print(2& 2)>>怎么了? 1个?

问题描述：

我只是想知道这段代码会发生什么. 为什么只在直接打印时结果不正确，为什么换行符会被忽略?

I am just wondering what happens with that piece of code. Why the result is incorrect only when printed directly, why is the newline ignored?

user@host_09:22 AM: perl
print 2 >> 1, "\n";
print 2 & 2, "\n";
print (2 & 2) >> 1, "\n";
1
2
2user@host_09:22 AM: perl
$a = (2 & 2) >> 1;
print "$a\n";
1

答

在打印带有警告的内容时，它会变得很清晰(er)

When you print it with warnings it becomes clear(er)

perl -we'print (2 & 2), "\n"'

说


print (...) interpreted as function at -e line 1.
Useless use of a constant ("\n") in void context at -e line 1.

它将print (2&2)作为对print ^†的函数调用，并正确打印2(没有换行符！)，然后继续评估，它也警告我们.

It works out print (2&2) as a function call to print^† and duly prints 2 (no newline!), and then it keeps evaluating the comma operator, with "\n" in void context next, which it also warns us about.

同时还有>> 1，print (2&2)的返回1(成功)已移至0，它消失在空白中，我们得到另一个"在无效上下文中对...的无用使用."

With >> 1 also there, the return 1 of print (2&2) (for success) is bit shifted to 0, which disappears into the void, and we get another "Useless use of ... in void context."

一种解决方法是添加+，因为其后必须是表达式

One fix is to add a + since what follows it must be an expression

perl -we'print +(2 & 2) >> 1, "\n"'

或者，适当地调用print，并在整个内容中加上括号

Or, make a proper call to print, with parenthesis around the whole thing

perl -we'print((2 & 2) >> 1, "\n")'

都用1打印一行.

在打印中已提及，并在术语和列表运算符和 perlop .有关另一个相关示例，请参见这篇文章.

This is mentioned in print, and more fully documented in Terms and List operators and in Symbolic Unary operators, both in perlop. For another, related, example see this post.

^† 还警告它，因为这很可能是错误的-在括号前留一个空格；没有空间，没有警告.

^† It also warns about it as it is likely an error -- with a space before parens; no space, no warning.

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