I am assigning a variable to the first element of a list.
If I make the assignment inside an if statement, I get a different value than if I assign it outside the statement.
Like this:
$ perl -E 'my @y = (0, 1) ; my $ x; if (($x) = @y) {say qq[yes ], $x} else {say q[no ], $x}'
yes 0
vs
$ perl -E 'my @y = (0, 1) ; my ($x) = @y; if ($x) {say qq[yes ], $x} else {say q[no ], $x}'
no 0
Why are the results different?
Is there a way to make the assignment inside the if statement and check its value?
The if-condition imposes a scalar context on your statement. The results are different because in the first one-liner you do:
if (2) # return value = @y (size)
And in the second case you do:
if (0) # return value = $x (value)
The return value of the statement ($x) = @y is not the value of $x, it becomes the size of @y. Like in the list counting idiom:
my $n = () = (list here);
Except in your case the first scalar assignment is implicit:
if-context = ($x) = @y
You finish with an additional question: Is there a way to make the assignment inside the if statement?
Yes, you can use the array value directly and not use list assignment.
if ( $x = $y[0] ) # the return value of the statement is 0
You can also use a subscript
if ( $x = (@y)[0] ) # the return value of the statement is 0
You can shift a value off the array, though this changes the array of course.
if ( $x = shift @y )