TypeScript：为什么我不能访问仅在一个 object 上定义的联合类型的属性？

Question

I want to have either an object with the property a which is a number or an object with property b which is a string. In plain JavaScript the variable test would be either a number or undefined if the other object has been passed. Why is this an error in TypeScript? Or is there a way to do this without an type guard?

function t(aOrB: {a:number} | {b:string}){
    const test: number | undefined = aOrB.a
}

https://www.typescriptlang.org/play?#code/FAMwrgdgxgLglgewgAhgCgIYHkBOAhALmQG8MCIwBbAIwFMcBfZAHxOoIGcYc4IBzBgEpiwZGORQkXVLS5EKNei2SQAJrRC9aq5AF5k2fADoMwBsCA

Answer 1

You need to do a check on if ('a' in aOrB) and if ('b' in aOrB) before returning.

Answer 2

you can use typeOf like this

 function t(aOrB:any){ const test =((typeof(aOrB) == "object")?aOrB.a:aOrB); console.log(test); } t({a:123}); t(123)