How can I pass an array to a function in Perl? - arrays

Question

Question 1:
I want to pass an array to a function. But the passed argument is changed in the function. Is it called by value?
Question 2:
#my ($name, $num, #array)= #_; <=1 )
my $name = shift; <=2 )
my $num = shift;
my #array = shift;
Case 1 and 2 has different output. Why did it occur?
#!/usr/bin/perl
use strict;
my #test1;
push #test1, ['a', 1];
push #test1, ['b', 1];
push #test1, ['c', 1];
push #test1, ['d', 1];
push #test1, ['e', 1];
for (my $i=0; $i< scalar(#test1); $i++) {
print "out1: $test1[$i][0] $test1[$i][1]\n";
}
test_func("test_func", 10, #test1);
sub test_func {
#my ($name, $num, #array)= #_; <=1)
my $name = shift; <=2)
my $num = shift;
my #array = shift;
print "$name\n";
print "$num\n";
for (my $i=0; $i< scalar(#test1); $i++) {
print "$array[$i][0] $array[$i][1]\n";
}
for (my $i=0; $i< scalar(#test1); $i++) {
if ($array[$i][0] eq 'a') {
$array[$i][0] = 'z';
}
}
for (my $i=0; $i< scalar(#test1); $i++) {
print "change: $array[$i][0] $array[$i][1]\n";
}
}
for (my $i=0; $i< scalar(#test1); $i++) {
print "out2: $test1[$i][0] $test1[$i][1]\n";
}
#
Below is the test output.
out1: a 1
out1: b 1
out1: c 1
out1: d 1
out1: e 1
test_func
10
a 1
b 1
c 1
d 1
e 1
change: z 1
change: b 1
change: c 1
change: d 1
change: e 1
out2: z 1 <= Why did it change?
out2: b 1
out2: c 1
out2: d 1
out2: e 1

Answer 1

I want to pass an array to a function [...] has different output. Why did it occur?
You cannot pass an array to a function sub. Subs can only take a list of scalars as arguments.
test_func("test_func", 10, #test1);
is the same as
test_func("test_func", 10, $test1[0], $test1[1], $test1[2], $test1[3], $test1[4]);
You are creating a new array in test_func when you do
my ($name, $num, #array) = #_;
shift returns the first element of #_, which is necessarily a scalar. #_ is an array, and elements of arrays are scalars. The equivalent would be
my $name = shift(#_);
my $num = shift(#_);
my #array = splice(#_);
To pass an array to a sub, one would normally pass a reference to it.
test_func("test_func", 10, \#test1);
my ($name, $num, $array) = #_;
my $name = shift;
my $num = shift;
my $array = shift;
say "#$array";
But the passed argument is changed in the function. Is it called by value?
Perl never passes by value. It always passes by reference. If you change any element of #_, it will change the corresponding argument in the caller.
$ perl -E'sub f { $_[0] = "def"; } my $x = "abc"; f($x); say $x;'
def
But that's not the issue. You don't change any elements of #_. What you are doing is changing the single array referenced by both $test[0] and $array[0].
This is what you are doing:
my $ref1 = [ 'a', 1 ]; # aka $test1[0]
my $ref2 = $ref1; # aka $array[0]
$ref2->[0] = 'z'; # Changes the single array (not $ref1 or $ref2).
It's short for
my #anon = ( 'a', 1 );
my $ref1 = \#anon; # aka $test1[0]
my $ref2 = $ref1; # aka $array[0]
$ref2->[0] = 'z'; # Changes #anon (not $ref1 or $ref2).
Storable's dclone can be used to make a "deep copy" of an array.
my $ref1 = [ 'a', 1 ];
my $ref2 = dclone($ref1); # clones the reference, the array, 'a' and 1.
$ref1->[0] = 'y'; # Changes the original array
$ref2->[0] = 'z'; # Changes the new array