Question

How to handle invalid argument in TypeScript

Answer 1

function processNumber(num: number): void {
    if (typeof num !== 'number' || num < 0) {
        console.error("Invalid input: Must be a positive number.");
        return;
    }
     
    console.log("input ok:", num);
}
 
processNumber(8);  
processNumber(-5); 
 
 
 
/*
run:
 
"input ok:",  8 
"Invalid input: Must be a positive number." 

*/

 

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Answer 2

function divide(a: number, b: number): number {
    if (b === 0) throw new Error("Division by zero is not allowed.");
    return a / b;
}
 
try {
    console.log(divide(8, 0));
} catch (error) {
    console.error("Error:", error.message);
}
 
 
 
/*
run:

Error: Division by zero is not allowed.
 
*/
 
 

 

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Answer 3

function say(name: string): void {
    name = name || "Guest"; // Default to "Guest" if falsy
    console.log("Hello, " + name + "!");
}
 
say(); 
say("Tom"); 
 
 
 
/*
run:
 
Hello, Guest!
Hello, Tom!

*/

 

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Answer 4

function addNumbers(a: number, b: number): number {
    if (typeof a !== "number" || typeof b !== "number") {
        throw new TypeError("Both arguments must be numbers.");
    }
    return a + b;
}
 
console.log(addNumbers(6, "abc")); // Throws TypeError
 
 
 
/*
run:
 
/usr/src/app/main.js:4
    throw new TypeError("Both arguments must be numbers.");
    ^

TypeError: Both arguments must be numbers.
    at addNumbers (/usr/src/app/main.js:4:11)
 
*/

 

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