如何将嵌套函数的访问范围定义为全局？

I have an R code which looks like:

我有一个R代码,看起来像:

func1 = function(a) {

  func2 = function(a) {
    return(a+2)
  }

  func3 = function(a) {
    return(a+3)
  }
  return(a+func2(a))
}

Is it possible that I be able to call func2 or func3 from outside func1? eg. How do I run:

我是否有可能从func1外部调用func2或func3?例如。我该怎么办:

x <- func2(10) #from the console?

2 个解决方案

#1

You could create a function closure:

你可以创建一个函数闭包:

##I've removed the brackets and return to shorten the function
func1 = function(a) {
  func2 = function(a) a+2
  func3 = function(a) a+3
  return(list(func2=func2, func3=func3))
}

You can use the closure to share variables:

您可以使用闭包来共享变量:

func1 = function(a) {
    a = a
    func2 = function() a + 2
    func3 = function() a + 3 
    return(list(func2=func2, func3=func3))
}

f = func1(50)
f$func2()
f$func3()

#2

No You can't.

不,你不能。

func2 is defined in the scope of func1.

func2在func1的范围内定义。

But The question is really ambiguous ! Any netsed function has access to the global scope.

但这个问题真的很模糊!任何netsed函数都可以访问全局范围。

a <- 10
function <- f(x){
   g <- function(y=x) x+a
   g(x)
}

here function g is nested and has a free variable a. the interpreter lokks in the scope of g and f and next looks for a value for a in the local frame of f(the global)

这里函数g是嵌套的并且有一个*变量a。解释器在g和f的范围内lokks,然后在f的本地帧中寻找a的值(全局)

Why defining a nested function as nested if you want access to it as global?

如果要将嵌套函数作为全局访问,为什么要将嵌套函数定义为嵌套函数?

#1