爲什麼我必須在下面的例子中使用rest參數和spread運算符？

Question

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

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

 
f(z);

預期輸出：

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

實際輸出：

"111" 
["a", "b", "c", "d", "e", "f"] 

只有當我改變f(z)到f(...z)我得到的預期輸出線。我是ECMAScript 2015的新手。請告訴我在這裏錯過了什麼。

Answer 1

f(z)意味着只有一個參數（在這種情況下，一個數組）傳遞給方法，而f(...z)意味着，數組中的值作爲參數傳遞給函數。

閱讀更多：

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);