Behavior of Scalar::Util::looks_like_number in Perl - perl

I am trying to find out if an input is number or string. I came across looks_like_number and cannot understand the values it returns.
use warnings;
use Scalar::Util qw(looks_like_number);
my $name = 11;
print looks_like_number ($name);
This code prints 1 if $name contains a string and a static number if $name contains an integer (i.e. 4352 for each integer).
I am using Perl on Windows.

You forgot to ask a question! Here are two possibilities.
Why doesn't it always return the same value for true?
Why not? It returns a true value as documented. It makes no difference which true value it is.
What is the value returned?
If the scalar contains a string, it uses grok_number which has specific document return values.
The type of the number is returned (0 if unrecognised), otherwise it is a bit-ORed combination of IS_NUMBER_IN_UV, IS_NUMBER_GREATER_THAN_UV_MAX, IS_NUMBER_NOT_INT, IS_NUMBER_NEG, IS_NUMBER_INFINITY, IS_NUMBER_NAN (defined in perl.h).
Otherwise, it uses
SvFLAGS(sv) & (SVf_NOK|SVp_NOK|SVf_IOK|SVp_IOK)
You can't tell which of the two was used, so you can't ascribe meaning to the value, which is why it's undocumented.

Don't rely on the exact numerical value. This is an abstraction leak, which the latest version of Scalar::Util (1.39) fixes. What is important is simply the truth of the result, not its exact numerical value.
See bug https://rt.cpan.org/Ticket/Display.html?id=94806

This is what the documentation says:
looks_like_number EXPR
Returns true if perl thinks EXPR is a number. See "looks_like_number" in perlapi.
The link to perlapi in this quote is not really helping us a lot unfortunately:
Test if the content of an SV looks like a number (or is a number). Inf
and Infinity are treated as numbers (so will not issue a non-numeric
warning), even if your atof() doesn't grok them. Get-magic is ignored.
I32 looks_like_number(SV *const sv)
In my case, your code will return an integer that is not 0, which is true.
I got 4352 when I used 11.
When I used '11' I got 1.
All of these are true, so that works.
When I put 'test' or 'foobar' I got 0, which is not true.
I never got a 1 for anything that did not look like a number.
I tried '1e1' and it printed 4, which is a true value, and the input looked like a number in scientific notation.
So, I'd say it always returns something true if Perl thinks the input looks like a number, though I do not know what exactly that true value represents. I cannot confirm that it also returns true with a name.

Related

How are scalars stored 'under the hood' in perl?

