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withDefaults
Like Function.prototype.bind() but for a function that receives their parameters wrapped in an object.
import { withDefaults } from "tsafe/lab/withDefaults";
function sum(params: { x: number; y: number; z: number }): number {
const { x, y, z } = params;
return x + y + z;
}
// sumWd is of type: (params: { y: number; z: number; })=> number
const sumWd = withDefaults(sum, { x: 10 });
console.log(sumWd({ y: 1, z: 2 })); // Prints "13" ( 10 + 1 + 2 )
console.log(sumWd({ y: 3, z: 4 })); // Prints "17" ( 10 + 3 + 4 )
console.log(
sumWd({
y: 3,
z: 4,
defaultsOverwrite: {
x: [20],
},
})
); // Prints "27" ( 20 + 3 + 4 )
If you have a function with a set of parameters wrapped in an object, and you wish to call this function multiple times with the same value for one or more of the parameters,
withDefaults enables you to instantly generate a new function with these parameters already set so that you do not have to fill them in at every call.Consider a function that takes two numbers as parameters and returns the sum of them.
function sum(params: { x: number; y: number }) {
const { x, y } = params;
return x + y;
}
Suppose we want to set the value of
x for example.const sumWd = withDefaults(sum, { x: 33 });
sumWd is a proxy to our original function with x set to 33. withDefaults first argument is the original function, and the second is an object with the parameters of the original function as properties. Naturally, the properties are inferred by typescript as shown below.
Now we can call
sumWd as many times as we want without having to set x. Its value will always be 33.const result = sumWd({ y: 10 }); //43
The value of
result will be 43. Typescript infers the remaining value to be set:
const result = sumWd({
y: 10,
defaultsOverwrite: { x: [23] },
}); // 33
The type of
x in defaultsOverwrite is [number] | undefined so that undefined cannot be assigned to x if that is not its type.Last modified 8mo ago