generating a character sequence

Rudif has asked for the wisdom of the Perl Monks concerning the following question:

Hi all

I need to generate simple sequences of $n characters, for any positive $n (within reason), starting at 'a' and wrapping back to 'a' after 'z'.
The first code line in my loop, below, generates the sequence that I need.

Question 1: Other / better / faste ways of generating this sequence, anyone?
Question 2: Why does my second code line generate something entirely different?


for my $n (25..28) {
    printf"%s\n",pack"C*",map{$_%26+ord'a'}0..$n-1; # ok
    printf"%s\n",pack"C*",map{ord'a'+$_%26}0..$n-1; # not ok
}

__END__

abcdefghijklmnopqrstuvwxy
0123456789111111111122222
abcdefghijklmnopqrstuvwxyz
01234567891111111111222222
abcdefghijklmnopqrstuvwxyza
012345678911111111112222220
abcdefghijklmnopqrstuvwxyzab
0123456789111111111122222201
[download]

Rudif

Comment on generating a character sequence Download Code

Replies are listed 'Best First'.
Re: generating a character sequence by I0 (Priest) on Aug 22, 2001 at 06:15 UTC
`printf"%s\n",pack"C*",map{(ord'a')+$_%26}0..$n-1;`	[reply] [d/l]
Re: Re: generating a character sequence by Hofmator (Curate) on Aug 22, 2001 at 15:01 UTC
Let me add a short explanation to that. This is a precedence problem, your statement is read as (I only show the map part here): `map { ord('a'+$_%26) } 0..$n-1;`Both things are added first and then passed to ord - btw, with use warnings you get a warning about 'a' not being numeric in addition ... this should have been sufficient as an explanation ;-) So I0's solution works nicely or you can also write `map { ord('a')+$_%26 } 0..$n-1;` according to the rule: 'What looks like a function, is a function!' -- Hofmator	[reply] [d/l] [select]
Re: generating a character sequence by little (Curate) on Aug 22, 2001 at 03:04 UTC
when `printf"%s\n",pack"C",map{ord'a'+$_%26}0..$n-1; # not ok` [download] then $_ is ord'a' and not one of the intended values of 0..(25..28)-1 Have a nice day All decision is left to your taste*	[reply] [d/l]
Re: generating a character sequence by runrig (Abbot) on Aug 22, 2001 at 03:05 UTC
I'm guessing pack is faster, but just for variety here's another option: `my $n = 42; my $str = join '',(("a".."z") x ($n/26 +1))[0..$n-1]; print $str,"\n"; print length($str),"\n";` [download]	[reply] [d/l]
Re: generating a character sequence by George_Sherston (Vicar) on Aug 22, 2001 at 03:23 UTC
Do these have anything to offer? They do the same thing in slightly different ways. Both let you start your sequence where you like. `$l = a; #or whatever letter you want to start with for my $n (1..28) {print $l; $l++; $l=substr($l,0,1)} $l = a; #or whatever letter you want to start with for my $n (1..28) {print $l; $l eq 'z' ? $l = 'a' : $l++ }` [download] § George Sherston	[reply] [d/l]


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