我想将以下变量传递给子例程mySubroutine,$name, $age
然后这个多维数组:
$name = "jennifer";
$age = 100;
$list[0][0] = "TEST NAME 2";
$list[0][1] = "TEST GROUP 2";
$[0][2] = 10;
$[1][0] = "TEST NAME 2";
$[1][1] = "TEST GROUP 2";
$[1][2] = 2;
子程序:
sub mySubroutine
{
}
我尝试了$_[0]
和@_
,但我似乎没有将变量正确地传递给子程序。
答案 0 :(得分:26)
有几种方法可以做到这一点(就像Perl中的大多数事情一样)。我个人这样做:
sub mySubroutine
{
# Get passed arguments
my ($name, $age, $refList) = @_;
# Get the array from the reference
my @list = @{$refList};
# Good to go
}
# You need to pass @list as reference, so you
# put \@list, which is the reference to the array
mySubroutine($name, $age, \@list);
答案 1 :(得分:6)
另一种方法,它通过引用传递数组,但随后复制它以避免在编辑时更改原始数据。
sub mySubroutine
{
## Retrieve name
my $name = shift;
## Retrieve age
my $age = shift;
## Retrieve list reference
my $refList = shift;
## De-reference the list's scalar
my @list = @{$refList};
# Good to go
}
## Function call
mySubroutine($name, $age, \@list);
为了更好地理解,请参考perlsub(Perl子程序)。
答案 2 :(得分:5)
另一个选项,只要您传递一个数组,就是通过值将其作为最后一个元素正常传递:
sub scalars_and_one_array {
my $name = shift;
my $age = shift;
foreach my $element (@_)
{
# Do something with the array that was passed in.
}
}
scalars_and_one_array($name,$age,@array);
但是,通过仅使用sub中的引用来避免数组的任何其他副本是最有效的。这确实意味着对sub中数组的更改会影响原始数据,但是:
sub array_by_ref {
my $array_ref = shift;
foreach my $element (@$array_ref)
{
# Changing $element changes @original_array!
}
}
array_by_ref(\@original_array);