typescript - 为什么这个未定义的检查不适用于 find 函数？

Question

有人可以解释为什么在下面的例子中，变量 z有类型 number ，但在电话内 f1(x.y) , x.y有类型 number | undefined

const f1 = (value: number) => !!value;

const f2 = (x: { y: number | undefined}): void => {
    if (x.y === undefined || !isFinite(x.y)) {
        throw new Error('Not a number');
    }

    const z = x.y; // z & x.y has type number 
    ['stuff', 'here'].find(item => f1(x.y)); // x.y has type number | undefined
};

Playground example

Answer 1

问题是在箭头函数内部，编译器不确定类型缩小是否仍然有效。

const f1 = (value: number) => !!value;

const f2 = (x: { y: number | undefined}): void => {
    if (x.y === undefined || !isFinite(x.y)) {
        throw new Error('Not a number');
  }
  // z is `number`, so use `z` again in the arrow function to "keep" the type narrowing
  const z = x.y;
  ['stuff', 'here'].find(item => f1(z));
};

TypeScript Playground