length of an array in a struct in an array of structs in a struct in an array of structs

[Tuan Bui]

Hello,

So I have this complex data structure I've built for myself, and I'm
trying to get this snippet to work:

$k <
scalar(@{@{@VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray}})

VisitArray is an array of VisitStructs. Each VisitStruct contains an
array of PSearchStructs called PSearchStructArray. Each PSearchStruct
contains an array called ClickPathArray. I'm trying to get the length
of ClickPathArray out. I've tried numerous variations of that line, and
none of them work. I do know that

scalar(@{@VisitArray[$i]->PSearchStructArray})

will return the length of PSearchStructArray well enough though.

I'm helplessly floundering here - hope one of you perl hackers can
help. Anyone who does help will earn my undying gratitude

Tuan

 

[it_says_BALLS_on_your forehead]

hmm, i'm unsure why you want to do everything in one line. try breaking
it up.

this is part of a double nested loop i'm assuming.

for my $i (0..$#VisitArray) {
my $PSearchArray_ref = $VisitArray[$i];
my @PSearchArray = @{$PSearchArray_ref};
for my $j (0..$#PSearchArray) {
$k < scalar(@{$PSearchArray[$j]});
}
}

i think that would work and is what you want--i could be wrong on both
counts though *shrug*

 

[Billy N. Patton]

Hello,

So I have this complex data structure I've built for myself, and I'm
trying to get this snippet to work:

$k <
scalar(@{@{@VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray}})

Ths should do it.
$l=scalar(@{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray});

VisitArray is an array of VisitStructs. Each VisitStruct contains an
array of PSearchStructs called PSearchStructArray. Each PSearchStruct
contains an array called ClickPathArray. I'm trying to get the length
of ClickPathArray out. I've tried numerous variations of that line, and
none of them work. I do know that

scalar(@{@VisitArray[$i]->PSearchStructArray})

will return the length of PSearchStructArray well enough though.

I'm helplessly floundering here - hope one of you perl hackers can
help. Anyone who does help will earn my undying gratitude

Tuan

 

[Paul Lalli]

I do know that

scalar(@{@VisitArray[$i]->PSearchStructArray})

will return the length of PSearchStructArray well enough though.

Perhaps I'm having a brain cramp here, but I can't especially see how
that could work. The arrow notation is used for either A)
dereferencing a hash or array reference to get at its elements or B)
calling a method of an object or class.

In other words, I can see that
$VisitArray[$i]->{PSearchStructArray}
might return the value of the hash referenced by $VisitArray[$i] at key
'PSearchStructArray', or
$VisitArray[$i]->PSearchStructArray()
might call a method of the object $VisitArray[$i], which could in turn
return a list of values.

I cannot, however, discern any way in which
$VisitArray[$i]->PSearchStructArray
could return "the length of PSearchStructArray".

Can you please enlighten me by showing the code that creates your
classes and objects?

(For why I replaced your '@' with '$' in each example, please see:
perldoc -q difference
)

Paul Lalli

 

[Tuan Bui]

I do know that

scalar(@{@VisitArray[$i]->PSearchStructArray})

will return the length of PSearchStructArray well enough though.

Perhaps I'm having a brain cramp here, but I can't especially see how
that could work. The arrow notation is used for either A)
dereferencing a hash or array reference to get at its elements or B)
calling a method of an object or class.

I am using Class::Struct, and indeed i am implicitly calling methods
from it. It indeed returns the correct lengths that I want.

In other words, I can see that
$VisitArray[$i]->{PSearchStructArray}
might return the value of the hash referenced by $VisitArray[$i] at key
'PSearchStructArray', or
$VisitArray[$i]->PSearchStructArray()
might call a method of the object $VisitArray[$i], which could in turn
return a list of values.

I cannot, however, discern any way in which
$VisitArray[$i]->PSearchStructArray
could return "the length of PSearchStructArray".

