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

Question

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?

Answer 1

Afaik:

• Like all "statement prefixes", do consumes a "blorst" (block or statement) on its right and treats it as a statement.

• do's only special power is that it evaluates to the value of the (last) statement. I think this is the only sense in which there's any sense of "expression" in its operation.

This interacts with the following behaviors, which are unrelated to do:

• When a "bare" block ({...}) is treated as a statement, it runs the block.

• Other blocks, treated as a statement, don't (or shouldn't), run.

I think the fix to the doc sentence is something like:

The simplest way to run a bare block where it cannot be is not behaving as a stand-alone statement is by writing do before it"

Though I don't know where/if the doc talks about the difference between bare blocks and other blocks.

Answer 2

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.