为什么我必须在下面的示例中同时使用rest参数和spread运算符？

Question

ts.forEachChild

预期产出：

function f(...y){
      y.forEach(v => {
        console.log("111");
        console.log(v);
      });
    }

var z=["a","b","c","d","e","f"];

f(z);

实际输出：

"111"
"a"
"111"
"b"
"111"
"c"
"111"
"d"
"111"
"e"
"111"
"f"

只有当我将行"111" ["a", "b", "c", "d", "e", "f"] 更改为f(z)时，才能获得预期的输出。我是ECMAScript 2015的新手。请告诉我这里缺少什么。

Answer 1

f(z)表示只有一个参数（在本例中为数组）传递给方法，而f(...z)表示，数组中的值作为参数传递给函数。

阅读更多：

1. Spread operator
2. Rest parameters

function f(...y){
      console.log(y.length);
    }

var z=["a","b","c","d","e","f"];
f(z);
f(...z);