-1 ** 0 == -1 ???

[kj]

I can't think of even the faintest justification for this:

% perl -le 'print( -1 ** 0 )'
-1

Absolutely horrible! Perl just showed itself wholly unsuitable
for any programming that involves numerical computation...

Still in shock...

Kynn

 

[kj]

OK, egg on my face. I see.

Sorry,

G.

I can't think of even the faintest justification for this:

% perl -le 'print( -1 ** 0 )'
-1

Absolutely horrible! Perl just showed itself wholly unsuitable
for any programming that involves numerical computation...

 

[Tim McDaniel]

OK, egg on my face. I see.

If you see, it's best to explain it so that other people can see.

I did a "man perlop", and the table is

Perl operators have the following associativity and precedence,
listed from highest precedence to lowest. Operators borrowed
from C keep the same precedence relationship with each other,
even where C's precedence is slightly screwy. (This makes
learning Perl easier for C folks.) With very few exceptions,
these all operate on scalar values only, not array values.

left terms and list operators (leftward)
left ->
nonassoc ++ --
right **
right ! ~ \ and unary + and -
left =~ !~
left * / % x
left + - .
left << >>
nonassoc named unary operators
nonassoc < > <= >= lt gt le ge
nonassoc == != <=> eq ne cmp
left &
left | ^
left &&
left ||
nonassoc .. ...
right ?:
right = += -= *= etc.
left , =>
nonassoc list operators (rightward)
right not
left and
left or xor

So what I think you see is that unary "-" has lower precedence than
"**", so
-1 ** 0
is equivalent to
-(1 ** 0)
Right?

 

[RedGrittyBrick]

OK, egg on my face. I see.
Sorry,

When startled, I try *not* to jump to the conclusion that whatever
language I am using is wholly unsuitable for any programming that
involves numerical computation.

I have done this for so long that it no longer requires much effort on
my part.

I assume that any programming language older than a few months with
widespread usage is suitable for numeric computation and that any odd
effects I am seeing are due to my own ignorance, stupidity or
foolishness - this has proved to be a good initial assumption. I commend
it to you

 

[Tim McDaniel]

I assume that any programming language older than a few months with
widespread usage is suitable for numeric computation

Um, you might be over-optimistic.

There's the famous feature of PL/I where 25 + 1/3 either throws a
fixed-point overflow error or evaluates to 5.33333... I've seen both
statements and I have no way to test it;
<http://publib.boulder.ibm.com/infocenter/comphelp/v7v91/topic/com.ibm.aix.pli.doc/ibml2d41004828.htm>
says it's an error but also mentions left-truncation.

The dc program was designed to do nothing but be a Reverse Polish
Notation calculator, but by default it does only pure integer
calculation (you can enter numbers with a decimal point: the
fractional part will be discarded silently). You can change that, but
it provides only a fixed number of decimal places.

For that matter, it surprises new programmers when they learn that 1/3
equals 0, or that 1.0/3.0*3.0 does not necessarily equal 1.0, in many
languages.

(I have a dim memory that older Visual Basics were loosey-goosey about
mixing booleans and integers, and that the "not" operator is a bitwise
"not" instead of a logical one, and FALSE is 0, and integers are
ones-complement, so for all integers I except -1, both I and NOT I
were considered true. But I'm not at all 100% certain of that, and
that's not really numeric.)


But that's really quibbling. Your general point is quite true.
It's really REALLY rare for someone to find a new fundamental flaw in
a long-standing language. By far the highest chance is that they're
misunderstanding something. There's a much smaller chance that
they've hit a known long-standing problem.

 

[neilsolent]

Hang on ....

perl -le 'print (-1)**0'
-1

perl -e 'print ((-1)**0)'
1

So I don't this is working. Isn't the latter just giving the number of
elements in the list ?
Mathematically, (-1) to the power 0 should be 1. So the first should
return 1 as well.

What am I missing ?

 

[Josef Moellers]

Hang on ....

perl -le 'print (-1)**0'
-1

perl -e 'print ((-1)**0)'
1

So I don't this is working. Isn't the latter just giving the number of
elements in the list ?
Mathematically, (-1) to the power 0 should be 1. So the first should
return 1 as well.

What am I missing ?

IIUC the first parses as "print(-1) ** 0" (blanks added for emphasis).

 

[sln]

Hang on ....

perl -le 'print (-1)**0'
-1

perl -e 'print ((-1)**0)'
1

So I don't this is working. Isn't the latter just giving the number of
elements in the list ?
Mathematically, (-1) to the power 0 should be 1. So the first should
return 1 as well.

What am I missing ?

I ran this script a thousand times. Still the same results:

use strict;
use warnings;

my $fv1 = -1**0;
my $fv2 = (-1)**0;

print "$fv1 , $fv2 \n";


__END__

-1 , 1

 

[John W. Krahn]

Hang on ....

perl -le 'print (-1)**0'
-1

perl -e 'print ((-1)**0)'
1

So I don't this is working. Isn't the latter just giving the number of
elements in the list ?
Mathematically, (-1) to the power 0 should be 1. So the first should
return 1 as well.

What am I missing ?

Turn on warnings to find out:

$ perl -wle 'print (-1)**0'
print (...) interpreted as function at -e line 1.
Useless use of exponentiation (**) in void context at -e line 1.
-1



John

 

[Tad J McClellan]

Turn on warnings to find out:

$ perl -wle 'print (-1)**0'
print (...) interpreted as function at -e line 1.
Useless use of exponentiation (**) in void context at -e line 1.
-1

And for the curious that would like more on this, it is covered
in the 3rd paragraph in:

perldoc perlfunc