什么时候由invisible()返回的对象停止不可见?

问题描述

?invisible说

该括号表示隐身不会永远持续下去，但是我找不到能解释消失的任何信息.我特别想知道这样的结构(来自这个旧的答案我的):

That parenthetical implies that the invisibility will not last forever, but I can't find anything that explains when it goes away. I'm particularly wondering about constructs like this one (from this old answer of mine):

printf <- function(...) invisible(print(sprintf(...)))

外面的invisible可能 不必要(因为print已经将其返回值标记为不可见). withVisible()报告说该函数的返回值是不可见的，但是我不知道该语言是否能保证该返回值，或者不知道它在当前实现中的工作方式.

where the outer invisible is probably unnecessary (because print already marked its return value invisible). withVisible() reports that this function's return value is invisible either way, but I don't know whether that is guaranteed by the language, or just the way it happens to work in the current implementation.

推荐答案

通过反复试验:

# invisible
withVisible(invisible())$visible
[1] FALSE

### passing the invisible value through a function seems to
# preserve the invisibility
withVisible(identity(invisible()))$visible
[1] FALSE

# the <- operator just returns its arguments, so it confirms the above
withVisible(i <- invisible())$visible
[1] FALSE
# but the assigned value is no longer invisible
withVisible(i)$visible
[1] TRUE

### passing an invisible value as argument keeps the invisibility
f <- function(x) withVisible(x)$visible
f(1)
[1] TRUE
f(invisible(1))
[1] FALSE

### every other operation seems to cancel the invisibility.
# e.g. assigning an invisible value cancels the it
i <- invisible()
withVisible(i)$visible
[1] TRUE

withVisible(invisible(1) + 1)$visible
[1] TRUE

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