print(2& 2)>>怎么了? 1个? [英] What is wrong with print (2 & 2) >> 1?

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

推荐答案

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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