Found this inside a loop. I've read up about splice but it just confused me more.
I'm not familiar with Perl, but am trying to translate an algorithm to another language.
my $sill = splice(#list,int(rand(#list)),1);
last unless ($sill);
To be more specific: What will be inside $sill if it doesn't exit the loop from the last?
Thanks for any help!
This randomly removes one element from the array #list. That value is assigned to $sill. If that was a false value, the enclosing loop (not shown) is broken out of.
splice takes an array, an offset, and a length, plus a replacement list. If the replacement is omitted, the elements are deleted.
The length is constant (1 element), but the offset is calculated as a random integer smaller between 0 and the length of #list.
That means :
remove a random element from the array list (0 -> numbers of element of the list) and
assign the sill variable with the removed element (pop() like) and
exit the loop if sill variable is false
See http://perldoc.perl.org/functions/splice.html
Related
I have following code:
foreach ($arg in $args) {
Write-Host "Arg: $arg";
$param1=$args[0]
}
Write-host "Number of args: " $args.Length
write-host Last Arg is: "$($args.count)"
I get this, when I run it:
./print_last_arg.ps1 a b c
Arg: a
Arg: b
Arg: c
Number of args: 3
Last Arg is: 3
What I would like to have is name of last argument, so:
Last Arg is: 3
should be:
Last Arg is: c
Sorry for such a stupid question but I am totally begginer in PS and cannot google the result...
PowerShell supports negative indices to refer to elements from the end of a collection, starting with -1 to refer to the last element, -2 to the penultimate (second to last) one, and so on.
Therefore, use $args[-1] to refer to the last argument passed.
For more information, see the conceptual about_Arrays help topic.
Note that you can also use the results of expressions as indices; e.g., the equivalent of $args[-1] is $args[$args.Count-1] (assuming the array has at least one element).
Additionally, you may specify multiple indices to extract a sub-array of arbitrary elements. E.g., $args[0, -1] returns a (new) array comprising the input array's first and the last element (assuming the array has at least two elements).
.., the range operator is particularly useful for extracting a range of contiguous elements. E.g., $args[0..2] returns a (new) array comprising the first 3 elements (the elements with indices 0, 1, and 2).
You can even combine individual indices with ranges, courtesy of PowerShell's + operator performing (flat) array concatenation.
E.g., $args[0..2 + -1] extracts the first 3 elements as well as the last (assumes at least 4 elements).
Note: For syntactic reasons, if a single index comes first in the index expression, you need to make it an array with the unary form of , the array constructor operator, to make sure that + performs array concatention; e.g., $args[,-1 + 0..2] extracts the last element followed by the first 3.
Pitfall: Combining a positive .. start point with a negative end point for up-to-the-last-Nth-element logic does not work as intended:
Assume the following array:
$a = 'first', 'middle1', 'middle2', 'last'
It is tempting to use range expression 1..-2 to extract all elements "in the middle", i.e. starting with the 2nd and up to the penultimate element - but this does not work as expected:
# BROKEN attempt to extract 'middle1', 'middle2'
PS> $a[1..-2]
middle1
first
last
middle2
The reason is that 1..-2, as a purely arithmetic range expression, expanded to the following array (whose elements happen to be used as indices into another array): 1, 0, -1, -2. And it is these elements that were extracted: the 2nd, the first, the last, the penultimate.
To avoid this problem, you need to know the array's element count ahead of time, and use an expression to specify the end of the range as a positive number:
# OK: extract 'middle1', 'middle2'
# Note that the verbosity and the need to know $a's element count.
PS> $a[1..($a.Count-2)]
middle1
middle2
Unfortunately, this is both verbose and inconvenient, especially given that you may want to operate on a collection whose count you do not know in advance.
GitHub issue #7940 proposes a future enhancement to better support this use case with new syntax, analogous to C#'s indices-and-ranges feature, so that the above could be written more conveniently with syntax such as $a[1..^1]
We can use range operator to get sequence of elements.
But is there any way to get say 3rd 7th and 11th element from an array in a single statement.
Yes, sure
my #selection = #array[2,6,10]
Basic syntax tutorials I followed do not make this clear:
Is there any practical/philosophical/context-dependent/tricky difference between accessing an array using the former or latter subscript notation?
$ perl -le 'my #a = qw(io tu egli); print $a[1], #a[1]'
The output seems to be the same in both cases.
$a[...] # array element
returns the one element identified by the index expression, and
#a[...] # array slice
returns all the elements identified by the index expression.
As such,
You should use $a[EXPR] when you mean to access a single element in order to convey this information to the reader. In fact, you can get a warning if you don't.
You should use #a[LIST] when you mean to access many elements or a variable number of elements.
But that's not the end of the story. You asked for practical and tricky (subtle?) differences, and there's one noone mentioned yet: The index expression for an array element is evaluated in scalar context, while the index expression for an array slice is evaluated in list context.
sub f { return #_; }
$a[ f(4,5,6) ] # Same as $a[3]
#a[ f(4,5,6) ] # Same as $a[4],$a[5],$a[6]
If you turn on warnings (which you always should) you would see this:
Scalar value #a[0] better written as $a[0]
when you use #a[1].
The # sigil means "give me a list of something." When used with an array subscript, it retrieves a slice of the array. For example, #foo[0..3] retrieves the first four items in the array #foo.
When you write #a[1], you're asking for a one-element slice from #a. That's perfectly OK, but it's much clearer to ask for a single value, $a[1], instead. So much so that Perl will warn you if you do it the first way.
The first yields a scalar variable while the second gives you an array slice .... Very different animals!!
I encountered this special variable ($#_) while browsing. Tried finding out what it means, but couldn't find any. Please let me know what this special variable mean.
In Perl, you get the index of the last element of #array with the syntax $#array. So $#_ is the index of the last element in the array #_. This is not the same thing as the number of elements in the array (which you get with scalar #array), because Perl arrays are normally 0-based.
I have some code where I am converting some data elements in a flat file. I save the old:new values to a hash which is written to a file at the end of processing. On subsequence execution, I reload into a hash so I can reuse previously converted values on additional data files. I also save the last conversion value so if I encounter an unconverted value, I can assign it a new converted value and add it to the hash.
I had used this code before (back in Feb) on six files with no issues. I have a variable that is set to ZCKL0 (last character is a zero) which is retrieved from a file holding the last used value. I apply the increment operator
...
$data{$olddata} = ++$dataseed;
...
and the resultant value in $dataseed is 1 instead of ZCKL1. The original starting seed value was ZAAA0.
What am I missing here?
Do you use the $dataseed variable in a numeric context in your code?
From perlop:
If you increment a variable that is
numeric, or that has ever been used in
a numeric context, you get a normal
increment. If, however, the variable
has been used in only string contexts
since it was set, and has a value that
is not the empty string and matches
the pattern /^[a-zA-Z][0-9]\z/ , the
increment is done as a string,
preserving each character within its
range.
As prevously mentioned, ++ on strings is "magic" in that it operates differently based on the content of the string and the context in which the string is used.
To illustrate the problem and assuming:
my $s='ZCL0';
then
print ++$s;
will print:
ZCL1
while
$s+=0; print ++$s;
prints
1
NB: In other popular programming languages, the ++ is legal for numeric values only.
Using non-intuitive, "magic" features of Perl is discouraged as they lead to confusing and possibly unsupportable code.
You can write this almost as succinctly without relying on the magic ++ behavior:
s/(\d+)$/ $1 + 1 /e
The e flag makes it an expression substitution.