R 中的闭包，如 Python

首先，考虑以下计算函数被调用次数的Python代码:

def counter(fn):
    count = 0
    def inner(*args, **kwargs):
        nonlocal count
        count +=1
        print('Function {0} was called {1} times'.format(fn.__name__, count))
        return fn(*args, **kwargs)
    return inner

def add(a,b):
    return a+b
def mult(a,b):
    return a*b
add = counter(add)
mult = counter(mult)
add(1,2)
add(2,3)
mult(1,5)
#output
Function add was called 1 times
Function add was called 2 times
Function mult was called 1 times

现在，我正try 在R中执行相同的方法，如下所示:

counter <- function(fn) {
  cnt <- 0
  inner <- function(...) {
    cnt <<- cnt + 1
    print(paste("Function", match.call(), "was called", cnt, "times\n"))
    return(fn(...))
  }
  return(inner)
  
}
add <- function(a, b) a + b
mult <- function(a, b) a*b
cnt_add <- counter(add)
cnt_add(1, 4) 
cnt_add(3, 9)
[1] "Function cnt_add was called 1 times\n"
[2] "Function 1 was called 1 times\n"   #<---- !!!!!!!!!!!!!!  L1  
[3] "Function 4 was called 1 times\n"   #<---- !!!!!!!!!!!!!!  L2 
[1] 5
[1] "Function cnt_add was called 2 times\n"
[2] "Function 3 was called 2 times\n"   #<---- !!!!!!!!!!!!!!  L3 
[3] "Function 9 was called 2 times\n"   #<---- !!!!!!!!!!!!!!   
[1] 12
cnt_mult<-counter(mult)
cnt_mult(1,6) 
[1] "Function cnt_mult was called 1 times\n"
[2] "Function 1 was called 1 times\n"   #<---- !!!!!!!!!!!!!!  L4  
[3] "Function 6 was called 1 times\n"   #<---- !!!!!!!!!!!!!!  L5  
[1] 6

a)我以为"Function？Is Call？Times\n"，但为什么打印L1、L2、L3、L4、L5？

当我try 时为b)(就像在Python中一样)

add <- counter(add)
add(3, 4)

我得到一个错误:Error: evaluation nested too deeply....

c)个 为了避免b中的错误，我try 了如下操作，但仍收到错误

cnt_add <- counter(add)
add <- cnt_add
add(6, 8)

我发现，如果我调用cnt_add函数一次，没有发生错误(除了控制台中的另外两行):

cnt_add <- counter(add)
cnt_add(1, 8)
[1] "Function cnt_add was called 1 times\n"
[2] "Function 1 was called 1 times\n"      
[3] "Function 8 was called 1 times\n"      
[1] 9
add <- cnt_add
add(6, 8)
[1] "Function add was called 2 times\n" "Function 6 was called 2 times\n"  
[3] "Function 8 was called 2 times\n"  
[1] 14

但是为什么"函数加被调用了2次"，我就调用了一次！为什么它至少需要一次通话才能正常工作？

如何解决这些问题？我不想要其他方法，因为这只是闭包中的一种做法.