The basic types in perl are different then most languages, with types being scalar, array, hash (but apparently not subroutines, &, which I guess are really just scalar references with syntactical sugar). What is most odd about this is that the most common data types: int, boolean, char, string, all fall under the basic data type "scalar". It seems that perl decides rather to treat a scalar as a string, boolean, or number based off of the operator that modifies it, implying the scalar itself is not actually defined as "int" or "String" when saved.
This makes me curious as to how these scalars are stored "under the hood", particularly in regards to it's effect on efficiency (yes I know scripting languages sacrifice efficiency for flexibility, but they still need to be as optimized as possible when flexibility concerns are not affected). It's much easier for me to store the number 65535 (which takes two bytes) then the string "65535" which takes 6 bytes, as such recognizing that $val = 65535 is storing an int would allow me to use 1/3 the memory, in large arrays this could mean fewer cache hits as well.
It's not just limited to saving memory of course. There are times when I can offer more significant optimizations if I know what type of scalar to expect. For instance if I have a hash using very large integers as keys it would be far faster to look up a value if I recognizing the keys as ints, allowing a simply modulo for creating my hash key, then if I have to run more complex hashing logic on a string that has 3 times the bytes.
So I'm wondering how perl handles these scalars under the hood. Does it store every value as a string, sacrificing the extra memory and cpu cost of constant converting string to int in the case that a scalar is always used as an int? Or does it have some logic for inference the type of scalar used to determine how to save and manipulate it?
Edit:
TJD linked to perlguts, which answers half my question. A scalar is actually stored as string, int (signed, unsigned, double) or pointer. I'm not too surprised, I had mostly expected this behavior to occur under the hood, though it's interesting to see the exact types. I'm leaving this question open though because perlguts is actually to low level. Other then telling me that 5 data types exist it doesn't specify how perl works to alternate between them, ie how perl decides which SV type to use when a scalar is saved and how it knows when/how to cast.
There are actually a number of types of scalars. A scalar of type SVt_IV can hold undef, a signed integer (IV) or an unsigned integer (UV). One of type SVt_PVIV can also hold a string[1]. Scalars are silently upgraded from one type to another as needed[2]. The TYPE field indicates the type of a scalar. In fact, arrays (SVt_AV) and hashes (SVt_HV) are really just types of scalars.
While the type of a scalar indicates what the scalar can contain, flags are used to indicate what a scalar does contain. This is stored in the FLAGS field. SVf_IOK signals that a scalar contains a signed integer, while SVf_POK indicates it contains a string[3].
Devel::Peek's Dump is a great tool for looking at the internals of scalars. (The constant prefixes SVt_ and SVf_ are omitted by Dump.)
$ perl -e'
use Devel::Peek qw( Dump );
my $x = 123;
Dump($x);
$x = "456";
Dump($x);
$x + 0;
Dump($x);
'
SV = IV(0x25f0d20) at 0x25f0d30 <-- SvTYPE(sv) == SVt_IV, so it can contain an IV.
REFCNT = 1
FLAGS = (IOK,pIOK) <-- IOK: Contains an IV.
IV = 123 <-- The contained signed integer (IV).
SV = PVIV(0x25f5ce0) at 0x25f0d30 <-- The SV has been upgraded to SVt_PVIV
REFCNT = 1 so it can also contain a string now.
FLAGS = (POK,IsCOW,pPOK) <-- POK: Contains a string (but no IV since !IOK).
IV = 123 <-- Meaningless without IOK.
PV = 0x25f9310 "456"\0 <-- The contained string.
CUR = 3 <-- Number of bytes used by PV (not incl \0).
LEN = 10 <-- Number of bytes allocated for PV.
COW_REFCNT = 1
SV = PVIV(0x25f5ce0) at 0x25f0d30
REFCNT = 1
FLAGS = (IOK,POK,IsCOW,pIOK,pPOK) <-- Now contains both a string (POK) and an IV (IOK).
IV = 456 <-- This will be used in numerical contexts.
PV = 0x25f9310 "456"\0 <-- This will be used in string contexts.
CUR = 3
LEN = 10
COW_REFCNT = 1
illguts documents the internal format of variables quite thoroughly, but perlguts might be a better place to start.
If you start writing XS code, keep in mind it's usually a bad idea to check what a scalar contains. Instead, you should request what should have been provided (e.g. using SvIV or SvPVutf8). Perl will automatically convert the value to the requested type (and warn if appropriate). API calls are documented in perlapi.
In fact, it can hold a string an either a signed integer or an unsigned integer at the same time.
