Difference between expr(mean(1:10)) and expr(mean(!!(1:10)) - r

Question

Going through metaprogramming sections of Hadley's book Advanced R 2nd ed, I have quite a bit of a tough time understanding the concept. I have been programming with R for a while but this is the first time I came across the concept of metaprogramming. This exercise question in particular confuses me
"The following two calls print the same, but are actually different:
(a <- expr(mean(1:10)))
#> mean(1:10)
(b <- expr(mean(!!(1:10))))
#> mean(1:10)
identical(a, b)
#> [1] FALSE
What’s the difference? Which one is more natural?"
when I eval them they both returns the same
> eval(a)
[1] 5.5
> eval(b)
[1] 5.5
when I look inside the a and b object the second object does print differently, but I am not sure what does this mean in terms of their difference:
> a[[2]]
1:10
> b[[2]]
[1] 1 2 3 4 5 6 7 8 9 10
also if I just run them without eval(expr(...)) then it will return differently:
mean(1:10)
[1] 5.5
mean(!!(1:10))
[1] 1
My guess is that without expr(...) !!(1:10) act as a double negation which with coercion essentially forcing all the numbers to be 1, hence mean of 1.
My questions are:
Why does the !! acts differently with and without the expr(...) ? I would expect eval(expr(mean(!!(1:10)))) to return the same as mean(!!(1:10)) but this is not so
I still do not quite fully grasp what is the difference between a object and b object ?
thank you in advance

Answer 1

!! here is used not as double negation, but the unquote operator from rlang.
Unquoting is one inverse of quoting. It allows you to selectively
evaluate code inside expr(), so that expr(!!x) is equivalent to x.
The difference between a and b is that the argument remains as an unevaluated call in a, while it is evaluated in b:
class(a[[2]])
[1] "call"
class(b[[2]])
[1] "integer"
The a behaviour may be an advantage in some circumstances because it delays evaluation, or a disadvantage for the same reason. When it is a disadvantage, it is of the the cause of much frustration. If the argument was a larger vector, the size of b would increase, while a would stay the same.
See section 19.4 of Advanced R for more details.

Answer 2

Here is the difference. When we negate (!) an integer vector, numbers other than 0 are converted to FALSE and 0 to TRUE. With another negate ie. double (!!), the FALSE are changed to TRUE and viceversa
!0:5
#[1] TRUE FALSE FALSE FALSE FALSE FALSE
!!0:5
#[1] FALSE TRUE TRUE TRUE TRUE TRUE
With the OP's example, it is all TRUE
!!1:10
#[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
and TRUE/FALSE can be otherwise 1/0
as.integer(!!1:10)
#[1] 1 1 1 1 1 1 1 1 1 1
thus the mean would be 1
mean(!!1:10)
#[1] 1
Regarding the 'a' vs. 'b'
str(a)
#language mean(1:10)
str(b)
#language mean(1:10)
Both are language objects and it will be evaluated to get the mean of numbers 1:10
all.equal(a, b)
#[1] TRUE
If we need to get the mean of 10 numbers, the first one is the correct way.
We could evaluate the second option correctly i.e. getting a mean value of 1, by quoteing
eval(quote(mean(!!(1:10))))
#[1] 1
eval(quote(mean(1:10)))
#[1] 5.5

Answer 3

!! has special meaing when used inside expr.
outside expr you will get different results because
!! is a double negation
even inside expr the two versions are different because
1:10 is an expression resulting in an integer vector
when evaluated, while !!(1:10) is the result of
evaluating this same expression.
an expression and its result after it is evaluated are
different things.