使用array.splice（）第一个参数未定义，仍然有效

Question

I have an array that I am calling .splice() on. By accident, sometimes the code runs with undefined as the first parameter eg.

array.splice(undefined, 1);

but my code still works as though it was being called like:

array.splice(0,1);

So my question is, why does this work? Why does the following work?

[0, 1,2,3].splice(undefined,1); //returns 0

Thanks

Answer 1

This occurs because Array.prototype.splice() treats falsey values the same as 0 , and undefined is falsey .

So you could also do:

[0, 1,2,3].splice(false,1); //returns 0
[0, 1,2,3].splice(null,1); //returns 0

Answer 2

Javascript developer are used to clean the value that you pass to the function in order to always have something to work with.

In this case, we can be sure that it's not an implicit coercion, because

Number(undefined); //NaN
Number("text"); //NaN

While in splice function they are coerced to 0 .

It's likely that they use a bitwise operator like >>0

"2">>0        //2
-8>>0         //-8
null>>0       //0
undefined>>0  //0
"text">>0     //0 

Answer 3

从docs arr.slice([begin[, end]]) ， 如果undefined begin ，则slice从索引0开始。