What I am having trouble understanding is how to make an array with an uknown length? I don't understand how that works sorry I am not trying to be complicated thanks =) | [reply] |
perlguru22:
The phrase "unknown length" simply means that you (the programmer) don't know how long the array is beforehand. The intent is to make you think of a way to accomplish the task that particular bit of information. Perl, however, will always know how many items are in the array. If you build that kind of knowledge into your code, you make your code less easy to reuse.
For example:
sub sum_5 {
# Returns the sum of array elements.
# ...if your array happens to contain 5 items...
my @array = @_;
my $sum=0;
# The programmer assume the array length right here:
# v
for my $i (0 .. 4) {
$sum += $array[$i];
}
return $sum;
}
sub sum_array {
# Return the sum of all array elements
my @array=@_;
# The programmer asks the array how long it is here:
vvvvvvv
for my $i (0 .. $#array) {
$sum += $array[$i];
}
}
my @a1 = (1,2,3,4,5);
my @a2 = (1,2,3,4,5,6,7,8,9,10);
print "** sum_5 **\n";
print "Sum of array 1 is ", sum_5(@a1), "\n";
print "Sum of array 2 is ", sum_5(@a2), "\n";
print "** sum_array **\n";
print "Sum of array 1 is ", sum_all(@a1), "\n";
print "Sum of array 2 is ", sum_all(@a2), "\n";
As you can see, the difference between the subroutines is that in the first one, the programmer assumes that all arrays it will be used with have exactly 5 elements. In the second example, the programmer instead asks the array how long it is.
Notes:
- I didn't test the example code, so it likely has a bug or two.
- This isn't the most "perlish" way of adding things up. There are better ways, but this seemed the most clear way of showing the difference between the two cases.
- Strictly speaking, $#array doesn't tell you how long the array is, it tells you the index of the last element in it. Since perl indexes arrays using 0-based numbers, it's one less than the number of items in the array.
...roboticus
When your only tool is a hammer, all problems look like your thumb. | [reply]
[d/l] |
You can use shift operator in this case..
shift operator takes an array, removes an element from the start and updates the existing array
If your array becomes empty, it returns undef, which you can handle
sub average {
$sum = shift @_;
while (defined($num = shift @_)) {
$sum += $num;
}
}
| [reply]
[d/l] |
#!/usr/bin/perl
@nums = ();
for(@nums)
{
$sum += $_;
}
$size = @nums;
$average = $sum / $size;
print "$average";
It works when I put elements in the @nums but I am just confused on the unknown length part. | [reply]
[d/l] |
Perl is a pretty much untyped, dynamic language. When declaring an array in Perl you don't need to specify how big you intend it to be or what it contains. It will expand and shrink at runtime depending what you do with it.
I imagine the problem is aimed at illustrating this and ensuring that you know how to handle it.
So, all arrays are of unknown length unless you know what you put in them or you find out with:
$size = @nums; or $last_index = $#nums
| [reply] |
It works when I put elements in the @nums but I am just confused on the unknown length part.
ofcourse, the array @nums = () is empty.
unknown length doesn't mean EMPTY array, does it?
Infact, your code will die with an error message, illegal division by zero at ... because the variable $size will be 0 since the array is empty.
If you tell me, I'll forget.
If you show me, I'll remember.
if you involve me, I'll understand.
--- Author unknown to me
| [reply]
[d/l]
[select] |
Perl will handle array size (memory) as needed while populating an array. The "unknown length" part becomes crucial when you have to know the number of array elements beforehand if you were to write a C, C++, or Java program to fill an array. (In latter two languages there are other data structures -- vector, set, etc. -- to avoid the need to know the number of elements beforehand).
An exercise of an array of "unknown length" would be to populate an array from a file of unknown lines (thus number of array elements) with each number listed on a line by itself.
| [reply] |
@nums = ();
for (0..$#nums)
So would this be going in the right direction. | [reply]
[d/l] |
I think what you want is
$size = $#nums;
| [reply]
[d/l] |
#!/usr/bin/perl
@nums = ();
for(0..$#nums)
{
$sum += $_;
}
$size = $#nums;
$average = $sum / $size;
print "$average";
This fix the illegal division of 0 and now it gives me the result of 0 | [reply]
[d/l] |