All scalars (including arrays and hashes, excluding one type of scalar that can only hold undef) have two memory blocks at their base. Pointers to the scalar point to its head, which contains the TYPE field and a pointer to the body. Upgrading a scalar replaces the body of the scalar. That way, pointers to the scalar aren't invalidated by an upgrade.
An undef variable is one without any uppercase OK flags set.
The formats used by Perl for data storage are documented in the perlguts perldoc.
In short, though, a Perl scalar is stored as a SV structure containing one of a number of different types, such as an int, a double, a char *, or a pointer to another scalar. (These types are stored as a C union, so only one of them will be present at a time; the SV contains flags indicating which type is used.)
(With regard to hash keys, there's an important gotcha to note there: hash keys are always strings, and are always stored as strings. They're stored in a different type from other scalars.)
The Perl API includes a number of functions which can be used to access the value of a scalar as a desired C type. For example, SvIV() can be used to return the integer value of a SV: if the SV contains an int, that value is returned directly; if the SV contains another type, it's coerced to an integer as appropriate. These functions are used throughout the Perl interpreter for type conversions. However, there is no automatic inference of types on output; functions which operate on strings will always return a PV (string) scalar, for instance, regardless of whether the string "looks like" a number or not.
If you're curious what a given scalar looks like internally, you can use the Devel::Peek module to dump its contents.
Others have addressed the "how are scalars stored" part of your question, so I'll skip that. With regard to how Perl decides which representation of a value to use and when to convert between them, the answer is it depends on which operators are applied to the scalar. For example, given this code:
my $score = 0;
The scalar $score will be initialised with an integer value. But then when this line of code is run:
say "Your score is $score";
The double quote operator means that Perl will need a string representation of the value. So the conversion from integer to string will take place as part of the process of assembling the string argument to the say function. Interestingly, after the stringification of $score, the underlying representation of the scalar will now include both an integer and a string representation, allowing subsequent operations to directly grab the relevant value without having to convert again. If a numeric operator is then applied to the string (e.g.: $score++) then the numeric part will be updated and the (now invalid) string part will be discarded.
This is the reason why Perl operators tend to come in two flavours. For example comparing values of numbers is done with <, ==, > while performing the same comparisons with strings would be done with lt, eq, gt. Perl will coerce the value of the scalar(s) to the type which matches the operator. This is why the + operator does numeric addition in Perl but a separate operator . is needed to do string concatenation: + will coerce its arguments to numeric values and . will coerce to strings.
There are some operators that will work with both numeric and string values but which perform a different operation depending on the type of value. For example:
$score = 0;
say ++$score; # 1
say ++$score; # 2
say ++$score; # 3
$score = 'aaa';
say ++$score; # 'aaa'
say ++$score; # 'aab'
say ++$score; # 'aac'
With regard to questions of efficiency (and bearing in mind standard disclaimers about premature optimisation etc). Consider this code which reads a file containing one integer per line, each integer is validated to check it is exactly 8 digits long and the valid ones are stored in an array:
my #numbers;
while(<$fh>) {
if(/^(\d{8})$/) {
push #numbers, $1;
}
}
Any data read from a file will initially come to us as a string. The regex used to validate the data will also require a string value in $_. So the result is that our array #numbers will contain a list of strings. However, if further uses of the values will be solely in a numeric context, we could use this micro-optimisation to ensure that the array contained only numeric values:
push #numbers, 0 + $1;
In my tests with a file of 10,000 lines, populating #numbers with strings used nearly three times as much memory as populating with integer values. However as with most benchmarks, this has little relevance to normal day-to-day coding in Perl. You'd only need to worry about that in situations where you a) had performance or memory issues and b) were working with a large number of values.
It's worth pointing out that some of this behaviour is common to other dynamic languages (e.g.: Javascript will silently coerce numeric values to strings).

