3

这是该文件的示例。

  powrup.asm            POWER_UP
                  ......EXTERNAL_RAM_ADDRESSING_CHECK    powrup.asm:461
                  ......EXRAM    powrup.asm:490
                  ......INRAM    powrup.asm:540
                  ......OUTPUT_TEST    powrup.asm:573
                  ............AD_READ    douttst.asm:276
                  ............AD_READ    douttst.asm:366
                  ......OUTPUT2_TEST    powrup.asm:584
                  ............AD_READ    douttst2.asm:253
                  ............AD_READ    douttst2.asm:342
                  ......OUTPUT3_TEST    powrup.asm:599
                  ............AD_READ    douttst3.asm:307
                  ............AD_READ    douttst3.asm:398
                  ......INPUT_TEST    powrup.asm:614
                  ......PROGRAM_PINS2_INPUT    powrup.asm:629
                  ......ARINC_TEST    powrup.asm:633
                  ............ARINC_LEVEL_TEST    artest.asm:178
                  ..................AD_READ    arltst.asm:204
                  ..................AD_READ    arltst.asm:250
                  ..................AD_READ    arltst.asm:300
                  ..................AD_READ    arltst.asm:346
                  ..................AD_READ    arltst.asm:396
                  ..................AD_READ    arltst.asm:442
                  ............ARINC_READ    artest.asm:209
                  ............ARINC_WORD_TXRX_TEST    artest.asm:221
                  ..................ARINC_OUT    artxrx.asm:207
                  ..................ARINC_READ    artxrx.asm:221
                  ............ARINC_READ    artest.asm:251
                  ............ARINC_WORD_TXRX_TEST    artest.asm:263
                  ..................ARINC_OUT    artxrx.asm:207
                  ..................ARINC_READ    artxrx.asm:221
                  ......PROGRAM_PINS2_INPUT    powrup.asm:640
                  ......PROGRAM_PIN_TEST    powrup.asm:642
                  ......PT_RCVR_BITE    powrup.asm:645
                  ............AD_READ10    ptbite.asm:225
                  ..................AD_READ    adread10.asm:141
                  ............AD_READ10    ptbite.asm:308
                  ..................AD_READ    adread10.asm:141
                  ............AD_READ10    ptbite.asm:384
                  ..................AD_READ    adread10.asm:141
                  ............AD_READ10    ptbite.asm:467
                  ..................AD_READ    adread10.asm:141
                  ............AD_READ10    ptbite.asm:542
                  ..................AD_READ    adread10.asm:141
                  ............AD_READ10    ptbite.asm:622
                  ..................AD_READ    adread10.asm:141
                  ......PROGRAM_PINS2_INPUT    powrup.asm:653
                  ......EXEC_INIT    powrup.asm:663

... 表示调用深度。行后的文件名表示文件名和在父级中调用它的行号。我可以解析文件。一旦我解析了文件，我想要做的就是将数据放在一个 n 叉树中。我正在进行数据耦合和控制耦合分析，并且已经收集了构建中所有变量的所有设置/使用数据。我现在需要能够遍历树并根据深度确定是否存在任何使用前设置的情况或任何设置但未使用的情况。我认为遍历树是最有意义的。

以下是收集的数据的示例：

$hash_DB{'alt_deviation_evaluation.asm->ALT_STATUS'} = [
   'alt_deviation_evaluation.asm',
   'ALT_STATUS',
   '1.1',
   1,
   "",
   "",
   "135,188,202,242",
   "130,144"
];

'alt_deviation_evaluation.asm->ALT_STATUS' is the file name and variable name.

