为什么在此Perl代码中不打印换行符? [英] Why aren't newlines being printed in this Perl code?

问题描述

我有一些简单的Perl代码:

I have some simple Perl code:

#!/usr/bin/perl

use strict;   # not in the OP, recommended
use warnings; # not in the OP, recommended

my $val = 1;
for ( 1 .. 100 ) {
    $val = ($val * $val + 1) % 8051;
    print ($val / 8050) . " \n";
}

但是当我运行它时，输出为:

But when I run it, the output is:

bash-3.2$ perl ./rand.pl
0.0002484472049689440.000621118012422360.003229813664596270.08409937888198760.92
... <snipped for brevity> ...
2919250.9284472049689440.3526708074534160.1081987577639750.2295652173913040.1839
751552795030.433540372670807bash-3.2$

我做错什么了吗?

推荐答案

C:\> perldoc -f print :

也请注意不要遵循 带左括号的print关键字 除非你想要相应的 右括号终止 打印参数-插入+ 或在所有 争论.

Also be careful not to follow the print keyword with a left parenthesis unless you want the corresponding right parenthesis to terminate the arguments to the print--interpose a + or put parentheses around all the arguments.

因此，您需要的是:

print( ($val / 8050) . "\n" );

或

print +($val / 8050) . "\n";

您拥有的语句打印$val / 8050的结果，然后将"\n"连接到print的返回值，然后丢弃结果值.

The statement you have prints the result of $val / 8050 and then concatenates "\n" to the return value of print and then discards the resulting value.

顺便说一句，如果您:

use warnings;

然后perl会告诉您:

   print (...) interpreted as function at t.pl line 5.
   Useless use of concatenation (.) or string in void context at t.pl line 5.

这篇关于为什么在此Perl代码中不打印换行符?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！