Range operator decrementing from largest to smallest: 10..1 [duplicate]

问题

This question already has answers here:

Alternate for making such a thing work in perl : `for(10…0)` (3 answers)

Closed 5 years ago.

Perl has a range operator, which when used in a foreach loop, does not create a temporary array:

foreach (1 .. 1_000_000) {
    # code
}

If the first integer is smaller than the second integer, then no iterations are run:

foreach (1_000_000 .. 1) {
    # code here never runs
}

I could use the reverse built-in, but would this keep the optimisation of not creating a temporary array?

foreach (reverse 1 .. 1_000_000) {
    # code
}

Is there a way that's as pretty and fast as the range operator for decreasing numbers, rather than increasing ones?

回答1:

Non pretty solution,

for (my $i=1_000_000; $i >= 1; $i--) {

   print "$i\n";
}

回答2:

While for is nice, while might be better suited.

my $n = 1_000_000;
while ($n >= 1) {
    ...
} continue {  # continue block will always be called, even if next is used
    $n--;
}

回答3:

You could just subtract the number you've got from one more than the top of the range:

foreach (1 .. 1_000_000) {
    my $n = 1_000_001 - $_;
    ...
}

and

for (-1_000_000 .. -1) {
    my $n = -$_;
    ...
}

来源：https://stackoverflow.com/questions/26505835/range-operator-decrementing-from-largest-to-smallest-10-1