Re: porting C code to Perl

haukex and syphilis have already provided some good answers but I thought that I would point out some things that I didn't see mentioned.

Perl's int is not the same as C's floor(). What you want is POSIX::floor. It has the identical behavior as C's floor().

Perl has a pre-decrement just like C.

    # for(int i = len; i > 0; --i) { # C
    # for(my $i = $len; $i > 0; $i--){ # incorrect Perl
    for(my $i = $len; $i > 0; --$i){ # correct Perl
[download]

YMMV as I haven't tried it out with these changes implemented.

Comment on Re: porting C code to Perl Download Code

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Re^2: porting C code to Perl by Discipulus (Canon) on Oct 23, 2017 at 19:55 UTC
Hello Mr. Muskrat and thanks for your reply, > Perl's int is not the same as C's floor().. I already admitted my superb ignorance but i read here that in C `floor` turns `1.6 1.2 2.8 2.3` into `1 1 2 2` or simply 'This function returns the largest integral value not greater than x'. This is identical to the `int` function where any decimal are stripped out: `say int $_ for qw(1.6 1.2 2.8 2.3)` > Perl has a pre-decrement just like C.. I have meditated a bit over this before trying to port the code and, sincerely I didnt read the C documents about post and pre increment, but as I have always understood the third part of the Perl's C style `for` loop is a statement on it's own: ie `++$var;` or `$var++;` are identical. Infact `for($i=0;$i<4;$i++){say $i}` and `for($i=0;$i<4;++$i){say $i}` produce the very same, expectable results. Switching between them in the above code does not change the result. L* There are no rules, there are no thumbs.. Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.	[reply] [d/l] [select]
Re^3: porting C code to Perl (updated) by haukex (Archbishop) on Oct 23, 2017 at 20:08 UTC
the third part of the Perl's C style `for` loop is a statement on it's own: ie `++$var;` or `$var++;` are identical. Yes, that's correct, when they're written standalone as in `for(;;$var++)`. `++$var` vs. `$var++` only makes a difference when its return value is used: `++$var` will first increment the variable and then return the new value, while `$var++` will return the old value and then increment the variable. `$ perl -le 'for ( my $x=0; $x<3; print $x++ ) { }' 0 1 2 $ perl -le 'for ( my $x=0; $x<3; print ++$x ) { }' 1 2 3` [download] Update: In fact, note what Perl does here (edited for brevity): `$ perl -MO=Deparse -e 'for ( my $x=0; $x<3; print $x++ ) { }' for (my $x = 0; $x < 3; print $x++) { (); } $ perl -MO=Deparse -e 'for ( my $x=0; $x<3; print ++$x ) { }' for (my $x = 0; $x < 3; print ++$x) { (); } $ perl -MO=Deparse -e 'for ( my $x=0; $x<3; ++$x ) { }' for (my $x = 0; $x < 3; ++$x) { (); } $ perl -MO=Deparse -e 'for ( my $x=0; $x<3; $x++ ) { }' for (my $x = 0; $x < 3; ++$x) { (); }` [download]	[reply] [d/l] [select]
Re^3: porting C code to Perl by soonix (Canon) on Oct 23, 2017 at 20:31 UTC
For this particular use case, it doesn't matter, but with negative numbers, int and floor give different results (this is mentioned explicitly in the first link)	[reply]


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