为什么我不能将这个perl sub分配给变量？

Question

我试图理解在HOP的第158页上的imap例程的复杂执行路径.

这段代码有效

# code from rng-iterator.pl
sub make_rand {
my $seed = shift || (time & 0x7fff);
print "\nin make_rand, at6: seed=$seed";
return sub 
    {   $seed = (29*$seed+11111) & 0x7fff;
print "\nin make_rand sub, at9: seed=$seed";
        return $seed; 
    }
}

# code adapted from HOP p.158, to make an iterator version of map
sub imap {
    my ($transform, $it) = @_;
print "\nin imap, at17";
    return sub 
    {   my $next = $it->();
print "\nin imap sub, at20, next=$next";
        return unless defined $next;
        $newVal = $transform->($next);
print "\nin imap sub, at23, newVal=$newVal";
        return $newVal;
    }
}

# to return random number 0 .. 1
$rng = imap(sub {$_[0] / 37268}, make_rand(1)); # set seed 
print "\nin main at30, rng=$rng";
while (<>) {    
    my $random = $rng->();  
    print "\nin main, at 32: random=$random";
}

将对子(imap)的引用返回到字符串$ rng并使用它指向imap的sub似乎没有问题.

我想将sub分配给imap的字符串INSIDE,并返回字符串,如下所示:

    $imapSub =  sub 
        {   my $next = $it->();
                print "\nin imap sub, at20, next=$next";
            return unless defined $next;
            $newVal = $transform->($next);
                print "\nin imap sub, at23, newVal=$newVal";
            return $newVal;
        }
    return $imapSub;

当我尝试返回或打印$ imapSub时,Perl报告了语法错误,甚至将其用作ref()的参数.当我将sub分配给变量时,它没有抱怨.

即使我明确地将子例程的引用转换为$ \&sub,它也会这样做.

当我尝试使用引用时,为什么会出现语法错误？

Answer 1

你在语句中的右大括号之后缺少一个分号$imapSub = sub { ... },所以你在之后放的是意外的并导致语法错误.