implicit_assignment_linter() misses assignments in later arguments of calls

[MichaelChirico]

lintr::lint("foo(a, x <- y)\n", lintr::implicit_assignment_linter())

[MichaelChirico]

@IndrajeetPatil shouldn't this lint?

lintr/tests/testthat/test-implicit_assignment_linter.R

Lines 92 to 95 in 8ac582c

	map(
	.x = 1:4,
	.f = ~ .x + 1 -> x
	)"),

Operator precedence means this is equivalent to

map(.x = 1:4, .f = x <- (~.x+1))

It makes sense to me that the version where the mapper is doing assignment shouldn't lint:

map(.x = 1:4, .f = ~(.x+1 -> x))

[MichaelChirico]

This also looks out of place:

lintr/tests/testthat/test-implicit_assignment_linter.R

Line 245 in 8ac582c

expect_lint("with_options(list(digits = 3L), x <- getOption('digits'))", NULL, linter)

If it's meant not to lint, shouldn't with_options() be part of the default except= instead?

[IndrajeetPatil]

@IndrajeetPatil shouldn't this lint?

lintr/tests/testthat/test-implicit_assignment_linter.R

Lines 92 to 95 in 8ac582c

	map(
	.x = 1:4,
	.f = ~ .x + 1 -> x
	)"),

Operator precedence means this is equivalent to

map(.x = 1:4, .f = x <- (~.x+1))

It makes sense to me that the version where the mapper is doing assignment shouldn't lint:

map(.x = 1:4, .f = ~(.x+1 -> x))

I am still not sure why this should lint with implicit assignment linter because there is no assignment inside a function call here. If anything, object_usage_linter() will produce a lint here for unused x variable.

If the issue is about operator precedence, I feel like this should be covered under a different linter? Am I missing something?

[MichaelChirico]

the linked test (lines 92-95) does have a top-level assignment in a call -- the .f argument is `<-`(x, ~.x+1)

[IndrajeetPatil]

I am still confused. I thought the lambda syntax for anonymous function creates a function that is passed as argument.
So, in the function call map(), the second argument being passed is the anonymous function, and not `<-`(x, ~.x+1) assignment inside that anonymous function. Am I mistaken about that?

identical(
  purrr::map(
    .x = 1:4,
    .f = ~ .x + 1 -> x
  ),
  purrr::map(
    .x = 1:4,
    .f = function(.x) .x + 1 -> x
  )
)
#> [1] TRUE

^{Created on 2023-09-12 with reprex v2.0.2}

[MichaelChirico]

after the ~ example x will exist in the calling environment; not so after the function one.