You used an if in a do block. So I'll give you an answer for a for loop, maybe in a function ;-)
perlsub:

If the last statement is a loop control structure like a foreach or a while, the returned value is unspecified

perlsyn:

Perceptive Perl hackers may have noticed that a for loop has a return value, and that this value can be captured by wrapping the loop in a do block. The reward for this discovery is this cautionary advice: The return value of a for loop is unspecified and may change without notice. Do not rely on it.

I quote those because I couldn't even find the relevant documentation for other kind of blocks. As far as I can tell it's not just unspecified, it's undocumented. The reason is quite clearer though, some freedom is left for perl to try and optimize your statement. So if (0) {SOMETHING} might be removed entirely, or replaced by a no-op. The condition in if (1) might be removed because the block is always executed. Actually this is confirmed by deparsing the code (perl -MO=Deparse yourfile.pl):

BEGIN { $/ = "\n"; $\ = "\n"; }
use strict;
use warnings;
my $uninitialized;
print do {
    $uninitialized
};
print do {
    ()
};
print do {
    do {
        ()
    }
};
print do {
    0
};
print '----------';
print scalar do {
    $uninitialized
};
print scalar do {
    ()
};
print scalar do {
    do {
        ()
    }
};
print scalar do {
    0
};
[download]

You can see that the if (1) {} is turned into do {}, and the if (0) is turned into 0. But this is not reliable behaviour, that's just a side effect of perl trying to replace some statements by the best equivalent it could find.