数据成员都匹配,也可能是内存位置必须匹配(意味着两个变量在内存中引用同一个对象),或者每个对象中只有一个数据成员必须匹配。

最近我开发了一个对象,它的构造函数在每次创建实例时都会创建一个新的id(从1开始,加1)。该对象有一个isEqual函数,用于将该id值与另一个对象的id值进行比较,如果匹配则返回true。

在这种情况下,我定义“相等”的意思是id值匹配。假设每个实例都有一个唯一的id,这可以用来加强匹配对象也占用相同内存位置的想法。尽管这是不必要的。

这里有一个非常基本的方法来检查对象的“值是否相等”。

var john = {
    occupation: "Web Developer",
    age: 25
};

var bobby = {
    occupation: "Web Developer",
    age: 25
};

function isEquivalent(a, b) {
    // Create arrays of property names

    var aProps = Object.getOwnPropertyNames(a);
    var bProps = Object.getOwnPropertyNames(b);

    // If number of properties is different, objects are not equivalent

    if (aProps.length != bProps.length) {
        return false;
    }

    for (var i = 0; i < aProps.length; i++) {
        var propName = aProps[i];

        // If values of same property are not equal, objects are not equivalent
        if (a[propName] !== b[propName]) {
           return false;
        }
    }

    // If we made it this far, objects are considered equivalent
    return true;
}

// Outputs: true
console.log(isEquivalent(john, bobby));

演示 - JSFiddle

如你所见,为了检查对象的“值是否相等”,我们本质上必须遍历对象中的每个属性,以查看它们是否相等。虽然这个简单的实现适用于我们的示例,但有很多情况它无法处理。例如:

如果其中一个属性值本身就是一个对象呢? 如果属性值之一是NaN(中唯一的值 JavaScript不等于它自己?) 如果a有一个值为undefined的属性,而b没有呢 这个属性(因此计算为undefined?)

对于一个健壮的检查对象“值是否相等”的方法,最好依赖于一个经过良好测试的库,它涵盖了各种边缘情况,如下划线。

var john = {
    occupation: "Web Developer",
    age: 25
};

var bobby = {
    occupation: "Web Developer",
    age: 25
};

// Outputs: true
console.log(_.isEqual(john, bobby));

演示 - JSFiddle

这是我的版本。它正在使用new Object。ES5中引入的keys特性以及+、+和+的想法/测试:

