为什么在递归 R 函数中“+”失败时“Sum()”会成功？

Question

I am experimenting with the functional programming paradigm in R. I have defined a function that sums a sequence of integers from n to m . When I use sum() the function returns the expected result:

sumRange <- function(n, m) {
    if (n <= m) { 
        return(sum(n, sumRange((n + 1), m)))
    }
}

sumRange(1, 10)
# [1] 55

However, when I use the + operator the function returns numeric(0) :

sumRange <- function(n, m) {
    if (n <= m) {
        return(n + sumRange((n + 1), m))
    }
}

sumRange(1, 10)
# numeric(0)

Why does the operator + not work in this recursive function? Is there a way to rewrite the function so that it does?

Answer 1

The issue is that you never specify an else condition, hence at the end of the recursion it appears that R is returning NULL when the if condition fails. Returning 0 as the else condition fixes your problem:

sumRange <- function(n, m) return(ifelse (n <= m, (n +  sumRange((n+1), m)), 0))
sumRange(1, 10)
[1] 55

Note that this is essentially defining a base case for your recursion. A base case, when hit, ends the recursion and causes the calls on the stack to be unwound.

To see the issue with the way you phrased your code, try writing out your function explicitly:

sumRange <- function(n, m) {
    if (n <= m) {
        return(n + sumRange((n+1), m))
    }
    // but what gets returned if n > m ?
    // this is undefined behavior
}

I'm not an R guru, but my understanding is that R was written in C, and C might allow a recursion like this with no else condition. But the behavior is not well defined and you should not be relying on it.

Demo

Answer 2

If there is no return (using a explicit or implicit return statement) is executed, then R functions seems to return a NULL object.

If you add numerical value to a this object, it will simply return numeric(0) .

So, what happens in the second case is that when n reaches 11, it returns a NULL object, and goes back adding values to it. But NULL + 10 + 9 .. = numeric(0) .

Check this with

no_ret <- function ()
{
 # just return nothing
}

obj <- no_ret()
obj
# NULL 
class(obj)
# "NULL
new_obj <- obj + 10
new_obj
# numeric(0)

When the first function is executed, the what the sum statement get is a vector with a NULL in it. For example, vec <- c(NULL, 10, 9,...) which is actually vec <- c(10, 9, ...) , so you get the expected outcome.

> c(NULL, 10:1)
 [1] 10  9  8  7  6  5  4  3  2  1
> sum(NULL, 10:1)
[1] 55

> NULL + 10:1
integer(0)