blocks|key|4407261|text|参数k的作用域是函数，去掉它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4407262|​|4407263|var+k+=+10;
function+getNumber(){
++if+((1<2)&&(k===10))+{
++++k+=+11;
++++return+4;
++}
++return+2;+
}
console.log(getNumber())
console.log(getNumber())|code-block|syntax|javascript|4407264|4407265|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>Parameter <em>k</em> is scoped to the function, remove that:</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>var k = 10;
function getNumber(){
  if ((1&lt;2)&amp;&amp;(k===10)) {
    k = 11;
    return 4;
  }
  return 2; 
}
console.log(getNumber())
console.log(getNumber())</code></pre>
</div>
</div>
</p>


blocks|key|5876006|text|未将k变量提供给getNumber()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|5876007|在函数作用域中，参数未定义|5876008|entityMap^0|2|1|8|B|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@$9|L|A|M|B|C]|$9|N|A|O|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|P|8|@]|E|@]|F|$]]|$1|I|3|-4|5|6|7|Q|8|@]|E|@]|F|$]]]|J|$]]

<p><strong>k</strong> variable is not given to <code>getNumber()</code>.</p>
<p>In function scope, parameter is undefined</p>


blocks|key|5876060|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5876061|为什么总是给2，忽略if+()？|blockquote|5876062|5876063|5876064|它不会忽略if()，它只是条件为false+-可以使用console.log进行检查|offset|length|style|CODE|5876065|​|5876066|var+k+=+10;
function+getNumber(k){
++console.log(k,+k+===+10);
++if+((1<2)&&(k===10))+{
++++++k+=+11;
++++++return+4;
++++++}
++return+2;+
}
console.log(getNumber())
console.log(getNumber())|code-block|syntax|javascript|5876067|5876068|正如您所看到的，函数内部的k是undefined，并且由于此变量是函数的局部变量，因此外部作用域中的k不是您希望通过将其设置为11来检查的值。|5876069|entityMap^0|0|0|0|0|5|4|G|5|R|B|0|0|0|0|D|1|F|9|1E|1|1R|2|0^^$0|@$1|2|3|-4|4|5|6|X|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|Y|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|10|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|11|7|@$H|12|I|13|J|K]|$H|14|I|15|J|K]|$H|16|I|17|J|K]]|8|@]|9|$]]|$1|L|3|M|4|5|6|18|7|@]|8|@]|9|$]]|$1|N|3|O|4|P|6|19|7|@]|8|@]|9|$Q|R]]|$1|S|3|M|4|5|6|1A|7|@]|8|@]|9|$]]|$1|T|3|U|4|5|6|1B|7|@$H|1C|I|1D|J|K]|$H|1E|I|1F|J|K]|$H|1G|I|1H|J|K]|$H|1I|I|1J|J|K]]|8|@]|9|$]]|$1|V|3|-4|4|5|6|1K|7|@]|8|@]|9|$]]]|W|$]]

<blockquote>
<p>Why does it always give 2, ignorning if ()?</p>
</blockquote>
<p>It doesn't ignore the <code>if()</code> its just the the condition is <code>false</code> - which can be checked using a <code>console.log</code></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>var k = 10;
function getNumber(k){
  console.log(k, k === 10);
  if ((1&lt;2)&amp;&amp;(k===10)) {
      k = 11;
      return 4;
      }
  return 2; 
}
console.log(getNumber())
console.log(getNumber())</code></pre>
</div>
</div>
</p>
<p>As you can see, <code>k</code> inside the function is <code>undefined</code> and as this variable is local to the function the <code>k</code> in outer scope is not the value you're working with that you expected to check by setting it to <code>11</code>.</p>


blocks|key|1647414|text|请参阅代码中的注释|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1647415|​|1647416|//+This+is+a+global+variable.+Similar+to
//+window.k+=+10;
var+k+=+10;

//+By+setting+k+as+parameter+to+the+function+you+define+it+as+
//+a+local+variable+in+the+function+scope
//+It's+somewhat+similar+to+this+line
//+var+k+=+arguments[0];
//+Since+you+call+getNumber()+without+arguments,+k+is+undefined+in+this+scope
function+getNumber(k)+{
++//+the+test+for+the+k+value+is+against+the+local+value,+
++//+which+is+undefined+and+not+10
++if+((1+<+2)+&&+(k+===+10))+{
++++//+This+code+is+never+executed
++++k+=+11;
++++return+4;
++}
++//+Always+returns+2
++return+2;
}
console.log(getNumber())
console.log(getNumber())|code-block|syntax|javascript|1647417|1647418|要使您的代码正常工作，请执行以下操作|1647419|1647420|var+k+=+10;

//+function+parameter+with+default+value+if+argument+is+not+given
function+getNumber(k+=+window.k)+{
++if+((1+<+2)+&&+(k+===+10))+{
++++//+set+global+k+instead+of+local
++++window.k+=+11;
++++return+4;
++}
++return+2;
}
console.log(getNumber())
console.log(getNumber())|1647421|1647422|顺便说一句，这不是我推荐的代码策略。应避免使用全局变量，尤其是在函数内部使用或更改全局变量。由于我将此视为演示代码，而不是真实的代码，因此我的代码更正仅用于演示目的，不应在生产代码中使用。|1647423|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|X|8|@]|9|@]|A|$]]|$1|L|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|O|3|C|5|6|7|10|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|11|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|S|$]]

<p>See the comments in the code</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>// This is a global variable. Similar to
// window.k = 10;
var k = 10;

// By setting k as parameter to the function you define it as 
// a local variable in the function scope
// It's somewhat similar to this line
// var k = arguments[0];
// Since you call getNumber() without arguments, k is undefined in this scope
function getNumber(k) {
  // the test for the k value is against the local value, 
  // which is undefined and not 10
  if ((1 &lt; 2) &amp;&amp; (k === 10)) {
    // This code is never executed
    k = 11;
    return 4;
  }
  // Always returns 2
  return 2;
}
console.log(getNumber())
console.log(getNumber())</code></pre>
</div>
</div>
</p>
<p>To make your code work, do this</p>
<p><div class="snippet" data-lang="js" data-hide="true" data-console="true" data-babel="false">
<div class="snippet-code snippet-currently-hidden">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var k = 10;

// function parameter with default value if argument is not given
function getNumber(k = window.k) {
  if ((1 &lt; 2) &amp;&amp; (k === 10)) {
    // set global k instead of local
    window.k = 11;
    return 4;
  }
  return 2;
}
console.log(getNumber())
console.log(getNumber())</code></pre>
</div>
</div>
</p>
<p>BTW This is not a code strategy that I'd recommend. Global variables should be avoided, especially the use or change of global variables inside functions. Since I regard this as a demonstration code and not real life code, my code correction is for this demonstration purpose alone and should not be used in production code.</p>


<p>I have a simple function:</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>var k = 10;
function getNumber(k){
  if ((1&lt;2)&amp;&amp;(k===10)) {
      k = 11;
      return 4;
      }
  return 2; 
}
console.log(getNumber())
console.log(getNumber())</code></pre>
</div>
</div>
</p>
<p>Why does it always give 2, ignorning if ()?</p>


Weird function behaviour

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。

页面排版不美观，阅读体验差。

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我有一个简单的函数：​var k = 10;function getNumber(k){  if ((1<2)&&(k===10)) {      k = 11;      return 4;      }  return 2; }console.log(getNumber())console.log(getNumber())​为什么它总是给出2，忽略if ()？

问奇怪的函数行为
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