   'alt_deviation_evaluation.asm', File name
   'ALT_STATUS', Variable name
   '1.1',   versions of file
   1,       indicates has been processed
   "",     not used (maybe in future)
   "",     not used  (maybe in future)
   "135,188,202,242", variable Set line numbers for this fileVariable
   "130,144"          Variable Use line number for this file/Variable

我还有一个包含所有变量名的数组。缩短示例：

our @vars_list = (
  'A429_TX_BUFFER_LENGTH',
  'A429_TX_INPUT_BUFFER',
  'A429_TX_INT_MASK',
  'ABS_ALT_DIFF',
  'ACTUAL_ALT',
  'ADDRESS_FAIL',
  'AD_CONV_FAIL',
  'AD_CONV_SIGNAL',
  'AD_DATA',
  'AD_FAIL',
  'AD_STATUS',
  'AIR_MODE',
  'AIR_MODE_COUNT',
  'AIR_MODE_LAST',
  'ALPHA_COR_SSM',
  'ALPHA_EC_SSM',
  'ALPHA_GRAD_SSM',
  'ALPHA_LE_SSM',
  'ALPHA_LG_SSM',
  'ALPHA_MAX_MC_SSM'
};

我最大的障碍是找出合适的数据结构和算法来完成这项任务。

我认为对 n 叉树进行深度优先搜索会给我想要的东西。

这是我的最终解决方案：

#!/usr/local/bin/perl
# !/usr/bin/perl
use Data::Dumper; #!!!
sub Create_Tree;
sub Treverse;

#for my $node (@TREE) {
#   print_node($node[0], 1);
#}


#Main

our @TREE;
Create_Tree("call_tree.small_01.txt");
my $str = Dumper @TREE;
$str =~ s/^(\s+)/' 'x(length($1)>>2)/meg;
#print "\n\n=======================================\n$str"; #!!!
#print "\n\n=======================================\n" . (Dumper @TREE); #!!!

#print "Arr = @TREE, SZ = $#TREE\n\n";
Treverse(\@TREE,1);

sub Create_Tree
{
  my ($call_tree) = @_;
  my @stack;
  my ($old_depth, $p_arr) = (0, \@TREE);

  open(IN, "< $call_tree" ) or die "Can not open '$call_tree' for input.\n";
  for (<IN>)
  {
    if (m/^(\s*)(\S+)\s*=>\s*(\S+):(\d+)/ or m/^(\s*)(\S+)()()/)
    {
      my ($depth, $callee_fn, $caller_fn, $ln, $diff) = ($1, $2, $3, $4, 0);
      $depth = int(length($depth) / 6);
      $diff  = $depth - $old_depth;

      if ($diff == 1)
      {
        push @stack, $p_arr;
        $p_arr = \@{$$p_arr[$#{$p_arr}]{children}};
      }
      elsif ($diff < 0)
      {
        $p_arr = pop @stack while ++$diff <= 0;
      }
      elsif ($diff > 1)
      {
        die "Incorrectly formated call tree:\n  $_\n";
      }

      push @$p_arr, {
          caller    => $caller_fn,
          called_by => $callee_fn,
          at_line   => $ln
      };

      $old_depth = $depth;
    }
  }

  close IN;
}
exit;

输出看起来像这样：

......file1

    ............file1    101:A
    ..................XXX.AA    102:AA
    ........................XXX.AAA    103:AAA
    ........................XXX.AAB    104:AAB
    ..............................XXX.AABA    105:AABA
    ..................XXX.AB    106:AB
    ........................XXX.ABA    107:ABA
    ............file1    108:B
    ..................XXX.BA    109:BA
    ........................XXX.BAA    110:BAA
    ........................XXX.BAB    111:BAB

从这个 call_tree.txt 文件：

file1
      A => file1:101
            AA => XXX.AA:102
                  AAA => XXX.AAA:103
                  AAB => XXX.AAB:104
                        AABA => XXX.AABA:105
            AB => XXX.AB:106
                  ABA => XXX.ABA:107
      B => file1:108
            BA => XXX.BA:109
                  BAA => XXX.BAA:110
                  BAB => XXX.BAB:111

使用这个子程序：

sub Treverse
{
  my ($p_arr, $level) = @_; 
  for (my $ind=0; $ind<=$#{$p_arr}; $ind++)
  {

        print "." x ($level*6);
        if ($$p_arr[$ind]{'caller'} ne "") {print "$$p_arr[$ind]{'caller'}" . " " x 4;}
        if ($$p_arr[$ind]{'at_line'} ne "") {print "$$p_arr[$ind]{'at_line' }" . ":";}
        if ($$p_arr[$ind]{'called_by'} ne "") {print "$$p_arr[$ind]{'called_by'}" . "\n";}


