Chomp将我的字符串更改为1

Question

在我的代码中,我将一个变量光盘分配给disc我的linux系统上的命令结果.这输出字符串RESEARCH

my $disc = `disc`;
print "$disc\n";
$disc = chomp($disc);
print "$disc\n";

但是,当我使用chomp从字符串中去除换行符时,它将字符串更改为1.这是输出

RESEARCH

1

到底是怎么回事？

Answer 1

来自perldoc -f chomp:

chomp VARIABLE
chomp( LIST )
chomp   This safer version of "chop" removes any trailing string that
        corresponds to the current value of $/ (also known as
        $INPUT_RECORD_SEPARATOR in the "English" module). It returns the
        total number of characters removed from all its arguments. 

正确的用法是简单地提供一个将在适当位置改变的变量或列表.你使用的返回值是它"扼杀"其参数列表的次数.例如

chomp $disc;

甚至:

chomp(my $disc = `disc`);

例如,您可以选择整个数组或列表,例如:

my @file = <$fh>;          # read a whole file
my $count = chomp(@file);  # counts how many lines were chomped

当然,使用单个标量参数,chomp返回值只能是1或0.