[JavaScript] [].some(x=>fn(x))与[].some(fn)不等价

1. 问题描述

以下函数为了判断两个集合是否有交集，可是r1和r2的值居然是不同的。

hasIntersection = (smallSet, bigSet) => {
    let fn = bigSet.includes.bind(bigSet);
    let r1 = smallSet.some(fn);  //false
    let r2 = smallSet.some(ele => fn(ele));  //true
    console.log(r1, r2);
}

hasIntersection([1, 2], [2, 3]);

2. [].some与[].includes

arr.some(callback[, thisArg])

1. callback: Function to test for each element, taking three arguments:
   (1) currentValue: The current element being processed in the array.
   (2) index: The index of the current element being processed in the array.
   (3) array: The array some() was called upon.
2. thisArg: Optional. Value to use as this when executing callback.

arr.includes(searchElement[, fromIndex])

1. searchElement: The element to search for.
2. fromIndex: Optional. The position in this array at which to begin searching for searchElement. A negative value searches from the index of array.length + fromIndex by asc. Defaults to 0.

3. 结论

所以，smallSet.some(fn)并不等价于smallSet.some(ele=>fn(ele))，
而是等价于：smallSet.some((...args)=>fn(...args))，
即，smallSet.some((currentValue,index,array)=>fn(currentValue,index,array))
这样index，就变成了includes的第二个参数fromIndex了。

4. 另外一个例子

["10", "10", "10", "10"].map(parseInt)
> [10, NaN, 2, 3]

["10", "10", "10", "10"].map(x=>parseInt(x))
> [10, 10, 10, 10]

参考

Array.prototype.some()
Array.prototype.includes()