perl - Perl 触发器运算符和行号

Question

我在看另一个问题时注意到了这一点......

如果我有这样的脚本：

while (<>) {
  print if 5 .. undef;
}

它跳过第 1..4 行，然后打印文件的其余部分。但是，如果我尝试这个：

my $start_line = 5;
while (<>) {
  print if $start_line .. undef;
}

它从第 1 行打印。谁能解释为什么？

实际上我什至不确定为什么第一个有效。

嗯，进一步研究我发现这有效：

my $start = 5;
while (<>) {
  print if $. ==  $start .. undef;
}

所以第一个版本神奇地使用$.了行号。但我不知道为什么它会因变量而失败。

Answer 1

The use of a bare number in the flip-flop is treated as a test against the line count variable, $.. From perldoc perlop:

If either operand of scalar ".." is a constant expression, that operand is considered true if it is equal (==) to the current input line number (the $. variable).

So

print if 5 .. undef;

is "shorthand" for:

print if $. == 5 .. undef;

---

The same is not true for a scalar variable as it is not a constant expression. This is why it is not tested against $..