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();
Answer from jpadvo on Stack OverflowSetting default value for TypeScript object passed as argument - Stack Overflow
With a functin that has two arguments with default values, what is the best way to call this function with the default value for the first argument but with a new value for the second argument?
How to prevent inferring of optional generic?
Interface separation or use optional parameters
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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');