Question

我正在尝试编写以下递归函数。问题是它永远不会结束，我无法理解为什么：

    sub do_smth(@first, @second){
    my @tmp_first = @first;
    $tmp = shift(@tmp_first);
    if (@tmp_first > 0){
        do_smth(@tmp_first, @second);
    }
    my @tmp_second = @second;
    $tmp = shift(@tmp_second);
    if (@tmp_second > 0){
        do_smth(@first, @tmp_second);
    }

}

Answer 1

此代码甚至无法编译。没有警告和严格，你会得到这些错误：

Type of arg 1 to shift must be array (not scalar dereference) at so.pl line 5, near "$tmp_first)"
Type of arg 1 to shift must be array (not scalar dereference) at so.pl line 10, near "$tmp_second)"
Execution aborted due to compilation errors.

并有警告和严格：

Illegal character in prototype for main::do_smth : @first,@second at so.pl line 4.
Global symbol "@first" requires explicit package name at so.pl line 5.
Global symbol "$tmp" requires explicit package name at so.pl line 6.
Global symbol "$tmp_first" requires explicit package name at so.pl line 6.
Type of arg 1 to shift must be array (not scalar dereference) at so.pl line 6, near "$tmp_first)"
Global symbol "@second" requires explicit package name at so.pl line 8.
Global symbol "@second" requires explicit package name at so.pl line 10.
Global symbol "$tmp" requires explicit package name at so.pl line 11.
Global symbol "$tmp_second" requires explicit package name at so.pl line 11.
Type of arg 1 to shift must be array (not scalar dereference) at so.pl line 11, near "$tmp_second)"
Global symbol "@first" requires explicit package name at so.pl line 13.
Execution aborted due to compilation errors.

我不知道你要做什么，但这里的代码是正确的语法：

use warnings;
use strict;

sub do_smth (\@\@);  # predeclaration needed since the prototyped sub
                     # is called recursively
sub do_smth (\@\@) {
    my ($first, $second) = @_;
    my @tmp_first = @$first;
    my $tmp = shift(@tmp_first);
    if (@tmp_first > 0){
        do_smth(@tmp_first, @$second);
    }
    my @tmp_second = @$second;
    $tmp = shift(@tmp_second);
    if (@tmp_second > 0){
        do_smth(@$first, @tmp_second);
    }
}

Answer 2

您正在转移（未定义的）标量 $ tmp_first和$ tmp_second。

没有再看了。

为什么我的Perl递归函数永远不会结束？

2 个答案: