为什么在使用超赋值 (<<-) 运算符时，在 function 中使用变量会更改 R 中返回的值？

Question

Why do bar and baz behave differently? When bar is called both the value of a printed and the value of a in the global scope are the same, but when baz is called the printed value and the value in the global scope are different. Seemingly, the only difference is that a is used (but not defined) in a parent environment.

a = 1:3
b = 4
foo <- function(a) {
  a[1] <<- b
  print(a)
}

bar <- function(a) {
  foo(a)
}

baz <- function(a) {
  a
  foo(a)
}

bar(a) # 4 2 3
a # 4 2 3 

a <- 1:3
baz(a) # 1 2 3
a # 1 2 3

Answer 1

Complex assignment operator <<- changes the value of a variable in the parent environment. When bar is called:

• It passes a to foo
• foo then changes the value of first element of a to 4 in global environment
• After that foo prints a

> bar(a) # 4 2 3
[1] 4 2 3

The only thing to note here is, since foo was created in global environment, it searches for the value of b through lexical scoping in the environment where it was created which is again global environment. Such is the case when foo prints a in the end, it again searches for its value in the environment where it was created which is global environment. So the a will change to c(4, 2, 3) .

However when you call baz ,

• It first prints a which is the original c(1, 2, 3)
• Then it passes it to foo where the same thing that I explained above happens

So that's why the first print is the original a and the second is the modified one.