Setting default value for TypeScript object passed as argument

Question

Actually, there appears to now be a simple way. The following code works in TypeScript 1.5:

function sayName({ first, last = 'Smith' }: {first: string; last?: string }): void {
  const name = first + ' ' + last;
  console.log(name);
}

sayName({ first: 'Bob' });

The trick is to first put in brackets what keys you want to pick from the argument object, with key=value for any defaults. Follow that with the : and a type declaration.

This is a little different than what you were trying to do, because instead of having an intact params object, you have instead have dereferenced variables.

If you want to make it optional to pass anything to the function, add a ? for all keys in the type, and add a default of ={} after the type declaration:

function sayName({first='Bob',last='Smith'}: {first?: string; last?: string}={}){
    var name = first + " " + last;
    alert(name);
}

sayName();

Typescript supports default parameters now:

https://www.typescriptlang.org/docs/handbook/functions.html

Also, adding a default value allows you to omit the type declaration, because it can be inferred from the default value:

function sayName(firstName: string, lastName = "Smith") {
  const name = firstName + ' ' + lastName;
  alert(name);
}

sayName('Bob');

Object destructuring the parameter object is what many of the answers above are aiming for and Typescript now has the methods in place to make it much easier to read and intuitively understand.

Destructuring Basics: By destructuring an object, you can choose properties from an object by key name. You can define as few or as many of the properties you like, and default values are set by a basic syntax of let {key = default} = object.

let {firstName, lastName = 'Smith'} = myParamsObject;

//Compiles to:
var firstName = myParamsObject.firstName, 
_a = myParamsObject.lastName, 
lastName = _a === void 0 ? 'Smith' : _a;

Writing an interface, type or class for the parameter object improves legibility.

type FullName = {
  firstName: string;
   
  /** @defaultValue 'Smith' */
  lastName ? : string;
}

function sayName(params: FullName) {

  // Set defaults for parameter object
  var { firstName, lastName = 'Smith'} = params;

  // Do Stuff
  var name = firstName + " " + lastName;
  alert(name);
}

// Use it
sayName({
  firstName: 'Bob'
});

No, TypeScript doesn't have a natural way of setting defaults for properties of an object defined like that where one has a default and the other does not. You could define a richer structure:

class Name {
    constructor(public first : string, 
        public last: string = "Smith") {

    }
}

And use that in place of the inline type definition.

function sayName(name: Name) {
    alert(name.first + " " + name.last);
}

You can't do something like this unfortunately:

function sayName(name : { first: string; last?:string } 
       /* and then assign a default object matching the signature */  
       = { first: null, last: 'Smith' }) {

} 

As it would only set the default if name was undefined.