What is wrong with this Perl subroutine?

懵懂的女人 提交于 2020-01-07 08:35:09

问题


I'm trying to implement a subroutine that calculates the d-neighbors of an input string. This is apart of an implementation of planted motif search, but my question is much more general. Here is the code:

#subroutine for generating d-neighbors
sub generate_d_neighbors{
    # $sequence is the sequence to generate d-neighbors from
    # $HD is the Hamming Distance
    my ($sequence, $HD) = @_;

    for(my $i = 0; $i=$HD; $i++){
        my @l = ['A', 'C', 'T', 'G'];
        my @t = splice(@l,$sequence[$i]);  
       #TODO
    }
}

The error is occurring at the last line, saying that:

Global symbol "@sequence" requires explicit package name (did you forget to declare "my @sequence"?

It was my understanding that Perl does not take parameters in the form subroutine(param1, param2) like in Java for example, but why is $sequence not being recognized as already having been initialized?


回答1:


There are some problems with your code:

sub generate_d_neighbors{
    my ($sequence, $HD) = @_;

    for(my $i = 0; $i=$HD; $i++){
        my @l = ['A', 'C', 'T', 'G'];
        my @t = splice(@l,$sequence[$i]);  
    }
}

First, let's look at

    for(my $i = 0; $i=$HD; $i++){

Assuming $HD is nonzero, this loop will never terminate because the condition will never be false. If you wanted $i to range from 0 to $HD, writing the statement as for my $i (0 .. $HD) would have been better.

Second, you have

        my @t = splice(@l,$sequence[$i]);  

where you seem to assume there is an array @sequence and you are trying to access its first element. However, $sequence is a reference to an array. Therefore, you should use

$sequence->[$i]

Third (thanks @Ikegami), you have

        my @l = ['A', 'C', 'T', 'G'];

in the body of the for-loop. Then @l will contain a single element, a reference to an anonymous array containing the elements 'A', 'C', 'T', and 'G'. Instead, use:

my @l = qw(A C T G);

I am not sure exactly what you want to achieve with splice(@l, $sequence->[$i]), but that can be better written as:

 my @t = @l[0 .. ($sequence->[$i] - 1)];  

In fact, you could reduce the two assignments to:

 my @t = qw(A C T G)[0 .. ($sequence->[$i] - 1)];



回答2:


It looks to me like you want

substring($sequence, 0, 1)

instead of

$sequence[0].

In Perl, strings are first class variables, not a type of array.

Or maybe you want splice(@l, $sequence->[0])?




回答3:


This list-assignment syntax:

my (@sequence, $HD) = @_;

doesn't do what you might want it to do (put the last argument in $HD and the rest in @sequence). The array always takes all the arguments it can, leaving none for whatever comes after it.

Reversing the order can work, for cases where there is only one array:

my ($HD, @sequence) = @_;

and you make the corresponding change in the caller.

To solve the problem more generally, use a reference:

my ($sequence, $HD) = @_;

and call the sub like this:

generate_d_neighbors(\@foo, $bar);

or this:

# Note the brackets, which make an array reference, unlike parentheses
# which would result in a flat list.
generate_d_neighbors([...], 42);

If you use a prototype:

sub generate_d_neighbors (\@$)

then the caller can say

generate_d_neighbors(@foo, $bar);

and the @foo automatically becomes a reference as if it had been\@foo.

If you use any of the reference-based solutions, you must alter the body of the function to use $sequence instead of @sequence, following these rules:

  1. Change @sequence to @$sequence
  2. Change $#sequence to $#$sequence
  3. Change $sequence[...] to $sequence->[...]
  4. Change @sequence[...] to @$sequence[...] (but be sure you really meant to use an array slice... if you're new to perl you probably didn't mean it, and should have used $sequence[...] instead)


来源:https://stackoverflow.com/questions/43545175/what-is-wrong-with-this-perl-subroutine

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