blocks|key|2727394|text|看起来像是addTo()函数将另一个函数返回到sum中，该函数将把您传递给它的任何内容添加到原始数字中(或者我假设这就是您要编写的内容；要做的第一件事就是更改add()中的语句，以使用%2B=而不只是%2B，以确保保存结果)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2727395|由于您希望将2到100之间的每个数字相加(因为您已经将1传递给了addTo())，因此可以尝试编写一个从2到100运行的for循环，将每个数字传递给sum()函数，以便将它们全部相加。下面是一个例子：|2727396|var+sum+=+addTo(1);
for+(var+i=2;+i<100;+i%2B%2B)+sum(i);
var+result+=+sum(100);|code-block|syntax|javascript|2727397|在这里，我在循环后添加了100，因为我想获取最终结果。您还可以在循环中添加100，并使用sum(0)来获得结果，而不需要在循环后更改它。|2727398|entityMap^0|5|7|N|3|27|5|2K|2|2Q|1|0|W|7|1O|3|22|5|0|0|18|6|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$]]

<p>It looks like the <code>addTo()</code> function returns another function into <code>sum</code> that will add whatever you pass it to the original number (or I assume that's what you meant to write; the first thing to do is change the statement inside of <code>add()</code> to use a <code>+=</code> instead of just <code>+</code> to make sure you save the result).</p>

<p>Since you want to add each number from 2 to 100 (since you already passed 1 into <code>addTo()</code>), try writing a <code>for</code> loop that runs from 2 to 100 passing each one into the <code>sum()</code> function to add them all together. Here's an example:</p>

<pre><code>var sum = addTo(1);
for (var i=2; i&lt;100; i++) sum(i);
var result = sum(100);
</code></pre>

<p>Here I added 100 after the loop since I wanted to grab the final result. You could also add 100 in the loop and use <code>sum(0)</code> to get the result without changing it after the loop.</p>


blocks|key|799933|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|799934|目前我在JS中有一个闭包，如下所示：|blockquote|799935|799936|799937|所有的函数都会创建闭包，只有当它们被利用时，它们才会显得特别。;-)|799938|var+addTo+=+function(num){|code-block|syntax|javascript|799939|我不知道为什么在声明更清晰的时候还要使用函数表达式：|799940|function+addTo(num)+{|799941|然后就是：|799942|++var+add+=+function(inner){
++++return+inner+%2B+num;
++}
++return+add;
}|799943|它(坚持使用表达式)可以是：|799944|++return+function+(inner)+{return+inner+%2B+num};
}|799945|然后用new调用它|799946|var+sum+=+new+addTo(1);|799947|这会导致addTo创建一个不使用的新对象，所以您也可以这样做：|799948|var+sum+=+addTo(1);|799949|产生完全相同的结果。所以：|799950|​|799951|function+addTo(num)+{
++++return+function+(inner)+{return+inner+%2B+num};
}

var+sum+=+addTo(1);

document.write(sum(3));|799952|799953|但是，这实际上只是Currying的一个版本，所以sum只是将提供的值与最初提供给addTo的值相加。|offset|length|799954|如果你想把从0到某个限制的所有数字相加，你只需要一个循环，不需要闭包：|799955|799956|function+sumTo(num)+{
++
++var+total+=+0;
++
++for+(var+i+=+0;+i+<=+num;+i%2B%2B)+{
++++total+%2B=+i;
++}
++
++return+total;
}

document.write(sumTo(5));+//+15|799957|799958|请注意，提供负数将导致无休止的循环，您应该防止出现这种情况(我将把它留给您自己解决)。|799959|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Currying^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|9|8|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1X|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1Y|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1Z|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|20|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|21|7|@]|8|@]|9|$]]|$1|H|3|I|4|J|6|22|7|@]|8|@]|9|$K|L]]|$1|M|3|N|4|5|6|23|7|@]|8|@]|9|$]]|$1|O|3|P|4|J|6|24|7|@]|8|@]|9|$K|L]]|$1|Q|3|R|4|5|6|25|7|@]|8|@]|9|$]]|$1|S|3|T|4|J|6|26|7|@]|8|@]|9|$K|L]]|$1|U|3|V|4|5|6|27|7|@]|8|@]|9|$]]|$1|W|3|X|4|J|6|28|7|@]|8|@]|9|$K|L]]|$1|Y|3|Z|4|5|6|29|7|@]|8|@]|9|$]]|$1|10|3|11|4|J|6|2A|7|@]|8|@]|9|$K|L]]|$1|12|3|13|4|5|6|2B|7|@]|8|@]|9|$]]|$1|14|3|15|4|J|6|2C|7|@]|8|@]|9|$K|L]]|$1|16|3|17|4|5|6|2D|7|@]|8|@]|9|$]]|$1|18|3|19|4|5|6|2E|7|@]|8|@]|9|$]]|$1|1A|3|1B|4|J|6|2F|7|@]|8|@]|9|$K|L]]|$1|1C|3|19|4|5|6|2G|7|@]|8|@]|9|$]]|$1|1D|3|1E|4|5|6|2H|7|@]|8|@$1F|2I|1G|2J|1|2K]]|9|$]]|$1|1H|3|1I|4|5|6|2L|7|@]|8|@]|9|$]]|$1|1J|3|19|4|5|6|2M|7|@]|8|@]|9|$]]|$1|1K|3|1L|4|J|6|2N|7|@]|8|@]|9|$K|L]]|$1|1M|3|19|4|5|6|2O|7|@]|8|@]|9|$]]|$1|1N|3|1O|4|5|6|2P|7|@]|8|@]|9|$]]|$1|1P|3|-4|4|5|6|2Q|7|@]|8|@]|9|$]]]|1Q|$1R|$4|1S|1T|1U|9|$1V|1W]]]]