Can you please enlighten me by showing the code that creates your
classes and objects?

struct( VisitStruct => {
VisitID => '$', # ID number of this visit
PSearchStructArray => '@', # array of psearch structures
(PSearchStruct type)

});

struct( PSearchStruct => {
ClickPathArray => '@', # array of click path elements (scalar type)
Before => EventStruct, # event that occured before psearch
(EventStruct type)
After => EventStruct, # event that occured after psearch
(EventStruct type)
ReachedItemScreen => '$', # bool that is true if user reached item
detail screen (scalar type)
ReachedCatalogPage => '$', # bool that is true if user reached
catalog screen (scalar type)
ReachedBookmarkScreen => '$', # bool that is true if user bookmarked
item (scalar type)
# note: in order to set this bool, we will have to go one
beyond the
# item detail screen when traversing events
});

struct( EventStruct => {
EventCode => '$', # code of event (scalar type)
EventDesc => '$', # short description of event (scalar type)
});

@VisitArray has VisitStructs pushed onto it elsewhere in the code.

(For why I replaced your '@' with '$' in each example, please see:
perldoc -q difference
)

Neat... thanks for pointing this out.

Paul Lalli

Thanks.
To everyone else who has responded, thank you also! I am testing out
your solutions as we type.

Tuan

 

[Paul Lalli]

I do know that

scalar(@{@VisitArray[$i]->PSearchStructArray})

will return the length of PSearchStructArray well enough though.

Perhaps I'm having a brain cramp here, but I can't especially see how
that could work. The arrow notation is used for either A)
dereferencing a hash or array reference to get at its elements or B)
calling a method of an object or class.

I am using Class::Struct, and indeed i am implicitly calling methods
from it. It indeed returns the correct lengths that I want.

Ahh, when you said you had structs, you actually meant structs, as in
objects of Class::Struct. My apolgies - I kinda through that word out
as a "doesn't mean what you think it means" when I scanned your
original post.

I'm reading up on Class::Struct as we speak. Much obliged.

Paul Lalli

 

[Paul Lalli]

Hello,

So I have this complex data structure I've built for myself, and I'm
trying to get this snippet to work:

$k <
scalar(@{@{@VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray}})

Ths should do it.
$l=scalar(@{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray});

Just to be pedantic, the scalar keyword and its parentheses are no-ops
there. The assignment to a scalar variable already imposes scalar
context on the array.

Assuming the above does what you want,

$l = @{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray};

should work just as well.

Paul Lalli

 

[Tuan Bui]

Hello,

So I have this complex data structure I've built for myself, and I'm
trying to get this snippet to work:

$k <
scalar(@{@{@VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray}})

Ths should do it.
$l=scalar(@{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray});

Just to be pedantic, the scalar keyword and its parentheses are no-ops
there. The assignment to a scalar variable already imposes scalar
context on the array.

Yeah, I'm a perl newbie and I'm using the constructs the llama tells me
to use

Assuming the above does what you want,

$l = @{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray};

should work just as well.

If by "just as well" you mean "syntax error," then yes It's one of
the first things I tried.

I'm pursuing the intermediate-step path right now:

my @PSARef = @{$VisitArray[$i]->PSearchStructArray};
for(my $k = 0; $k < scalar(@{$PSARef[$j]->ClickPathArray}); $k++) {

but for some reason it's returning abnormal lengths that are invariant
with different iterations of the innermost loop.

Paul Lalli

Back to the grind,
Tuan

 

[Tuan Bui]

I'm pursuing the intermediate-step path right now:

my @PSARef = @{$VisitArray[$i]->PSearchStructArray};
for(my $k = 0; $k < scalar(@{$PSARef[$j]->ClickPathArray}); $k++) {

but for some reason it's returning abnormal lengths that are invariant
with different iterations of the innermost loop.

Update: This approach actually works. The constant-length problem
seems to stem from a bug elsewhere in my code.

Thanks for all the help, peepz!

Tuan

 

[it_says_BALLS_on_your forehead]

that's great news .

one suggestion:

using C-type loops in Perl is inefficient.
The fastest For Loop would loop by value, i.e.:
for my $value (@array) {
...do something with $value...
}

or you could use the default variable $_:

for (@array) {
$_ is the element...
}


when you need an iterator, this is much faster than a C-style for loop:
for my $i (0..$#array) {
blah $array[$i];
}


hope this helps!