Perl variables defined with * vs $

What's the difference between defining a variable with a * vs a $? For example:
local $var;
local *var;
The initial character is known as a sigil, and says what sort of value the identifier represents. You will know most of them. Here's a list
Dollar $ is a scalar value
At sign # is an array value
Percent % is a hash value
Ampersand & is a code value
Asterisk * is a typeglob
You are less likely to have come across the last two recently, because & hasn't been necessary when calling subroutines since Perl 5.0 was released. And typeglobs are a special type that contains all of the other types, and are much more rarely used.
I'm considering how much deeper to go into all of this, but will leave my answer as it is for now. I may write more depending on the comments that arise.
$var is a scalar. *var is a typeglob. http://perldoc.perl.org/perldata.html#Typeglobs-and-Filehandles
It's not a variable in the strictest sense. You shouldn't generally be using it.

How does this Perl one-liner actually work?

So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?
This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.
The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);

Float comparison issues in Perl [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
How do I fix this Perl code so that 1.1 + 2.2 == 3.3?
I'm working on a Perl script that compares strings representing gene models and prints out a summary of the comparison. If the gene models match perfectly, I print out a very terse summary, but if they are different, the summary is quite verbose.
The script looks at the value of a variable to determine whether it should do the terse or verbose summary--if the variable is equal to 1, it should print the terse summary; otherwise, it should print the verbose summary.
Since the value is numeric (a float), I've been using the == operator to do the comparison.
if($stats->{overall_simple_matching_coefficient} == 1)
{
print "Gene structures match perfectly!\n";
}
This worked correctly for all of my tests and even for most of the new cases I am running now, but I found a weird case where the value was equal to 1 but the above comparison failed. I have not been able to figure out why the comparison failed, and stranger yet, when I changed the == operator to the eq operator, it seemed to work fine.
I thought the == was for numerical comparison and eq was for string comparison. Am I missing something here?
Update: If I print out the value right before the comparison...
printf("Test: '%f', '%d', '%s'\n", $stats->{overall_simple_matching_coefficient}, $stats->{overall_simple_matching_coefficient}, $stats->{overall_simple_matching_coefficient});
...I get this.
Test: '1.000000', '0', '1'
The first thing any computer language teacher should teach you about any computer language is that YOU CANNOT COMPARE FLOATS FOR EQUALITY. This is true of any language. Floating point arithmetic is not exact, and two floats that look like they're the same will be different in the insignificant digits somewhere where you can't see it. Instead, you can only compare that they are close to each other - like
if (abs(stats->{overall_simple_matching_coefficient)-1) < 0.0001)
What do you get if you print the value of $stats->{overall_simple_matching_coefficient} just before the comparison? If it's 1, try printf with a format of "%20.10f". I strongly suspect you have some rounding error (less then 1e-6) accumulated in the variable and it's not comparing equal numerically. However when converted to string, since the error is right of the 6th decimal place, and the default string format is to six places, it compares equal.

Why can't I properly encode a boolean from PostgreSQL via JSON::XS via Perl?

I have a query on a PostgreSQL system returning a boolean:
my $sth = $dbh->prepare("select 'f'::boolean");
$sth->execute;
my #vals = $sth->fetchrow_array;
According to the DBD::Pg docs,
The current implementation of
PostgreSQL returns 't' for true and
'f' for false. From the Perl point of
view, this is a rather unfortunate
choice. DBD::Pg therefore translates
the result for the BOOL data type in a
Perlish manner: 'f' becomes the number
0 and 't' becomes the number 1. This
way the application does not have to
check the database-specific returned
values for the data-type BOOL because
Perl treats 0 as false and 1 as true.
You may set the pg_bool_tf attribute
to a true value to change the values
back to 't' and 'f' if you wish.
So, that statement should return a 0, which it does, so long as pg_bool_tf returns 0, which it does. However, somewhere along the way JSON::XS (and plain JSON) interprets the returned 0 as a string:
use JSON::XS qw(encode_json);
my $options =
{
layout => 0,
show_widget_help => $vals[0] // 1,
};
die encode_json($options);
...dies with:
{"layout":0,"show_widget_help":"0"}
...which would be fine, except that my JavaScript is expecting a boolean there, and the non-empty string "0" gets evaluated to true. Why is the latter 0 quoted and the former not?
According to the JSON::XS docs, this is a main feature:
round-trip integrity
When you serialise a perl data
structure using only data types
supported by JSON, the deserialised
data structure is identical on the
Perl level. (e.g. the string "2.0"
doesn't suddenly become "2" just
because it looks like a number). There
minor are exceptions to this, read the
MAPPING section below to learn about
those.
...which says:
Simple Perl scalars (any scalar that
is not a reference) are the most
difficult objects to encode: JSON::XS
will encode undefined scalars as JSON
null values, scalars that have last
been used in a string context before
encoding as JSON strings, and anything
else as number value.
But I never use #vals[0] in a string context. Maybe DBD::Pg uses its boolean 0 as a string somewhere before returning it?
The JSON::XS doc says the following will be converted to true/false
references to the integers 0 and 1, ie. \0 and \1
JSON::XS::true and JSON::XS::false
Using one of these should solve your problem