blocks|key|2876230|text|Safari更新:+2020年1月1日，是唯一支持尾部调用优化的浏览器。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2876231|铬团队明确表示，尾部调用优化并未在积极开发中，可以通过here进行跟踪。|2876232|火狐的实现可以在here上找到|2876233|原始帖子|2876234|是的，ES2015在严格模式下提供了尾部调用优化。Axel+Rauschmayer博士在下面的链接中很好地展示了这一点，所以我不在这里重复他的话。|2876235|注意:+ES+5不会优化尾部调用。|2876236|http://www.2ality.com/2015/06/tail-call-optimization.html|2876237|entityMap|0|LINK|mutability|MUTABLE|url|https://bugs.chromium.org/p/v8/issues/detail?id=4698|1|https://bugzilla.mozilla.org/show_bug.cgi?id=723959|2^0|K|G|0|R|4|0|0|8|4|1|0|0|4|0|0|0|0|1L|2|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|H|3|I|5|6|7|19|8|@]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|J|3|K|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1G|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1H|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1I|8|@]|D|@$9|1J|A|1K|1|1L]]|E|$]]|$1|R|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]|Z|$5|U|V|W|E|$X|10]]|11|$5|U|V|W|E|$X|Q]]]]

Update: As of January 1, 2020 Safari is the only browser that supports tail call optimization.

The chromium team explicitly states that Tail Call Optimization is not under active development and can be tracked <a href="https://bugs.chromium.org/p/v8/issues/detail?id=4698" rel="noreferrer">here</a>.

The implementation for Firefox can be tracked <a href="https://bugzilla.mozilla.org/show_bug.cgi?id=723959" rel="noreferrer">here</a>

Original Post 

Yes, ES2015 offers tail call optimization in strict mode. Dr. Axel Rauschmayer lays it out beautifully at the link below so I shall not repeat his words here.

Note: ES 5 does not optimize tail calls.

<a href="http://www.2ality.com/2015/06/tail-call-optimization.html" rel="noreferrer">http://www.2ality.com/2015/06/tail-call-optimization.html</a>

blocks|key|117580|text|理论上是这样的。正如另一个答案所说的那样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|117581|但实际上，截至2017年7月，No.只有Safari支持它。|117582|Javascript+ES6+(ES2015)兼容性：https://kangax.github.io/compat-table/es6/|offset|length|117583|entityMap|0|LINK|mutability|MUTABLE|url|https://kangax.github.io/compat-table/es6/^0|0|0|R|16|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

In theory yes. As the other answer states.

In practice though, as of July 2017, No. Only Safari supports it.

Javascript ES6 (ES2015) compatability:
<a href="https://kangax.github.io/compat-table/es6/" rel="noreferrer">https://kangax.github.io/compat-table/es6/</a>

blocks|key|2665379|text|正如其他答案所说，在实践中并非如此。但是，您可以定义一个实用程序来提供帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2665380|class+Tco+{
++constructor(func)+{
++++this.func+=+func;
++}
++execute()+{
++++let+value+=+this;
++++while+(value+instanceof+Tco)
++++++value+=+value.func();
++++return+value;
++}
}

const+tco+=+(f)+=>+new+Tco(f);|code-block|syntax|javascript|2665381|function+factorial+(n)+{
++const+fact+=+(n,+acc)+=>+tco(()+=>+{
++++if+(n+<+2)+{
++++++return+acc;
++++}+else+{
++++++return+fact(n-1,+n+*+acc);
++++}
++});

++return+fact(n,+1).execute();
}

console.log(factorial(2000000));+//+Infinity|2665382|正如您所看到的，这允许您编写语法上只有很小差异的尾递归函数，而不会遇到最大的调用堆栈错误。|2665383|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|O|8|@]|9|@]|A|$E|F]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

As the other answers have said, not in practice. However, you can define a utility to help out.

<pre class="lang-js prettyprint-override"><code>class Tco {
 constructor(func) {
 this.func = func;
 }
 execute() {
 let value = this;
 while (value instanceof Tco)
 value = value.func();
 return value;
 }
}

const tco = (f) =&gt; new Tco(f);
</code></pre>

<pre class="lang-js prettyprint-override"><code>function factorial (n) {
 const fact = (n, acc) =&gt; tco(() =&gt; {
 if (n &lt; 2) {
 return acc;
 } else {
 return fact(n-1, n * acc);
 }
 });

 return fact(n, 1).execute();
}

console.log(factorial(2000000)); // Infinity
</code></pre>

As you can see, this allows you to write tail recursive functions with only a small difference in syntax, without running into a max call stack error.

blocks|key|2876291|text|Safari是唯一支持尾部调用优化的浏览器。ES2015在严格模式下提供尾部调用优化|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2876292|++++function+factorial(n,+returnVal=+1)+{
++++++'use+strict';
+++++++if+(n+<=+1)+return+returnVal;
++++++++return+factorial(n+-+1,+n+*+returnVal);
}


factorial(555)|code-block|syntax|javascript|2876293|遵循LINK|offset|length|2876294|entityMap|0|LINK|mutability|MUTABLE|url|https://2ality.com/2015/06/tail-call-optimization.html^0|0|0|2|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Safari is the only browser that supports tail call optimization. ES2015 offers tail call optimization in strict mode
<pre><code> function factorial(n, returnVal= 1) {
 'use strict';
 if (n &lt;= 1) return returnVal;
 return factorial(n - 1, n * returnVal);
}


factorial(555)
</code></pre>
Follow the <a href="https://2ality.com/2015/06/tail-call-optimization.html" rel="nofollow noreferrer">LINK</a>

I have been trying to understand <code>Tail call optimization</code> in context of JavaScript and have written the below recursive and tail-recursive methods for <code>factorial()</code>.

Recursive:

<pre><code>function factorial (n) {
 if (n &lt; 2) {
 return 1;
 } else {
 return n * factorial(n-1);
 }
}
</code></pre>

Tail-recursive:

<pre><code>function factorial (n) {
 function fact(n, acc) {
 if (n &lt; 2) {
 return acc;
 } else {
 return fact(n-1, n * acc);
 }
 }

 return fact(n, 1)
}
</code></pre>

But I am not sure if the <code>tail-recursive</code> version of the function will be optimised by JavaScript compiler as it is done in other languages like Scala etc. Can someone help me out on this one?

Are functions in JavaScript tail-call optimized?

编译器

我一直在尝试在JavaScript上下文中理解Tail call optimization，并为factorial()编写了下面的递归和尾递归方法。递归：function factorial (n) { if (n < 2) { return 1; } else { return n * factori...

问JavaScript尾部调用中的函数是优化的吗？
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