blocks|key|2386207|text|您不能重新分配const。它被命名为const+(+constant的缩写)是有原因的。这意味着，一旦你定义了它，它就不能重新分配。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2386208|必须重新分配并需要重新分配的变量应该声明为let。|2386209|let+a+=+'a';
a+=+'new+value';+//+this+is+allowed

const+x+=+'x';
x+=+'new+value';+//+this+is+not+allowed|code-block|syntax|javascript|2386210|var还允许您重新分配一个值，但它是一种旧的表示法，除非用于罕见的特定情况，否则通常不需要使用var。|2386211|在问题的代码中|2386212|const+{+count+}+=+this.state;|2386213|计数值从状态中解构，并分配给一个名为count的变量，这意味着它是值this.state.count的副本，而不是this.state.count本身。|2386214|entityMap^0|7|5|I|5|Q|8|0|L|3|0|0|0|3|0|0|0|I|5|Y|G|1L|G|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|V|$]]

<p>You cannot reassign <code>const</code>. There's a reason it's named as <code>const</code>, short form for <code>constant</code>. It means, once you define it, it cannot be reassigned.</p>

<p>Variable that has to and needs to be reassigned should be declared as <code>let</code>.</p>

<pre><code>let a = 'a';
a = 'new value'; // this is allowed

const x = 'x';
x = 'new value'; // this is not allowed
</code></pre>

<p><code>var</code> will also let you reassign a value, but it's an old notation and you'll mostly won't need to use var unless for rare specific cases.</p>

<p>In the code from the question</p>

<pre><code>const { count } = this.state;
</code></pre>

<p>the count value is destructured from state and assigned to a variable named <code>count</code>, meaning it is a copy of the value <code>this.state.count</code> and not <code>this.state.count</code> itself.</p>


blocks|key|2386251|text|代码没有覆盖const的行为。增量正在发生，不是通过更改const+count变量的值，而是通过调用setState来替换(异步)组件的整个状态。然后下一次调用增量或递减时，count将通过this.state的解构赋值接收新值--这不是对变量值的更改，而是每次调用increment时一个新变量(增量函数的局部变量)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2386252|entityMap^0|6|5|S|B|1E|8|2F|5|2N|A|3P|9|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|G|$]]

<p>The code is not overriding the behavior of <code>const</code>. The increment is happening, not by changing the value of the <code>const count</code> variable, but by calling <code>setState</code> which will replace (asynchronously) the entire state of the component. Then the next time increment or decrement is called, <code>count</code> will receive the new value via the deconstructing assignment from <code>this.state</code> -- that isn't a change to the variable's value, it is a new variable (local to the increment function) each time <code>increment</code> is called.</p>


blocks|key|2386276|text|你不能推翻康斯特，另一个意思是你不能重新分配康斯特。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2386277|const+a+=+1;
a+=+2;+//+you+will+get+TypeError.++***+numbers+and+strings+are+also+immutable|code-block|syntax|javascript|2386278|但是，您可以修改常量而不重新分配类型。|2386279|const+b+=+{};
b.a+=+'new+property';+//+{+a:+"new+property"+}

const+c+=+[];
c[0]+=+"new+stack";+//+["new+stack"]|2386280|您可以修改，但不能重新分配const。|2386281|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<p>You can't override const another meaning is you can't reassign const.</p>

<pre><code>const a = 1;
a = 2; // you will get TypeError.  *** numbers and strings are also immutable
</code></pre>

<p>but you can modify the constant without reassigning the type</p>

<pre><code>const b = {};
b.a = 'new property'; // { a: "new property" }

const c = [];
c[0] = "new stack"; // ["new stack"]
</code></pre>

<p>You can modify but you can't reassign const.</p>


blocks|key|585645|text|当您执行const+{+count+}+=+this.state;时，实际上是在创建副本+of+this.state.count，而不是直接操作this.state.count。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|585646|除非你需要，否则const是不会意外地操纵它的。|585647|下面的代码片段演示了更新count的“副本”不会更新state.count|585648|(单击Run+Code+Snippet查看结果)。|585649|​|585650|const+state+=+{+count:+0+};
let+{count}+=+state;
count+%2B=+999;

console.log(`state.count`,+state.count);
console.log(`count`,+count);|code-block|syntax|javascript|585651|585652|而这将失败(不幸)。|585653|585654|const+state+=+{+count:+0+};
const+{count}+=+state;
count+%2B=+999;

console.log(`state.count`,+state.count);
console.log(`count`,+count);|585655|585656|entityMap^0|4|T|1C|G|20|G|16|2|0|8|5|0|C|5|Q|B|0|3|G|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|E|@]|F|$]]|$1|I|3|J|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|E|@]|F|$]]|$1|K|3|L|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|E|@]|F|$]]|$1|M|3|N|5|6|7|1M|8|@]|E|@]|F|$]]|$1|O|3|P|5|Q|7|1N|8|@]|E|@]|F|$R|S]]|$1|T|3|N|5|6|7|1O|8|@]|E|@]|F|$]]|$1|U|3|V|5|6|7|1P|8|@]|E|@]|F|$]]|$1|W|3|N|5|6|7|1Q|8|@]|E|@]|F|$]]|$1|X|3|Y|5|Q|7|1R|8|@]|E|@]|F|$R|S]]|$1|Z|3|N|5|6|7|1S|8|@]|E|@]|F|$]]|$1|10|3|-4|5|6|7|1T|8|@]|E|@]|F|$]]]|11|$]]



<p>When you do <code>const { count } = this.state;</code>, you are actually creating a <strong>copy</strong> of <code>this.state.count</code>, not manipulating <code>this.state.count</code> directly.</p>

<p>And <code>const</code> is there as not to accidentally manipulate it unless you need to.</p>

<p>Following code snippet demos that updating "copy" of <code>count</code> doesn't update the <code>state.count</code></p>

<p>(click on <code>Run Code Snippet</code> to see the result).</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>const state = { count: 0 };
let {count} = state;
count += 999;

console.log(`state.count`, state.count);
console.log(`count`, count);</code></pre>
</div>
</div>
</p>

<p>And this will fail (miserably).</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>const state = { count: 0 };
const {count} = state;
count += 999;

console.log(`state.count`, state.count);
console.log(`count`, count);</code></pre>
</div>
</div>
</p>




<p>So I was reading an <a href="https://css-tricks.com/an-overview-of-render-props-in-react/" rel="nofollow noreferrer">article</a> and stumbled and saw this <a href="https://codepen.io/kinsomicrote/pen/pxyvwX" rel="nofollow noreferrer">CodePen</a> and saw that in <code>increment</code> method it is setting <code>count</code> as <code>const</code> and I'm quite unable to process this, that How can you override <code>const</code>? Can anyone explain what I'm missing?</p>

<pre class="lang-js prettyprint-override"><code>  state = {
    count: 0
  };

  // Increase count
  increment = () =&gt; {
    const { count } = this.state;
    return this.setState({ count: count + 1 });
  };
</code></pre>

<p>Why declare <code>const</code> when you can use <code>let</code> or <code>var</code> because usage of <code>const</code> is quite confusing knowing the deconstructed variable is going to be manipulated?</p>


Overriding the value of const

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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所以我读了一个，无意中看到了这个，并看到在increment方法中，它将count设置为const，而我完全无法处理这个问题，您怎么能重写const呢？有人能解释我错过了什么吗？  state = {    count: 0  };  // Increase count  increment = () => {    ...

问重写const的值
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