为什么在这个Perl示例中定义的sdf返回true？

perl -Mstrict -e "print qq{definedn} if defined sdf"

Bareword "sdf" not allowed while "strict subs" in use at -e line 1.
Execution of -e aborted due to compilation errors.

为了更多的乐趣,试试吧

perl -Mstrict -e "print sdf => qq{n}"

见Strictly speaking about use strict：

The subs aspect of use strict disables the interpretation of “bare words” as text strings. By default,a Perl identifier (a sequence of letters,digits,and underscores,not starting with a digit unless it is completely numeric) that is not otherwise a built-in keyword or previously seen subroutine definition is treated as a quoted text string:

06003

However,this is considered to be a dangerous practice,because obscure bugs may result:

06004

Can you spot the bug? Yes,the 10th entry is not the string ‘oct’,but rather an invocation of the built-in oct() function,returning the numeric equivalent of the default $_ treated as an octal number.

更正:(谢谢@ysth)

E:Home> perl -we "print sdf"
Unquoted string "sdf" may clash with future reserved word at -e line 1.
Name "main::sdf" used only once: possible typo at -e line 1.
print() on unopened filehandle sdf at -e line 1.

如果提供裸字以在间接对象槽中打印,则将其作为要打印的文件句柄.由于没有提供其他参数,因此打印默认为将$_打印到filehandle sdf.由于sdf尚未打开,因此失败.如果在没有警告的情况下运行此操作,则看不到任何输出.另请注意：

E:Home> perl -MO=Deparse -e "print sdf"
print sdf $_;

作为对此观察的确认.另请注意：

E:Home> perl -e "print asdfg,sadjkfsh"
No comma allowed after filehandle at -e line 1.

E:Home> perl -e "print asdfg => sadjkfsh"
asdfgsadjkfsh

后者打印两个字符串,因为=>如果它们仅由’word’字符组成,则自动引用LHS上的字符串,删除第一个参数的文件句柄解释.

所有这些例子表明,使用裸字会带来许多惊喜.你应该使用严格来避免这种情况.