为什么在Perl的printf输出后得到一个尾随'1'？

Question

当我实现下面的代码时,我得到了正确的日期:

10/05/2008
10/05/2009

当我使用printf而不是sprintf,我得到以下内容:

10/05/200910/05/20081
1

关于为何printf打印拖尾的任何想法1？

---

#!/usr/bin/perl

use strict; use warnings;

my ($from_date, $to_date) = to_from_dates();

print "$from_date\n";
print "$to_date\n";

sub to_from_dates {
    my ($day, $month, $year) = (localtime)[3,4,5];
    my $to_date   = sprintf "%02d/%02d/%04d", $month+1, $day, $year+1900;
    my $from_date = sprintf "%02d/%02d/%04d", $month+1, $day, $year+1899;
    return ($from_date, $to_date);
}

Answer 1

仅返回可打印的值.打印值并返回以告诉您输出成功.sprintf printf 1

如果您只是s从调用开头删除-es,那么您显示的输出正是我期望的输出.

sub to_from_dates {
    my ($day, $month, $year) = (localtime)[3,4,5];
    my $to_date=printf("%02d/%02d/%04d", $month+1, $day, $year+1900);
    # printed: 10/05/2009 (no carriage return)
    # $to_date = '1';
    my $from_date=printf("%02d/%02d/%04d", $month+1, $day, $year+1899);
    # printed: 10/05/2008 (no carriage return)
    # $from_date = '1';
    return ($from_date,$to_date);
}

($from_date,$to_date)=to_from_dates(); # returns ( 1, 1 )
# output: 10/05/200910/05/2008
print $from_date."\n";                 # prints "1\n"; <- first line feed
# output: 10/05/200910/05/20081\n
print $to_date."\n";                   # prints "1\n"; <- second line feed.
# output: 10/05/200910/05/20081\n1\n

Answer 2

如果你使用printf,Perl打印该行并返回操作的结果,即1,然后你打印这个结果,这是来自的地方.你可以做任何一个printf或print sprintf