<blockquote>
  <p>Currently I have a closure in JS that looks like the following:</p>
</blockquote>

<p>All functions create closures, they're only remarkable when advantage is taken of them. ;-)</p>

<pre><code>var addTo = function(num){
</code></pre>

<p>I don't know why function expressions are used when declarations are clearer (to me):</p>

<pre><code>function addTo(num) {
</code></pre>

<p>then there's:</p>

<pre><code>  var add = function(inner){
    return inner + num;
  }
  return add;
}
</code></pre>

<p>Which (sticking with an expression) can be:</p>

<pre><code>  return function (inner) {return inner + num};
}
</code></pre>

<p>Then you call it with <em>new</em>:</p>

<pre><code>var sum = new addTo(1);
</code></pre>

<p>which causes <em>addTo</em> to create a new object that is not used, so you might as well do:</p>

<pre><code>var sum = addTo(1);
</code></pre>

<p>which produces exactly the same result. So:</p>

<p><div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function addTo(num) {
    return function (inner) {return inner + num};
}

var sum = addTo(1);

document.write(sum(3));</code></pre>
</div>
</div>
</p>

<p>However, this is really just a version of <a href="https://en.wikipedia.org/wiki/Currying" rel="nofollow">Currying</a>, so that <em>sum</em> will just add the supplied value to whatever was initially supplied to <em>addTo</em>.</p>

<p>If you want to add all the numbers from 0 to some limit, you just need a loop, no closure required:</p>

<p><div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function sumTo(num) {
  
  var total = 0;
  
  for (var i = 0; i &lt;= num; i++) {
    total += i;
  }
  
  return total;
}

document.write(sumTo(5)); // 15</code></pre>
</div>
</div>
</p>

<p>Note that supplying a negative number will result in an endless loop, you should protect against that (I'll leave it up to you to work out how).</p>


blocks|key|4193383|text|也许你想使用递归而不是循环？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4193384|​|4193385|function+addTo(initial)+{

++function+add(adder)+{
++++if+(initial+<+100)+{
++++++initial%2B=1
++++++return+add(adder%2Binitial)
++++}
++++else+{
++++++return+adder
++++}
++}

++return+add(initial)
}

document.write(addTo(1))|code-block|syntax|javascript|4193386|4193387|只要初始值不超过100，它就会在%2B本身%2B+1之前将所有计算的总和相加。|4193388|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>Maybe you want to use recursive instead of loops?</p>

<p><div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function addTo(initial) {

  function add(adder) {
    if (initial &lt; 100) {
      initial+=1
      return add(adder+initial)
    }
    else {
      return adder
    }
  }

  return add(initial)
}

document.write(addTo(1))</code></pre>
</div>
</div>
</p>

<p>As long as the initial values don't go over 100, it would just add with sum of all calculation before + itself + 1.</p>


blocks|key|412193|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|412194|​|412195|function+sum(x)+{
++var+input+=+x;

++function+add(y)+{
++++return+input+%2B+y;
++}
++return+add;
}
//var+sum1+=+sum(2);
//console.log(sum1(3));+//+5
console.log(sum(2)(3));+//5|code-block|syntax|javascript|412196|412197|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Try</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function sum(x) {
  var input = x;

  function add(y) {
    return input + y;
  }
  return add;
}
//var sum1 = sum(2);
//console.log(sum1(3)); // 5
console.log(sum(2)(3)); //5</code></pre>
</div>
</div>
</p>


<p>Currently I have a closure in JS that looks like the following:</p>
<pre><code>var addTo = function(num){

   var add = function(inner){
      return inner + num;
   };
   return add;
};

var sum = new addTo(1);
</code></pre>
<p>My goal is to use the above closure to compute the sum from <code>1</code> all the way to <code>100</code> (i.e. <code>sum = 1+2+3+...+99+100</code>). Any help? I know a loop is needed, but am unsure of what should go inside the loop and how to use closure to achieve the goal. Thanks guys.</p>


Use closure to sum numbers

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

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目前，我在JS中有一个闭包，如下所示：var addTo = function(num){   var add = function(inner){      return inner + num;   };   return add;};var sum = new addTo(1);我的目标是使用上面的闭包来计算从1...

问使用闭包对数字求和
EN

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