以下是该文件的示例。
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
...代表呼叫深度。该行后面的文件名表示文件名 以及在父级中调用它的行号。我可以解析文件。解析文件后，我尝试做的是将数据放入n-ary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计算出在使用情况之前是否有任何设置或任何设置但未使用的情况。我认为树遍历最有意义。
以下是收集数据的示例：
$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。缩短的例子：
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正确的数据结构和算法来完成这项任务。
我认为深度优先搜索n-ary树会给我我想要的东西。
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 = \@{$$p_arr[$#{$p_arr}]{children}};
}
elsif ($diff < 0)
{
$p_arr = pop @stack while ++$diff <= 0;
}
elsif ($diff > 1)
{
die "Incorrectly formated call tree:\n $_\n";
}
push @$p_arr, {
caller => $caller_fn,
called_by => $callee_fn,
at_line => $ln
};
$old_depth = $depth;
}
}
close IN;
}
exit;
OUTPUT看起来像这样：
......file1
............file1 101:A
..................XXX.AA 102:AA
........................XXX.AAA 103:AAA
........................XXX.AAB 104:AAB
..............................XXX.AABA 105:AABA
..................XXX.AB 106:AB
........................XXX.ABA 107:ABA
............file1 108:B
..................XXX.BA 109:BA
........................XXX.BAA 110:BAA
........................XXX.BAB 111:BAB
来自此call_tree.txt文件：
file1
A => file1:101
AA => XXX.AA:102
AAA => XXX.AAA:103
AAB => XXX.AAB:104
AABA => XXX.AABA:105
AB => XXX.AB:106
ABA => XXX.ABA:107
B => file1:108
BA => XXX.BA:109
BAA => XXX.BAA:110
BAB => XXX.BAB:111
使用此子程序：
sub Treverse
{
my ($p_arr, $level) = @_;
for (my $ind=0; $ind<=$#{$p_arr}; $ind++)
{
print "." x ($level*6);
if ($$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
答案 0 :(得分：6)
以下是如何使用Wes的答案打印您构建的结构。
处理完数据后，您最终会得到以下结果：
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 },
],
}
]
}
);
结构是递归的，children键指向另一个哈希的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
答案 1 :(得分：2)
对于数据结构，perl的最大权力之一是结构的任意嵌套。因此，不是只有一个包含所有笔记的数据的变量，而是在父母的内部可以有“子节点”。
假设您有一个条目的哈希值：
%node1 = ( name => 'ALPHA_MAX_MC_SSM', file => 'arltst.asm', line => 42 );
上面的代码将创建一个很好的简单节点来存储数据。但是你实际上可以在其中存储更多数据。 “子节点”：
%node2 = (
name => 'ACTUAL_ALT',
file => 'foo.asm',
line => 2001
);
$node1{children}[0] = \%node2;
然后你在第一个节点中有一个子节点（'children'），它是所有孩子的数组。您可以直接访问孩子的日期，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。获取概念需要一些新的perl程序员，但是一旦你得到它，你就可以做一些非常强大的快速程序来收集复杂的分层数据并进行处理。