 

[Paul Lalli]

Ths should do it.

$l=scalar(@{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray});

Assuming the above does what you want,

$l = @{$VisitArray[$i]->PSearchStructArray[$j]->ClickPathArray};

should work just as well.

If by "just as well" you mean "syntax error," then yes It's one of
the first things I tried.

Wait. Do you actually mean that the first line of code above, from
Billy Patton, does work, but that the second line of code, from me,
does not?

What syntax error does the second give that the first does not?

Paul Lalli

 

[Tuan Bui]

Wait. Do you actually mean that the first line of code above, from

Billy Patton, does work, but that the second line of code, from me,
does not?

Yes, that's right.

What syntax error does the second give that the first does not?

syntax error at test.pl line 396, near "->PSearchStructArray["

Paul Lalli

Now I'm trying to puzzle my way through how Perl handles pointers to
structs declared with "new."

Tuan

 

[Tuan Bui]

Oops, sorry - the code from balls works. I never tried the code from
Billy.

 

[T Beck]

that's great news .

one suggestion:

using C-type loops in Perl is inefficient.

-snip-

I don't know where that got started... It could be in the camel, but I
don't remember reading that. What's more, on my machine, C-Style is
almost universally faster (if not the same speed) than (0..n) for
simple loops. For instance:
~>time perl -e '$a=1; for ($i=0; $i<10001; $i++){$a*=100}'
0.02u 0.01s 0:00.02 150.0%
~>time perl -e '$a=1; for (0..10000){$a*=100}'
0.05u 0.03s 0:00.07 114.2%

IMHO, the difference isn't a big deal one way or the other, so I say if
someone feels good writing C-Style loops, then TIMTOWTDI, indeed!

--T Beck

 

[it_says_BALLS_on_your forehead]

last year I attended a couple of lectures by Damian Conway ( he wrote,
among othe things, the Object Oriented Perl book, and he is a major
contributor to the regex engine for Perl ). he handed out his new book,
"Perl Best Practices". in it, he describes what the best "for loops"
are. i think he's pretty familiar with how Perl works behind the
scenes. I benchmarked all 3 methods, and got this:

[nagano 25] ~> benchForLoop.pl
Benchmark: running array by c iterator, array by perl iterator, array
by value, array by value using default var, each for at least 1 CPU
seconds...
array by c iterator: 1 wallclock secs ( 1.09 usr + 0.00 sys = 1.09
CPU) @ 76.15/s (n=83)
array by perl iterator: 2 wallclock secs ( 1.05 usr + 0.00 sys =
1.05 CPU) @ 122.86/s (n=129)
array by value: 1 wallclock secs ( 1.04 usr + 0.00 sys = 1.04 CPU) @
146.15/s (n=152)
array by value using default var: 2 wallclock secs ( 1.09 usr + 0.00
sys = 1.09 CPU) @ 146.79/s (n=160)

....here is the code:

$x = 0;
for my $index (0..3000) {
push( @numbers, $index );
}

sub perlIter {
my ($data_ref) = @_;
my @data = @{$data_ref};

for my $i (0..$#data) {
$data[$i] = 2;
}
}

sub cIter {
my ($data_ref) = @_;
my @data = @{$data_ref};

for (my $i = 0; $i < scalar(@data); $i++) {
$data[$i] = 2;
}
}

sub byValue {
my ($data_ref) = @_;
my @data = @{$data_ref};

for my $num (@data) {
$num = 2;
}
}

sub byDefaultValue {
my ($data_ref) = @_;
my @data = @{$data_ref};

for (@data) {
$_ = 2;
}
}

timethese(-1, {
'array by perl iterator' => '\&perlIter(\@numbers)',
'array by c iterator' => '\&cIter(\@numbers)',
'array by value' => '\&byValue(\@numbers)',
'array by value using default var' => '\&byDefaultValue(\@numbers)',
});