R语言中的函数组合：基于闭包扩展示例的问题排查

Let's break down the issues step by step:

1. Missing Required Argument: When you call square <- power(exponent = 3), the power function expects two arguments (exponent and base_f), but you only provided one. R will throw an error about the missing base_f parameter right here.
2. Scope of exponent: Even if you passed base_f to power, the global base_f references exponent, which isn't defined in its environment (the global workspace). Since base_f was created outside power, it doesn't have access to the exponent variable inside power—closures depend on the inner function being defined within the enclosing function to capture its environment.
3. Undefined x in power: The power function tries to call base_f(x), but x isn't defined anywhere inside power. You need a way to pass x through to the function when it's eventually called.

If you want to keep base_f as a standalone function while still using closure logic to bind the exponent, here's a clean approach:

First, define base_f to accept both x and exponent as parameters—this makes it reusable with any exponent value:

base_f <- function(x, exponent) {
  x ^ exponent
}

Then, rewrite power to take an exponent and return a new function that remembers that exponent. This returned function will take x as its argument and call base_f with the stored exponent:

power <- function(exponent) {
  # Return a closure that captures the exponent and takes x
  function(x) {
    base_f(x, exponent)
  }
}

Now you can use it exactly like your original working code:

# Create a cube function (since we're using exponent=3)
cube <- power(exponent = 3)
# Call it with x=2
cube(x = 2) # Returns 8

This setup keeps base_f independent while leveraging closures to "bake in" the exponent value when you create your specialized function (like cube). The closure returned by power retains access to the exponent variable even after power has finished running.

内容的提问来源于stack exchange，提问作者Fernando Hoces De La Guardia