Does "print $ARGV" alter the argument array in any way?

Here is the example:

$a = shift; 
$b = shift; 
push(@ARGV,$b); 
$c = <>; 

print "\$b: $b\n"; 
print "\$c: $c\n"; 
print "\$ARGV: $ARGV\n"; 
print "\@ARGV: @ARGV\n"; 

And the output:

$b: file1 
$c: dir3 

$ARGV: file2 
@ARGV: file3 file1 

I don't understand what exactly is happening when printing $ARGV without any index. Does it print the first argument and then remove it from the array? Because I thought after all the statements the array becomes:

file2 file3 file1

Invocation:

perl port.pl -axt file1 file2 file3 

file1 contains the lines:

dir1 
dir2 

file2:

dir3 
dir4 
dir5 

file3:

dir6 
dir7