for loops

[root]

For loops allow you to do something a certain number of times in their most basic forms;

They take on the form:

for(initial_expression; test_expression; change_expression){
   do_something;
}

This is equivalent to a while statement that looks like

initial_expression;
while(test_expression){
   do_something;
   change_expression;
}

This for loop will print the numbers 1 through 100 on separate lines;

for($i=1; $i<=100; $i++){
   print "$i\n";
}

First the variable is set to 1, then it is tested to see if it is less than or equal to 100. Then it is printed. Then the change expression is evaluated and $i is incremented by 1. Then the text expression is evaluated again since $i is equal to 2 it is less than or equal to 100 so the loop is repeated until 100 is printed, $i is incremented to 101, $i is no longer less than or equal to 100 so the for loop is finished.

Now for foreach loops

[nop]

When using multiple initializations in a for loop, I found that scoping didn't work as I expected it would. For example, in this fake snippet

use strict;
### this code contains an error
for (my $i=1,
     my $j = $i + 100;
     $i<50;
     $i++, $j++) {
    print "i=$i j=$j hello!\n";
}

the line

     my $j = $i + 100;

generates an error because $i doesn't exist yet. Not a big deal, and trivial to work-around, but can someone explain why scoping works this way here? nop

[mirey]

I do not think you are looking at a scope issue. You are including code in the defenition of your for loop. Try this: for($i; $i < 50; $i++) { $j = $i + 100; $j++; print "i=$i j=$j hello!\n"; } Another question would be why do you have $j = $i + 100; and $j++; instead combine them the following way $j = $i + 101;

[hossman]

can someone explain why scoping works this way here?

That question implies that the scoping works differently elsewhere -- it doesn't. you get that error regardless of the for loop...

laptop:~> perl -Mstrict -e 'my $i=1, my $j=$i+1; print $i + $j;'
Global symbol "$i" requires explicit package name at -e line 1.
Execution of -e aborted due to compilation errors.

...the behavior of the comma oporator is not changed by being used in a for loop.

[apotheon]

Because the value of $j has no bearing on the test expression, it doesn't belong in the initial and change expressions of the for statement. As such, they should be included in the set of code executed by the loop.

- apotheon
CopyWrite Chad Perrin

[Falthor]

just wondering here... if a value is not assigned to a variable at the startup does that mean that the value is automatically assigned to 0 ? Jason Temple AKA Falthor Assistant Operartions Manager Neurogames

[davorg]

No. An unassigned variable contains the special value undef. This will evaluate as false under some circumstances. 0 (and, indeed, '0') also evaluates as false, which is where the confusion arises.

--
<http://www.dave.org.uk>

European Perl Conference - Sept 22/24 2000
<http://www.yapc.org/Europe/>

[Anonymous Monk]

Now how does it know to print on the next line ? do we specify anything or it just does it ?

[Anonymous Monk]

Suggest you re-visit the String scalars section. The \n is a newline character to be printed. Thus after each value, a newline is printed (CR/LF) so the next print will start on a new line

[Willman023]

I think it would be useful to show an additional way the for loop can be used. Although the C-like for loop can do the same thing, I find myself using the one below more and more.

for $countdown (10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 'Liftoff!') {
    print $countdown, "\n";
}

for $y (0..$x) {
    print $array[$y], "\n";
}

for (@array) { s/value1/value2/ } # Substitues value2 for value 1.

bW

[xalphabetx]

okay so just trying to understand the logic of these new commands (for me) for($i=1; $i<=100; $i++) okay so now I understand how this line is read and the syntax but if the line starts over at the end doesnt $i return to the value of 1? I know already that this code works obviously but howcome? in my brain it says it should be starting $i back to a value of $i=1 and the evaulating the rest again so I guess Im just not getting it any help??

[amarceluk]

$i++ is the equivalent of

$i=$i+1;

where the = is an operator. It's not just sitting there, like it does in math, saying "these two things are equal"; it tells Perl to make the two things equal. So here it's saying "take the current value of $i and add 1, and make that the new value of $i." Then the for loop continues with the new value of $i. This happens each time the loop runs, until $i reaches 100.

_________
"Abby-somebody. Abby-normal."
—Young Frankenstein

[apotheon]

On second and following iterations, the for loop ignores the initial assignment of value to $i and uses the incremented value (or decremented, if you use a decrement operator instead of an increment operator) instead. The $i=1 assignment happens once, and once only, in for($i=1; $i<=100; $i++).

- apotheon
CopyWrite Chad Perrin