I have a function that returns multiple types (union type) but then I have another function later that uses the value returned from the first function but only accepts a single type. Right now typescript is giving me an error saying that there is a type mismatch because the first function could potentially return something that would not be acceptable for the second function, but in the way that I am calling it I know it will not have a conflict. How can I inform typescript that the return value of the first method will work in the second?
Example:
function returnNumberOrString (variable): string | number {
if (variable === 'hello') return 'hello to you';
if (variable === 123) return 456;
}
function onlyNumbers (number : number) {
return number + 1;
}
let number = returnNumberOrString(123);
onlyNumbers(number); // This shows a warning because returnNumberOrString could possibly return a string
The accepted solution only hides the compiler error. You will still be able to call onlyNumbers with a string even if you use Type Assertion to tell the compiler it is a number. If you perform the check I've written below, you won't ever pass anything but a number to your onlyNumbers function. This is a safer operation if you're doing anything that requires strictly numbers (as your function name implies)
function returnNumberOrString (variable): string | number {
if (variable === 'hello') return 'hello to you';
if (variable === 123) return 456;
}
function onlyNumbers (number : number) {
return number + 1;
}
// Can pass string or number, will only ever get a number in val
let val = typeof returnNumberOrString(123) === "number" ? 456 : -1; // I used -1, you can use any value to mean "error" though.
console.log(val); // 456
console.log(onlyNumbers(val)); // 457
let val2 = typeof returnNumberOrString("hello") === "number" ? 456 : -1;
console.log(val2); // -1
console.log(onlyNumbers(val2)); // 0
Collected from the Internet
Please contact [email protected] to delete if infringement.
Comments