Perl 在字符串上下文中的条件运算符 [英] Perl's conditional operator in string context

问题描述

让我们采用以下简约脚本:

Let's take the following minimalistic script:

#!/usr/bin/perl

#
# conditional_operator.pl
#

use strict;

print ( 1 ? "true" : "false" )." statement\n";

exit;

我希望输出总是真实的陈述".但是当我执行这个片段时，我看到...

I expect the output always to be "true statement". But when I execute this snippet, I see ...

deviolog@home:~/test$ perl conditional_operator.pl
true

" statement\n" 连接似乎被忽略了.

我的 perl 版本是 v5.14.2.我阅读了 perlop 手册关于条件运算符 并认为，字符串连接应该是可能.

My perl version is v5.14.2. I read the perlop manual about the conditional operator and think, a string concatenation should be possible.

有人可以解释这种行为吗?

Can somebody explain this behaviour?

推荐答案

Always include 使用警告； 在每个脚本的顶部.

Always include use warnings; at the top of every script.

要获得您想要的行为，只需添加括号，以便使用整个参数而不是第一部分调用 print:

To get your desired behavior, just add parenthesis so print is called with the entire argument instead of just the first part:

print(( 1 ? "true" : "false" )." statement\n");

如果您打开了warnings，您就会收到此警报:

If you'd had warnings turned on, you would've gotten this alert:

Useless use of concatenation (.) or string in void context

您还可以通过以空白连接开头来避免不受欢迎的行为，或者您可以在括号前放置一个加号:

You can also avoid the undesired behavior by leading with a blank concatenation, or you could put a plus sign before the parenthesis:

print +( 1 ? "true" : "false" )." statement\n";
print ''.( 1 ? "true" : "false" )." statement\n";

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