问题
In following reduction + map operations, no. 3 is puzzling me. Can anyone please explain why
// 1
[1,2,3,4,5].filter(x => x==3).reduce((x, y) => y) // -> 3, all good
// 2
[1,2,3,4,5].filter(x => x<=3).reduce((x, y) => 0) // -> 0, still good
// 3
[1,2,3,4,5].filter(x => x==3).reduce((x, y) => 0) // -> 3, hello?
In other words: how come the reduction on the array of one element ignores the map to 0 operation? This would ultimately be used on an array of objects, as in .reduce((x,y) => y.attr) which also returns y instead of y.attr for single element arrays.
回答1:
The filtered array contains only one element so reduce will return that value.
Read the docs :
If the array has only one element (regardless of position) and no initialValue was provided, or if initialValue is provided but the array is empty, the solo value would be returned without calling callback.
For more : https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
来源：https://stackoverflow.com/questions/37143814/array-prototype-reduce-on-arrays-of-one-element