您如何检查Perl中打开(文件)是否成功? [英] How do you check the success of open (file) in Perl?

问题描述

以下(不是很Perl的)代码

The following (not very Perl-ish) code

#!/usr/bin/perl

if (! -e "mydir/")
{
  print "directory doesn't exist.\n";
}
open (my $fh, ">", "mydir/file.txt");
if ($fh)
{
  print "file opened.\n";
  print $fh;
  print $fh "some text\n" or die "failed to write to file.\n";
  close ($fh);
}
else
{
  print "failed to open file.\n";
}

产生这样的输出

directory doesn't exist.
file opened.
failed to write to file.
GLOB(0x...some-hex-digits...)

为什么在公开通话后 $ fh 不等同于false?由于 mydir/不存在，我希望打开文件的尝试失败.

Why is $fh not equivalent to false following the open call? As mydir/ does not exist, I'd expect the attempt to open the file to fail.

如果目录和文件存在，但文件是只读的，我得到类似的结果.

I get similar results if the directory and file exist, but the file is read-only.

我已经在Windows 7 x64的Perl 5.10.1和Fedora-11 Linux的Perl 5.10.0上进行了尝试.

I've tried this with Perl 5.10.1 on Windows 7 x64, and with Perl 5.10.0 on Fedora-11 Linux.

我猜我的文件句柄测试是错误的.我尝试了Google搜索，但是没有运气.我希望这很明显，但是任何提示或链接都将不胜感激.

I'm guessing my file handle test is wrong. I've tried Googling this without luck. I expect it's something obvious, but any hints or links would be much appreciated.

谢谢，罗布.

推荐答案

$ fh 未被设置为零位值，而是被设置为 GLOB ，如代码所示.这与 open 返回的内容不同，这就是习惯用法所在的原因

$fh isn't being set to a zero-ish value, it is being set to a GLOB as your code shows. This is different from what open returns which is why the idiom is

open(...) or die ... ;

或

unless(open(...)) {
    ...
}

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