什么语法错误导致函数的 'text' 在此 JS 闭包中返回，而不是让它递增？

Question

我只是想创建一个递增函数，但我的运气为零。我只能设法得到这种类型的输出，或者我得到一个不递增的整数。

const count = () => {
  let countUp = function(n) {
    return function() {
      n += 1;
      return n;
    }
  };
  return countUp;
};

console.log(count());

输出：

function(n) {
    return function() {
      n += 1;
      return n;
    }
  }

Answer 1

下面是一个 ES5 脚本，它会在每次调用时递增一个计数器。下面的代码使用了 immediately invoked function expressions and closures.

//The number we want to start counting up from
var startAt = 0;

//Immediately invoke the anonymous outer function - this will return a function into our "count" variable. The closure will retain access to local variable j which is passed as an argument.
var count = (function(i) {
  var j = i;
  return function() {
    j += 1;
    return j;
  }
}(startAt));

//Call count 10 times
var k;
for (k = 0; k < 10; k++) {
  console.log(count());
}

下面是使用更简洁的 ES6 语法编写的相同代码：

const count = ((i) => {
  var j = i;
  return () => ++j;
})(0);

for (let k = 0; k < 10; k++) {
  console.log(count());
}

什么语法错误导致函数的 'text' 在此 JS 闭包中返回，而不是让它递增？

What syntax errors are causing the 'text' of the function to be returned in this JS closure, rather than having it increment?

javascript

closures

increment