给定一个数组[1,2,3,4],如何求其元素的和?(在这种情况下,总数为10。)

我认为每个美元可能有用,但我不确定如何实现它。


当前回答

//Try this way

const arr = [10,10,20,60]; 
const sumOfArr = (a) =>{
    let sum=0;
    for(let i in a) { 
        sum += a[i];
    }
    return sum;
}
console.log(sumOfArr(arr))

其他回答

    <!DOCTYPE html>
    <html>
    <body>

      <p>Click the button to join two arrays.</p>
      <button onclick="myFunction()">Try it</button>
      <p id="demo"></p>
    <script>
var hege = [1, 2,4,6,7,8,8];
var stale = [1, 2,4,5];
function myFunction() {
    console.log((hege.length > stale.length))    
    var children  = (hege.length > stale.length)? abc1() :abc2();       document.getElementById("demo").innerHTML = children;
}
function abc1(){
    console.log(hege,"Abc1")    
    var abcd=hege.map(function (num, idx) {
        console.log(hege.length , idx)
        return stale.length>idx?num + stale[idx]:num;
    })
    return abcd;
}

function abc2(){

    console.log(hege,"Abc2",stale)    
    var abcd=stale.map(function (num, idx) {
        console.log(hege.length , idx)
        return hege.length>idx?num + hege[idx]:num;
    })
    return abcd;
}
</script>

</body>
</html>

我看到所有答案都是“减少”解决方案

var array = [1,2,3,4]
var total = 0
for (var i = 0; i < array.length; i++) {
    total += array[i]
}
console.log(total)

是否有理由不首先过滤数组以删除非数字?看起来很简单:

[1, 2, 3, null, 'a'].filter((x) => !isNaN(x)).reduce((a, b) => a + b)

对于真正大量的人来说,循环或减少可能是过程密集型的。使用高斯怎么样?

sum = (n * (n+1))/2;

来自mathcentral。

Vanilla JavaScript是您所需要的一切:

> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(function(e){sum += e}); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(e => sum += e); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e in a) sum += e; sum
"00123foobar"
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e of a) sum += e; sum
10