How to extract if condition to function and not break narrowed types in TypeScript

Question

given the following if statement:

interface MyData {
    prop1?: number,
    prop2?: number,
}

const x: MyData = {
    prop1: 1,
    prop2: 2,
}

function fun(a: number, b: number) {
    console.log(a, b);
}

if (x.prop1 && x.prop2) {
  fun(x.prop1, x.prop2);
}

the compiler is able to infer that inside the if body, x.prop1 is defined and x.prop2 is defined.

I want to give name to the condition by extracting it to a function.

function bothPropsDefined(prop1: number | undefined, prop2: number | undefined): boolean {
    return !!prop1 && !!prop2;
}

if (bothPropsDefined(x.prop1, x.prop2)) {
  fun(x.prop1, x.prop2);
}

but the compiler fails to conclude that in the if body, both props are defined.

// Argument of type 'number | undefined' is not assignable to parameter of type 'number'.
   Type 'undefined' is not assignable to type 'number'.(2345)

How can I give the compiler a hint that if bothPropsDefined returns true, than prop1 and prop2 are defined?

I tried type guard:

function bothPropsDefinedGuard(prop1: number | undefined, prop2: number | undefined): prop1 is number {
    return !!prop1 && !!prop2;
}

if (bothPropsDefinedGuard(x.prop1, x.prop2)) {
  fun(x.prop1, x.prop2);
}

but I only managed to add guard to one property.

Update

for the question above, I managed to write:

function bothPropsDefinedGuard2(x: {prop1?: number, prop2?: number}): x is {prop1: number, prop2: number} {
    return !!x.prop1 && !!x.prop2;
}

if (bothPropsDefinedGuard2(x) {
  fun(x.prop1, x.prop2);
}

but my real case is a bit harder, as props come from multiple objects:

interface MyData1 {
    prop1?: number,
}

interface MyData2 {
    prop2?: number,
}

const x1: MyData1 = {
    prop1: 1,
}

const x2: MyData2 = {
    prop2: 2,
}

Answer 1

You are looking for custom type guards. Here is a solution:

const x: { prop1: number | undefined, prop2: number | undefined } = {
  prop1: 1,
  prop2: 2,
}

function fun(a: number, b: number) {
  console.log(a, b);
}

// Custom type guard
function propDefined(prop: number | undefined): prop is number {
  return typeof prop === 'number';
}

// Use
if (propDefined(x.prop1) && propDefined(x.prop2)) {
  fun(x.prop1, x.prop2);
}

Note: You cannot have a type guard that changes the types for two arguments. A type guard can only narrow one argument eg prop is number as shown.

Answer 2

My goal was to name the condition, so answer by @basarat is not ideal.

I ended up

• creating an artificial combined object out of individual data pieces
• making a type guard on the combined object
• using combined object in the if body

This is not ideal as well, due to artificially added combined object, but lets me name the condition:

interface MyData1 {
    prop1?: number,
}

interface MyData2 {
    prop2?: number,
}

const x1: MyData1 = {
    prop1: 1,
}

const x2: MyData2 = {
    prop2: 2,
}

const xCombined: MyData = {
    prop1: x1.prop1,
    prop2: x2.prop2,
}

if (bothPropsDefinedGuard2(xCombined)) {
  fun(xCombined.prop1, xCombined.prop2);
}