    Treverse(\@{$$p_arr[$ind]{children}}, $level +1) if defined $$p_arr[$ind]{children};

  }
}
# END of Treverse
4

2 回答 2

6

以下是如何打印您使用 Wes 的答案构建的结构。

处理完数据后，您将得到如下结果：

my @nodes = (
    {   name => 'ARINC_TEST', file => 'powrup.asm', line => 633,
        children => [
            {   name => 'ARINC_LEVEL_TEST', file => 'artest.asm', line => 178,
                children => [
                    { name => 'AD_READ', file => 'arltst.asm', line => 204 },
                    { name => 'AD_READ', file => 'arltst.asm', line => 250 },
                    { name => 'AD_READ', file => 'arltst.asm', line => 300 },
                    { name => 'AD_READ', file => 'arltst.asm', line => 346 },
                    { name => 'AD_READ', file => 'arltst.asm', line => 396 },
                    { name => 'AD_READ', file => 'arltst.asm', line => 442 },
                ],
            },
            {   name => 'ARINC_READ', file => 'artest.asm', line => 209,
                children => [],
            },
            {   name => 'ARINC_WORD_TXRX_TEST', file => 'artest.asm', line => 221,
                children => [
                    { name => 'ARINC_OUT',  file => 'artxrx.asm', line => 207 },
                    { name => 'ARINC_READ', file => 'artxrx.asm', line => 221 },
                ],
            }
        ]
    }
);

该结构是递归的，children关键点指向另一个哈希的arrayref。要打印出来，您需要递归代码：

for my $node (@nodes) {
    print_node($node, 1);
}

sub print_node {
    my ($node, $level) = @_;

    # the node itself
    print "." x ($level*6)
        , $node->{name}, " " x 4
        , $node->{file}, ":"
        , $node->{line}, "\n";

    # recurse for children
    if(defined $node->{children}) {
        for my $child (@{ $node->{children} }) {
            print_node($child, $level + 1);
        }
    }
}

对于上面的数据，代码输出

......ARINC_TEST    powrup.asm:633
............ARINC_LEVEL_TEST    artest.asm:178
..................AD_READ    arltst.asm:204
..................AD_READ    arltst.asm:250
..................AD_READ    arltst.asm:300
..................AD_READ    arltst.asm:346
..................AD_READ    arltst.asm:396
..................AD_READ    arltst.asm:442
............ARINC_READ    artest.asm:209
............ARINC_WORD_TXRX_TEST    artest.asm:221
..................ARINC_OUT    artxrx.asm:207
..................ARINC_READ    artxrx.asm:221
于 2011-02-04T12:57:43.460 回答
2

对于数据结构，perl 的最大功能之一是结构的任意嵌套。因此，与其拥有一个包含所有笔记的所有数据的变量，不如在它们的父节点中拥有“子节点”。

假设您有一个条目的哈希：

%node1 = (
  名称 => 'ALPHA_MAX_MC_SSM',
  文件 => 'arltst.asm',
  线 => 42
);

上面的代码将创建一个很好的简单节点来存储数据。但实际上您可以在该节点本身存储更多数据。一个“子节点”：

%node2 = (
  名称 => 'ACTUAL_ALT',
  文件 => 'foo.asm',
  线 => 2001
);

$node1{孩子}[0] = \%node2;

然后你在第一个节点中有一个子节点（'children'），它是一个包含所有子节点的数组。您可以直接访问孩子中的日期，例如：

$node1{'children'}[0]{'name'};

要理解这一点及其工作原理，您需要阅读 perl 引用和 perl 数据类型。作为一个新的 perl 程序员需要一点时间来了解这些概念，但是一旦你掌握了它，你就可以编写出非常强大的快速程序来抓取复杂的分层数据并进行处理。

于 2011-02-03T22:34:41.963 回答