Raku Books / Perl 6 at a Glance / Operators / Prefixes

?, so

? is a unary operator casting the context to a Boolean one by calling the Bool method on an object.

say ?42; # True

The second form, so, is a unary operator with lower precedence.

say so 42;   # True
say so True; # True
say so 0.0;  # False

CodeBlockPlaceholder3raku my Str $a = ~42; say $a.WHAT; # (Str) CodeBlockPlaceholder4raku my $x = 41; say ++$x; # 42 CodeBlockPlaceholder5

CodeBlockPlaceholder6raku my $f = “file001.txt”;

++$f; say $f; # file002.txt

++$f; say $f; # file003.txt CodeBlockPlaceholder7raku my $x = 42; say –$x; # 41 CodeBlockPlaceholder8raku my $x = 10; my $y = +^$x; say $y; # -11 (but not -10) CodeBlockPlaceholder9

Compare this operator with the following one.

?^ is a logical negation operator. Please note that this is not a bitwise negation. First, the argument is converted to a Boolean value, and then the result is negated.

my $x = 10;
my $y = ?^$x;
say $y;       # False
say $y.WHAT;  # (Bool)

CodeBlockPlaceholder11raku .print for ^5; # 01234 CodeBlockPlaceholder12

This code is equivalent to the following, where both ends of the range are explicitly specified:

.print for 0..4; # 01234

| flattens the compound objects into a list. For example, this operator should be used when you pass a list to a subroutine, which expects a list of scalars:

sub sum($a, $b) {
    $a + $b
}

my @data = (10, 20);
say sum(|@data); # 30

Without the | operator, the compiler will report an error, because the subroutine expects two scalars and cannot accept an array as an argument:

Calling sum(Positional) will never work with declared signature ($a, $b)

temp creates a temporary variable and restores its value at the end of the scope (like it does the local built-in operator in Perl 5).

my $x = 'x';
{
    temp $x = 'y';
    say $x; # y
}
say $x;     # x

CodeBlockPlaceholder17raku my $var = ‘a’; try { let $var = ‘b’; die; } say $var; # a ```

With a die, this example code will print the initial value a. If you comment out the call of a die, the effect of the assignment to b will stay, and the variable will contain the value b after the try block.

The let keyword looks similar to the declarators like my and our, but it is a prefix operator.

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