How to check if property exists if obj is one of types

Question

Let's say I have few interfaces A, B, C implementing common Base.

interface Base {
    x: number;
    y: number;
    z: number;
}

interface A extends Base {
    a: true;
}

interface B extends Base {
    b: true;
}

interface C extends Base {
    C: true;
}

And function with if statements:

function foo(arg: A|B|C){
    if(arg.a!==undefined){//throws type error
        //do stuff for type a
    } else if(arg.b !== undefined){//throws type error
        //do stuff for type b
    } else if(arg.c !== undefined){ //throws type error
        //do stuff for type c
    }
}

How to correctly check if property exists? I don't wan't to use any type. Is //@ts-ignore only option?

Answer 1

Typescript will only allow access to common properties. Since the properties you test are not common to all members of the union, typescript will not let you access them.

You can use an in type guard instead to test for the presence of the property.

interface Base {
    x: number;
    y: number;
    z: number;
}

interface A extends Base {
    a: true;
}

interface B extends Base {
    b: true;
}

interface C extends Base {
    C: true;
}

function foo(arg: A|B|C){
    if('a' in arg){
        arg.a
    } else if('b' in arg){
        arg.b
    } else { 
        arg.C
    }
}

Answer 2

You can use a type guard:

function isA(arg: A | B | C): arg is A {
    return (<A>arg).a !== undefined;
}

function isB(arg: A | B | C): arg is B {
    return (<B>arg).b !== undefined;
}

function foo(arg: A | B | C) {
    if (isA(arg)) {
        // do stuff for type a
    } else if (isB(arg)) {
        // do stuff for type b
    } else {
        // do stuff for type c
    }
}