Type Assertion in TypeScript without assigning a new variable

Question

I have a function that returns values with one of two interfaces, like so:

interface K {
    A:number;
    NUM:number;
}

interface L {
    A:string;
    STR:string;
}

function FUN():K|L { }

var x:K|L = FUN();

but I can't access the non-shared property using a type-guard like this one:

if(typeof x.A === 'string'){
    console.log(x.STR) // Property 'STR' does not exist on type 'K | L'
}

even though the if clause does distinguish between the two types. I know I can use as operator with a variable assignment to indicate the type to the compiler like so:

if(typeof x.A === 'string'){
    var y = x as L;
    console.log(y.STR) // OK 
}

but that seems a bit kludgy as this add a unnecessary var statement and assignment in the resulting js code:

function FUN() { }
var x = FUN();
if (typeof x.A === 'string') {
    var y = x; // unnecessary...
    console.log(y.STR); 
}

Is there any way to indicate the type in this situation without the extra var y=x; in the JS code?

rgvassar · Accepted Answer · 2016-03-12 07:44:57Z

3

Yes, TypeScript supports type casting. You should be able to do:

if(typeof x.A === 'string'){
    console.log((<L>x).STR);
}

edited Mar 12, 2016 at 7:44

answered Mar 12, 2016 at 7:00

rgvassar

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4 Comments

Jthorpe Over a year ago

Hmmm. That doesn't seem to work. Casting would work if I had a variable that was already declared as type L, as in: var x:L; x = <L>FUN(); . But doesn't re-cast x once it's already defined (var'd).

rgvassar Over a year ago

Oops, you're right. I needed some parenthesis. I edited my answer. Give that a try.

Jthorpe Over a year ago

Thanks so much! Solved my problem and I learned an new usage for the type assertion!

rgvassar Over a year ago

No problem. Glad I was able to help you out.

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