blocks|key|1378548|text|&&操作符“短路”--也就是说，如果左边的条件是假的，它就不会费心计算右边的条件。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1378549|类似地，如果左条件为真，则%7C%7C操作符短路。|1378550|编辑:不过，在您对性能进行基准测试并确定这是一个问题之前，您不应该担心性能。过早的微优化是可维护性的祸根。|1378551|entityMap^0|0|2|0|D|2|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

The <code>&amp;&amp;</code> operator "short-circuits" - that is, if the left condition is false, it doesn't bother evaluating the right one.

Similarly, the <code>||</code> operator short-circuits if the left condition is true.

EDIT: Though, you shouldn't worry about performance until you've benchmarked and determined that it's a problem. Premature micro-optimization is the bane of maintainability.

blocks|key|1163438|text|它在看到b不等于1后退出。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1163439|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It exits after seeing that b does not equal one.

blocks|key|1378592|text|It短路-在您的示例中只比较a和b。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1378593|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It short circuits - only a and b will be compared in your example.

blocks|key|1163496|text|它只会测试所有的条件，如果第一个条件是真的，那就自己测试吧：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1163497|javascript:+alert+(false+&&+alert("A")+&&+false);|code-block|syntax|javascript|1163498|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It will only test all the conditions if the first ones are true, test it for yourself:

<pre><code>javascript: alert (false &amp;&amp; alert("A") &amp;&amp; false);
</code></pre>

blocks|key|1163509|text|这就是为什么你可以在javascript代码中这样做的原因|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1163510|var+x+=+x+%7C%7C+2;|code-block|syntax|javascript|1163511|这将意味着如果x是未定义的或否则为'false‘，则缺省值为2。|1163512|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

That's why you can do in javascript code like 

<pre><code>var x = x || 2;
</code></pre>

Which would mean that if x is undefined or otherwise 'false' then the default value is 2.

blocks|key|1163552|text|从性能的角度来看，这不是微优化。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1163553|如果我们有3个布尔变量，a，b，c，那就是微优化。|1163554|如果我们调用3个返回布尔变量的函数，每个函数可能会花费很长时间，而且不仅要知道这种短路，而且要知道这种短路的顺序。例如：|1163555|if+(takesSeconds()+&&+takesMinutes())|code-block|syntax|javascript|1163556|远好于|1163557|if+(takesMinutes()+&&+takesSeconds())|1163558|如果两者都同样有可能返回false。|1163559|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

From a performance standpoint, this is not a micro-optimization.

If we have 3 Boolean variables, a, b, c that is a micro-optimization.

If we call 3 functions that return Boolean variables, each function may take a long time, and not only is it important to know this short circuits, but in what order. For example:

<pre><code>if (takesSeconds() &amp;&amp; takesMinutes())
</code></pre>

is much better than

<pre><code>if (takesMinutes() &amp;&amp; takesSeconds())
</code></pre>

if both are equally likely to return false.

blocks|key|1378711|text|左边有1个或多个参数时停止求值的另一个原因。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1378712|if+(response.authResponse+&&+(response.authResponse.accessToken+!=+user.accessToken)){+...+}|1378713|第二个评估依赖于第一个条件为真，如果response.authResponse为null或未定义等，则不会抛出编译错误，因为第一个条件失败。|1378714|其他语言在早期就有这个问题，我认为这是现在构建编译器的标准方法。|1378715|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Another reason why stopping evaluation with 1 or more parameters to the left.

if (response.authResponse &amp;&amp; (response.authResponse.accessToken != user.accessToken)){
...
}

the second evaluation relies on the first being true and won't throw a compile error if response.authResponse is null or undefined etc because the first condition failed.

Other languages had this problem in the early days and I think it's a standard approach in building compilers now.

blocks|key|1163658|text|对于任何对这个问题感到困惑的人，因为他们没有看到在使用%7C%7C和?操作符时的短路行为，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1163659|x+=+1+%7C%7C+true+?+2+:+3++//+value+of+x+will+be+2,+rather+than+1+as+expected|1163660|似乎短路规则不起作用了。为什么它要评估%7C%7C的第二项(真的?2:+3)，而第一项是真的？它被证明是一个操作顺序问题，因为上面的等价物是|1163661|x+=+(1+%7C%7C+true)+?+2+:+3|1163662|首先评估%7C%7C，然后评估?。你可能想要的是：|1163663|x+=+1+%7C%7C+(true+?+2+:+3)|1163664|entityMap^0|R|2|U|1|0|0|21|0|J|2|0|0|N|0|4|2|B|1|0|0|N|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|P|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|Q|$]]

For anyone on this question confused because they're not seeing the short-circuit behaviour when using an <code>||</code> in conjunction with an <code>?</code> operator like so:

<code>x = 1 || true ? 2 : 3 // value of x will be 2, rather than 1 as expected</code>

it seems like the short circuit rule isn't working. Why is it evaluating the second term of the <code>||</code> (true ? 2 : 3) when the first is true? It turns out to be an order of operations problem because the above is the equivalent of 

<code>x = (1 || true) ? 2 : 3</code>

with the <code>||</code> evaluated first and the <code>?</code> evaluated second. What you likely want is:

<code>x = 1 || (true ? 2 : 3)</code>

blocks|key|1378772|text|对于这种情况：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1378773|a+=+1
b+=+2
c+=+1

if+(a==1+&&+b==1+&&+c==1)|code-block|syntax|javascript|1378774|您可以使用：|1378775|if+([a,+b,+c].every(x+=>+x+==+1))
++++//+do+something+if+a,+b+and+c+are+equal+to+1.|1378776|如果想知道是否至少有一个等于1，请使用some()方法而不是every()：|1378777|if+([a,+b,+c].some(x+=>+x+==+1))
++++//+do+something+if+a,+b+or+c+is+equal+to+1.|1378778|every()和some()是数组方法。|1378779|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

For this case:
<pre><code>a = 1
b = 2
c = 1

if (a==1 &amp;&amp; b==1 &amp;&amp; c==1)
</code></pre>
You can use:
<pre><code>if ([a, b, c].every(x =&gt; x == 1))
 // do something if a, b and c are equal to 1.
</code></pre>
If you want to know if at least one is equal to 1, use the some() method instead of every():
<pre><code>if ([a, b, c].some(x =&gt; x == 1))
 // do something if a, b or c is equal to 1.
</code></pre>
every() and some() are array methods.

In javascript, when using an if statement with multiple conditions to test for, does javascript test them all regardless, or will it bail before testing them all if it's already false?

For example:

<pre><code> a = 1
 b = 2
 c = 1

 if (a==1 &amp;&amp; b==1 &amp;&amp; c==1)
</code></pre>

Will javascript test for all 3 of those conditions or, after seeing that b does not equal 1, and is therefore false, will it exit the statement?

I ask from a performance standpoint. If, for instance, I'm testing 3 complex jQuery selectors I'd rather not have jQuery traverse the DOM 3 times if it's obvious via the first one that it's going to return FALSE. (In which case it'd make more sense to nest 3 if statements). 

ADDENDUM: More of a curiosity, what is the proper term for this? I notice that many of you use the term 'short circuit'. Also, do some languages do this and others dont?

Does a javascript if statement with multiple conditions test all of them?

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在javascript中，当使用带有多个条件的if语句进行测试时，javascript是否会不管怎样地全部测试它们，或者如果已经为false，它会在测试它们之前停止吗？例如： a = 1 b = 2 c = 1 if (a==1 && b==1 && c==1)javascript是否会测试所有3个条件，或者在看到b不...

问有多个条件的javascript if语句会测试所有条件吗？
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