在搜索是否可能在我使用的框架中记录所有已使用的子例程时,我发现了这个问题。顶部的答案使用子程序DB::DB记录所有已使用的子例程。这在某种程度上起作用,与调用者()一起使用,以找出程序采用的路径。但是我有一个问题,当程序从一个函数返回并进入一个新的函数时,一个子例程被“称为”第二次。
My::DB在/etc/perl/Devel/AllSubs.pm中
package Devel::AllSubs;
use Data::Dumper;
$Data::Dumper::Sortkeys = 1;
my $LastSub = '::';
sub DB::DB {
my ($Package, $Filename, $Line, $Subroutine) = caller(1);
if ($Subroutine ne $LastSub ){
print STDERR Data::Dumper->Dump(
[
$Package,
$Filename,
$Line,
$Subroutine
],
['Package-1', 'Filename-1', 'Line-1', "Subroutine-1"]
);
COUNT:
for ( my $Count = 2; $Count < 30; $Count++ ) {
my ( $NextPackage, $NextFilename, $NextLine, $NextSubroutine ) = caller( $Count );
last COUNT if !$NextLine;
print STDERR Data::Dumper->Dump(
[
$NextPackage,
$NextFilename,
$NextLine,
$NextSubroutine
],
["Package-$Count", "Filename-$Count", "Line-$Count", "Subroutine-$Count",]
);
}
say STDERR "";
$LastSub = $Subroutine;
}
}
1;我想检查perl -d:AllSubs AllTest.pl的程序
&One();
sub One {
&Two();
}
sub Two {
&Three();
&Six();
}
sub Three {
&Four();
&Five();
}
sub Four {}
sub Five {}
sub Six {}
1;上面提到的答案声称,每个子例程都调用DB::DB,所以我的预期结果是:
# One
# Two
# Three
# Four
# Five
# Six我得到的是:
# One
# Two
# Three
# Four
# Three
# Five
# Two
# Six完整的翻车输出如下:
$Package-1 = undef;
$Filename-1 = undef;
$Line-1 = undef;
$Subroutine-1 = undef;
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 30;
$Subroutine-1 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 34;
$Subroutine-1 = 'main::Two';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 30;
$Subroutine-2 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 38;
$Subroutine-1 = 'main::Three';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 34;
$Subroutine-2 = 'main::Two';
$Package-3 = 'main';
$Filename-3 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-3 = 30;
$Subroutine-3 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 43;
$Subroutine-1 = 'main::Four';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 38;
$Subroutine-2 = 'main::Three';
$Package-3 = 'main';
$Filename-3 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-3 = 34;
$Subroutine-3 = 'main::Two';
$Package-4 = 'main';
$Filename-4 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-4 = 30;
$Subroutine-4 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 38;
$Subroutine-1 = 'main::Three';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 34;
$Subroutine-2 = 'main::Two';
$Package-3 = 'main';
$Filename-3 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-3 = 30;
$Subroutine-3 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 44;
$Subroutine-1 = 'main::Five';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 38;
$Subroutine-2 = 'main::Three';
$Package-3 = 'main';
$Filename-3 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-3 = 34;
$Subroutine-3 = 'main::Two';
$Package-4 = 'main';
$Filename-4 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-4 = 30;
$Subroutine-4 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 34;
$Subroutine-1 = 'main::Two';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 30;
$Subroutine-2 = 'main::One';
$Package-1 = 'main';
$Filename-1 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-1 = 39;
$Subroutine-1 = 'main::Six';
$Package-2 = 'main';
$Filename-2 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-2 = 34;
$Subroutine-2 = 'main::Two';
$Package-3 = 'main';
$Filename-3 = '/opt/frameworks/test/scripts/AllTest.pl';
$Line-3 = 30;
$Subroutine-3 = 'main::One';
$Package-1 = undef;
$Filename-1 = undef;
$Line-1 = undef;
$Subroutine-1 = undef;是否有任何方法可以跳过DB::DB被调用的情况,即使它在代码中没有第二次调用?
编辑:我取得了一些进展。DB::DB为代码的每一行调用。DB:另一方面,潜艇是我所需要的。如果在这里使用调用者(),就会得到调用方(0)的前一个子。当前的潜艇在$DB::sub中。但我也需要文件名和这个潜艇被调用的行。它说这里,$DB::filename应该包含文件名,但是它是空的。我还在这个perl4书中找到了一些信息,但在这一点上还不足以帮助我。
发布于 2021-11-01 11:59:12
下面是一个似乎有效的例子:
lib/Devel/MyDebugger.pm
package Devel::MyDebugger;
package DB;
use feature qw(say);
use warnings;
our $sub;
our $dbline;
our $dbpack;
our $dbfile;
our $START_DEBUG = 0;
sub DB {
($dbpack, $dbfile, $dbline) = caller;
}
sub sub {
if ("$sub" eq "main::One") {
$START_DEBUG = 1;
}
if ($START_DEBUG) {
say "";
say "[sub = $sub, lineno = $dbline, pack = $dbpack, file = $dbfile]";
for ( my $frame = 0; $frame < 30; $frame++ ) {
my @info = my ($package, $filename, $line, $subroutine) = caller $frame;
last if !$line;
print_info($frame, @info);
}
}
&$sub;
}
sub print_info {
my ($frame, $package, $filename, $line, $subroutine) = @_;
my $indent = " " x $frame;
say "${indent}Package-$frame: $package";
say "${indent}Filename-$frame: $filename";
say "${indent}Line-$frame: $line";
say "${indent}Subroutine-$frame: $subroutine";
}p.pl
#! /usr/bin/env perl
use strict;
use warnings;
&One();
sub One {
&Two();
}
sub Two {
&Three();
&Six();
}
sub Three {
&Four();
&Five();
}
sub Four { }
sub Five { }
sub Six { }运行调试器如下:
$ perl -I./lib -d:MyDebugger p.pl输出
[sub = main::One, lineno = 6, pack = main, file = p.pl]
[sub = main::Two, lineno = 8, pack = main, file = p.pl]
Package-0: main
Filename-0: p.pl
Line-0: 6
Subroutine-0: main::One
[sub = main::Three, lineno = 11, pack = main, file = p.pl]
Package-0: main
Filename-0: p.pl
Line-0: 8
Subroutine-0: main::Two
Package-1: main
Filename-1: p.pl
Line-1: 6
Subroutine-1: main::One
[sub = main::Four, lineno = 15, pack = main, file = p.pl]
Package-0: main
Filename-0: p.pl
Line-0: 11
Subroutine-0: main::Three
Package-1: main
Filename-1: p.pl
Line-1: 8
Subroutine-1: main::Two
Package-2: main
Filename-2: p.pl
Line-2: 6
Subroutine-2: main::One
[sub = main::Five, lineno = 16, pack = main, file = p.pl]
Package-0: main
Filename-0: p.pl
Line-0: 11
Subroutine-0: main::Three
Package-1: main
Filename-1: p.pl
Line-1: 8
Subroutine-1: main::Two
Package-2: main
Filename-2: p.pl
Line-2: 6
Subroutine-2: main::One
[sub = main::Six, lineno = 12, pack = main, file = p.pl]
Package-0: main
Filename-0: p.pl
Line-0: 8
Subroutine-0: main::Two
Package-1: main
Filename-1: p.pl
Line-1: 6
Subroutine-1: main::Onehttps://stackoverflow.com/questions/69790622
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