在Typescript中返回动态编译时间类型？

Question

I've written a method to imitate the Null conditional operator / Elvis operator , including autocomplete and array support :

Say I have this object :

var o = { a: { b: { c:[{a:1},{a:2}] , s:"hi"} } };

I can access the second element in the array and fetch a :

if ( n(o, (k) => k.a.b.c[1].a) == 2 ) 
 {
    alert('Good'); //true
 }

What I've done was to cause the expression to be sent as a function , which then I can parse as string :

function n<T>(o :T , action : (a:T)=>any):any { 
    let s = action.toString().toString().split('return')[1].split('}')[0].split(';')[0];
    s = s.replace(/\[(\w+)\]/g, '.$1'); //for  array access
    s = s.replace(/^\./, '');    //remove first dot        
    var a = s.split('.');
    for (var i = 1, n = a.length; i < n; ++i) {   //i==0 is the k itself  ,aand we dont need it
        var k = a[i];
        if ( o && o.hasOwnProperty(k)) {
            o = o[k];
        } else {
            return null;
        }
    }
    return o;
}

It does work as expected , but I have a small problem.

The signature of the method returns any :

function n<T>(o :T , action : (a:T)=>any) :any 
                                           ^^^ 

Question:

Is there any option that the return value will more specific ( or even exact) as the prop I'm trying to access ?

So n(o, (k) => kabc[1].a) will be :number and n(o, (k) => kabs) will be :string

Is it possible ? if not , is there a way to make return value to be more "typi" ?

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Answer 1

You can just add an extra type parameter for the return value and let the compiler figure out that the type parameter is the return type of the expression:

var o = { a: { b: { c: [{ a: 1 }, { a: 2 }], s: "hi" } } };
function n<T, TValue>(o: T, action: (a: T) => TValue): TValue | null {
    let s = action.toString().toString().split('return')[1].split('}')[0].split(';')[0];
    s = s.replace(/\[(\w+)\]/g, '.$1'); //for  array access
    s = s.replace(/^\./, '');    //remove first dot        
    var a = s.split('.');
    let result: any = o
    for (var i = 1, n = a.length; i < n; ++i) {   //i==0 is the k itself  ,aand we dont need it
        var k = a[i];
        if ( result && result.hasOwnProperty(k)) {
            result = result[k];
        } else {
            return null;
        }
    }
    return result;
}


let nr = n(o, (k) => k.a.b.c[1].a) // is number | null

var str = n(o, (k) => k.a.b.s) // is string | null

Note The |null part is optional but makes sense in the context of your implementation.