爲什麼`a :: - > func;`有效？

Question

package a; 
sub func { 
print 1; 
} 
package main; 
a::->func; 

海事組織已經足夠擁有a::func,a->func。

a::->func;看起來很奇怪，爲什麼Perl支持這種奇怪的外觀語法？

Answer 1

要引用chromatic最近關於Modern Perl blog主題的最新博客帖子：「爲了避免裸語解析含糊不清。」

爲了說明爲什麼這樣的語法是很有用的，這裏是從您的樣品演變的例子：

package a; 
our $fh; 
use IO::File; 
sub s { 
    return $fh = IO::File->new(); 
} 

package a::s; 
sub binmode { 
    print "BINMODE\n"; 
} 

package main; 
a::s->binmode; # does that mean a::s()->binmode ? 
       # calling sub "s()" from package a; 
       # and then executing sub "open" of the returned object? 
       # In other words, equivalent to $obj = a::s(); $obj->binmode(); 
       # Meaning, set the binmode on a newly created IO::File object? 

a::s->binmode; # OR, does that mean "a::s"->binmode() ? 
       # calling sub "binmode()" from package a::s; 
       # Meaning, print "BINMODE" 

a::s::->binmode; # Not ambiguous - we KNOW it's the latter option - print "BINMODE" 

Answer 2

a::是一個字符串字面量產生串a。都是一樣的：

a->func() # Only if a doesn't exist as a function. 
"a"->func() 
'a'->func() 
a::->func() 
v97->func() 
chr(97)->func() 

等

>perl -E"say('a', a, a::, v97, chr(97))" 
aaaaa