Question

我希望完成此练习，请写一个名为join的函数，就像内置联接一样，但是使用reduce！

＆＃13;

Let's write a function called join that works just like the built-in join, but using reduce! If you're unfamiliar with the built-in version of join, this is how it works:

["the", "quick", "brown", "fox"].join(" "); // => "the quick brown fox"
["one", "two", "three"].join(":"); // => "one:two:three"
Part One: First, write a function called joinWith that takes three arguments: the string to join onto, the string that will be joined, and a separator. That is, it should work like this:

function joinWith(onto, next, separator) {
  // YOUR CODE HERE
}
joinWith("the quick", "brown", " "); // => "the quick brown"
joinWith("one:two", "three", ":"); // => "one:two:three"
Part Two: Now, using joinWith and reduce, write join:

function join(array, separator) {
  // YOUR CODE HERE
}
join(["the", "quick", "brown", "fox"], " "); // => "the quick brown fox"
join(["one", "two", "three"], ":"); // => "one:two:three"

＆＃13;

我写了以下代码：

＆＃13;

var joinWith = function(start, toAdd, separt){

	 return reduce(arguments, function(start, toAdd){
	 	 return start + separt + toAdd; 
	 	  
	 }, " ");
}

＆＃13;

＆＃13; 但是，我不确定为什么我会得到以下结果。

＆＃13;

joinWith("one: two", "three", ":");
" :one: two:three::"

＆＃13;

有人可以告诉我这里发生了什么，以及如何从字符串的前面和分隔符不应出现的区域中删除分隔符。

我非常感谢你的帮助。

Answer 1

目前正在发生什么：

您没有显示reduce()函数的实现，但假设它的工作方式与标准Array.prototype.reduce()相同，那么您的输出可以解释如下：

arguments是一个类似于数组的对象，包含以下值：

["one: two", "three", ":"]

当您在reduce()上致电arguments时，它会依次访问每个项目：

第一次迭代：start为" "，从最后一个参数reduce()开始，toAdd为"one: two"。从该迭代中返回字符串" :one: two"。

第二次迭代：start为" :one: two"（前一次迭代的返回值），toAdd为"three"，您返回" :one: two:three"。< / p>

第3次迭代：start为" :one: two:three"（上一次迭代的返回值），toAdd为":"，您返回" :one: two:three::"

你应该做什么：

你知道这个函数有三个参数，所以只需使用：

function joinWith(start, toAdd, separt) {
  return start + separt + toAdd;
}

但确实想要使用reduce()：

当你的规范只加入两个字符串时，它确实是错误的工作工具，但如果你必须，你想在joinWith()的最后一个参数之外的所有字符串上使用它 - 注意如下所示，只要分隔符是最后一个参数，就可以传递任意数量的字符串来加入：

＆＃13;

var joinWith = function(){
  if (arguments.length < 3) return arguments[0];
  var separt = arguments[arguments.length-1];
  return reduce(Array.prototype.slice.call(arguments,0,-1), function(start, toAdd){
    return start + separt + toAdd;
  }); // note: do not pass an initial value
}
    
console.log(joinWith("one: two", "three", ":"));
console.log(joinWith("one","two","three","four","five",":"));
console.log(joinWith("a","b","c","d","e",", "));
console.log(joinWith("x","y"));
    
// Dodgy simulation of your reduce() function for demo purposes only: 
function reduce(arr, cb) {
  return Array.prototype.reduce.call(arr, cb);
}

＆＃13;

实现正在传递3个参数的连接函数

1 个答案: