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Array.lengtheck issue: 这一比较似乎无意中,因为第一类和第二类没有重叠[重复]。
原标题:Array.length check issue: This comparison appears to be unintentional because the types 1 and 0 have no overlap [duplicate]
  • 时间:2024-05-14 05:09:29
  •  标签:
  • typescript


const x = [1];

if (x.length !== 1)
throw "error";


if (x.length === 0)
throw "error";

compliler Weirdly在if(x.length=0)上提出申诉。 这一比较看来是无意的,因为第一类和第二类没有重叠。

似乎在第一个条件之后,它总是假定时间为1,而阵列修改并没有改变。 我不理解理由,也不知道我们如何解决这一问题?



This is one of these cases when TypeScript s type narrowing isn t the smartest. TypeScript doesn t follow .length changes through methods, so its way of narrowing the .length type to a union often does not make much sense.

不幸的是,你需要工作。 其中一项是确定新期限,而不是偷窃:

if (x.length > 0) { // This is needed, as an error would be thrown for an empty array
    x.length -= 1; // works like x.pop() minus returning the last item

The following code

if (x.length !== 1) throw "error";

narrows down the type of x.length to be 1.


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