问为什么说found.remove()不是一个函数

EN

const names = ['Milan', 'Poudel', 'Soudel', 'Academia']
const filteredNames = names.filter((name) => !name.includes('ou'))

console.log(filteredNames)

let names = ['Milan', 'Poudel', 'Soudel', 'Academia']    
names = names.filter(function(name) {
  return !(name.includes('ou'));
})
console.log(names);

// Array of strings is initialized
const names = ['Milan', 'Poudel', 'Soudel', 'Academia']

/**
* Filter the names array and return 
* a new array of strings that INCLUDE
* strings with 'ou' which will
* be stored in a global variable 'found'
*/
found = names.filter(function(name) {
  return name.includes('ou')
})

// Result ['Poudel', 'Soudel']
console.log(found);


// Exception thrown because Array.remove is not a function
found.remove()

/**
* names is still in-tact and in it's original
* state since you returned the result of
* your names.filter function into a global
* variable called 'found'
*/
console.log(names)

Array.prototype.remove = function (name) {
  return this.filter( e => !e.includes(name));
}

let names = ['Milan', 'Poudel', 'Soudel', 'Academia']

names = names.remove('ou')
console.log(names)