为什么！1给我什么在Perl？ [英] Why does !1 give me nothing in Perl?

问题描述

这是奇怪的。
以下内容：

This is strange. The following:

$sum = !0;
print $sum;

打印出1如你所愿。但这种

prints out 1 as you would expect. But this

$sum = !1;
print $sum;

打印出什么。为什么呢？

prints out nothing. Why?

推荐答案

注意：你写的不是做你觉得它在做什么。请记住，Perl有没有真正的布尔数据类型。它有标量，散列，列表和引用。它处理真/假值的方式，那么，是上下文。一切都计算为true在Perl除了不确定变量，则空单，空字符串，而数字0。

Be careful: what you've written isn't doing what you think it's doing. Remember, perl has no real boolean datatype. It's got scalars, hashes, lists, and references. The way it handles true/false values, then, is contextual. Everything evaluates to "true" in perl except for undefined variables, the empty list, the empty string, and the number 0.

你的code是干什么的，然后，走的是一条价值的计算结果为假的倒数，它可以是任何东西这是不是在上面的列表中。按照惯例，并为简单起见，PERL返回1（虽然你不应该依赖于;它很可能返回一个包含一系列的随机号码清单，因为这将评估为真为好）

What your code is doing, then, is taking the inverse of a value that evaluates to "false", which can be anything which is not in the list above. By convention and for simplicity's sake, perl returns 1 (though you should not rely on that; it could very well return a list containing a series of random numbers, because that will evaluate to "true" as well.)

当你问一个值的计算结果为逆类似的事情发生真。什么是真正被打印出来，不是什么，它是空字符串（''），其中，正如我所说，在评估前布尔pressions为假。您可以检查这一点：

A similar thing happens when you ask for the inverse of a value that evaluates to "true." What's actually being printed out is not "nothing," it's the empty string (''), which, as I mentioned, evaluates to "false" in boolean expressions. You can check this:

print "This evaluates to false\n" if( (!1) eq '');

如果你问为什么perl的吐出来的是空字符串，而不是其他的假值中的一个，那么，这可能是因为是用perl作出处理字符串，这是一个完全合理的字符串交还。

If you're asking for why perl spits out the empty string instead of one of the other "false" values, well, it's probably because perl is made to handle strings and that's a perfectly reasonable string to hand back.

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