function objectEquals(x, y) { 'use strict'; if (x === null || x === undefined || y === null || y === undefined) { return x === y; } // after this just checking type of one would be enough if (x.constructor !== y.constructor) { return false; } // if they are functions, they should exactly refer to same one (because of closures) if (x instanceof Function) { return x === y; } // if they are regexps, they should exactly refer to same one (it is hard to better equality check on current ES) if (x instanceof RegExp) { return x === y; } if (x === y || x.valueOf() === y.valueOf()) { return true; } if (Array.isArray(x) && x.length !== y.length) { return false; } // if they are dates, they must had equal valueOf if (x instanceof Date) { return false; } // if they are strictly equal, they both need to be object at least if (!(x instanceof Object)) { return false; } if (!(y instanceof Object)) { return false; } // recursive object equality check var p = Object.keys(x); return Object.keys(y).every(function (i) { return p.indexOf(i) !== -1; }) && p.every(function (i) { return objectEquals(x[i], y[i]); }); } /////////////////////////////////////////////////////////////// /// The borrowed tests, run them by clicking "Run code snippet" /////////////////////////////////////////////////////////////// var printResult = function (x) { if (x) { document.write('<div style="color: green;">Passed</div>'); } else { document.write('<div style="color: red;">Failed</div>'); } }; var assert = { isTrue: function (x) { printResult(x); }, isFalse: function (x) { printResult(!x); } } assert.isTrue(objectEquals(null,null)); assert.isFalse(objectEquals(null,undefined)); assert.isFalse(objectEquals(/abc/, /abc/)); assert.isFalse(objectEquals(/abc/, /123/)); var r = /abc/; assert.isTrue(objectEquals(r, r)); assert.isTrue(objectEquals("hi","hi")); assert.isTrue(objectEquals(5,5)); assert.isFalse(objectEquals(5,10)); assert.isTrue(objectEquals([],[])); assert.isTrue(objectEquals([1,2],[1,2])); assert.isFalse(objectEquals([1,2],[2,1])); assert.isFalse(objectEquals([1,2],[1,2,3])); assert.isTrue(objectEquals({},{})); assert.isTrue(objectEquals({a:1,b:2},{a:1,b:2})); assert.isTrue(objectEquals({a:1,b:2},{b:2,a:1})); assert.isFalse(objectEquals({a:1,b:2},{a:1,b:3})); assert.isTrue(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assert.isFalse(objectEquals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}},{1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); Object.prototype.equals = function (obj) { return objectEquals(this, obj); }; var assertFalse = assert.isFalse, assertTrue = assert.isTrue; assertFalse({}.equals(null)); assertFalse({}.equals(undefined)); assertTrue("hi".equals("hi")); assertTrue(new Number(5).equals(5)); assertFalse(new Number(5).equals(10)); assertFalse(new Number(1).equals("1")); assertTrue([].equals([])); assertTrue([1,2].equals([1,2])); assertFalse([1,2].equals([2,1])); assertFalse([1,2].equals([1,2,3])); assertTrue(new Date("2011-03-31").equals(new Date("2011-03-31"))); assertFalse(new Date("2011-03-31").equals(new Date("1970-01-01"))); assertTrue({}.equals({})); assertTrue({a:1,b:2}.equals({a:1,b:2})); assertTrue({a:1,b:2}.equals({b:2,a:1})); assertFalse({a:1,b:2}.equals({a:1,b:3})); assertTrue({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}})); assertFalse({1:{name:"mhc",age:28}, 2:{name:"arb",age:26}}.equals({1:{name:"mhc",age:28}, 2:{name:"arb",age:27}})); var a = {a: 'text', b:[0,1]}; var b = {a: 'text', b:[0,1]}; var c = {a: 'text', b: 0}; var d = {a: 'text', b: false}; var e = {a: 'text', b:[1,0]}; var i = { a: 'text', c: { b: [1, 0] } }; var j = { a: 'text', c: { b: [1, 0] } }; var k = {a: 'text', b: null}; var l = {a: 'text', b: undefined}; assertTrue(a.equals(b)); assertFalse(a.equals(c)); assertFalse(c.equals(d)); assertFalse(a.equals(e)); assertTrue(i.equals(j)); assertFalse(d.equals(k)); assertFalse(k.equals(l)); // from comments on stackoverflow post assert.isFalse(objectEquals([1, 2, undefined], [1, 2])); assert.isFalse(objectEquals([1, 2, 3], { 0: 1, 1: 2, 2: 3 })); assert.isFalse(objectEquals(new Date(1234), 1234)); // no two different function is equal really, they capture their context variables // so even if they have same toString(), they won't have same functionality var func = function (x) { return true; }; var func2 = function (x) { return true; }; assert.isTrue(objectEquals(func, func)); assert.isFalse(objectEquals(func, func2)); assert.isTrue(objectEquals({ a: { b: func } }, { a: { b: func } })); assert.isFalse(objectEquals({ a: { b: func } }, { a: { b: func2 } }));

如果您有一个方便的深度复制函数,您可以使用下面的技巧来使用JSON。Stringify同时匹配属性的顺序:

function equals(obj1, obj2) {
    function _equals(obj1, obj2) {
        return JSON.stringify(obj1)
            === JSON.stringify($.extend(true, {}, obj1, obj2));
    }
    return _equals(obj1, obj2) && _equals(obj2, obj1);
}

演示:http://jsfiddle.net/CU3vb/3/

理由是:

由于obj1的属性被逐个复制到克隆中,因此它们在克隆中的顺序将被保留。当obj2的属性被复制到克隆对象时,因为obj1中已经存在的属性将被简单地覆盖,它们在克隆对象中的顺序将被保留。