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Reputation: 9600

Does the `do` keyword run a block or treat it as an expression?

The docs state that "The simplest way to run a block where it cannot be a stand-alone statement is by writing do before it" and provide the following example:

# This dies half of the time 
do { say "Heads I win, tails I die."; Bool.pick } or die; say "I win.";

However, do doesn't seem to cause all blocks to run. In particular, it doesn't seem to run blocks with a signature:

do -> $a = 42 { say "ran with $a"; 0 } or die; say 'done'; # OUTPUT: «done»

So would it be better to say that do treats a block as an expression which sometimes causes it to be run? Or is Rakudo incorrect in its behavior here? Or is my understanding incorrect?

Upvotes: 9

Views: 242

Answers (1)

Brad Gilbert
Brad Gilbert

Reputation: 34110

The do keyword turns a statement into an expression.

my @b = do for @a { $_² if $_ %% 2 }

my $c = do if @b > 4 { +@b } else { 3 }

A bare block can either be a statement:

my $a = 1;
say $a;
{
  my $a = 2;
  say $a;
}
say $a;

or it can be an expression:

my $code = { … }

When you use do with a bare block it acts as the statement form, but also returning the last value evaluated.

my $result = do { say 'hi'; 5 }
# hi

say $result;
# 5

A pointy block on the other hand is always an expression.

my $code = -> $ { … }

Even if you don't assign it anywhere

say 'a';
-> { say 'hi' }
do -> { say 'bye' }
say 'b';

# a
# b

It is pointless to use do with something that is already an expression.

Note that the do keyword in the compiler takes a blorst (block or statement) instead of just a statement because it doesn't make much sense to turn the block into a statement just to undo that and turn it into an expression-like block so that it will return its last value.

Basically blorst only ever captures a statement, but it captures the block type of statement directly to simplify the internals of the compiler.

Upvotes: 2

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