Javascript 方法和变量(初学者)

Question

我在理解代码时遇到了严重的问题。

function replace(str, before, after) {

    var rtt = before;
    var rt;

    if(before[0] === before[0].toUpperCase()) {
        var aa = after.split("");
        var u = after[0].toUpperCase();
        aa.splice(0,1,u);
        rt =  aa.join("");    
    }
    else {
       rt = after;
    }

    var r = str.split(" ");  
    var i = r.indexOf(rtt);
    r.splice(i,1, rt);
    var join = r.join(" ");
    return join;
}

replace("A quick brown fox jumped over the lazy dog", "jumped", "leaped");

我不明白为什么我不能在 r.splice(i,1,rt) 之后执行 r.join("") 然后返回r？ 为什么我必须声明新的变量 join？否则 return 将不会返回正确的语句。不过代码是正确的。

Answer 1

根据 MDN ,

The splice() method changes the content of an array by removing existing elements and/or adding new elements.

这个方法返回什么？

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

就这个问题而言

Why I must declare new variable join ?

实际上，您不必声明新变量。你可以只返回这个值

return r.join(" ");

function replace(str, before, after) {

    var rtt = before;
    var rt;

    if(before[0] === before[0].toUpperCase()) {
        var aa = after.split("");
        var u = after[0].toUpperCase();
        aa.splice(0,1,u);
        rt =  aa.join("");    
    }
    else {
       rt = after;
    }

    var r = str.split(" ");  
    var i = r.indexOf(rtt);
    r.splice(i,1, rt);
    return r.join(" ");
}

var replaced = replace("A quick brown fox jumped over the lazy dog", "jumped", "leaped");

document.write(replaced);