I am attempting to invoke a JavaScript function from within another function, but unfortunately, it does not seem to be functioning

Question

I am attempting to invoke a JavaScript function from within another function, but unfortunately, it does not seem to be functioning

I encountered an issue that is causing me to receive the following error message:

TypeError: Cannot read property 'sum' of undefined

How can this be resolved?

function calculator(firstNumber) {
    var result = firstNumber;

    function sum() {
    for (let i = 0; i <= arguments.lenght; i++) {
        result += arguments[i];
    }

    return result;
    }

    function dif() {
    for (let i = 0; i <= arguments.lenght; i++) {
        result -= arguments[i];
    }

    return result;
    }
}

var myCalculator = calculator(10);
console.log(myCalculator.sum(2, 3));
console.log(myCalculator.dif(1, 2, 5));

The expected output should be:

15

2

javascript

Answer 1

Answer №1

To retrieve the desired functions, access the length property of arguments and loop through < arguments.length, returning an object with the specified functions.

Keep in mind that using a variable named result retains its value from previous calculations. In this case, the initial result is 15 and later becomes 7, rather than the expected value of 2.

function calculator(firstNumber) {

    function sum() {
        for (let i = 0; i < arguments.length; i++) {
            result += arguments[i];
        }
        return result;
    }

    function dif() {
        for (let i = 0; i < arguments.length; i++) {
            result -= arguments[i];
        }
        return result;
    }

    var result = firstNumber;
    return { sum: sum, dif: dif };
}

var myCalculator = calculator(10);
console.log(myCalculator.sum(2, 3));
console.log(myCalculator.dif(1, 2, 5));

If you want the result to always include the original firstNumber, initialize the result variable with firstNumber.

function calculator(firstNumber) {

    function sum() {
        var result = firstNumber;
        for (let i = 0; i < arguments.length; i++) {
            result += arguments[i];
        }
        return result;
    }

    function dif() {
        var result = firstNumber;
        for (let i = 0; i < arguments.length; i++) {
            result -= arguments[i];
        }
        return result;
    }

    return { sum: sum, dif: dif };
}

var myCalculator = calculator(10);
console.log(myCalculator.sum(2, 3));
console.log(myCalculator.dif(1, 2, 5));

Answer 2

To retrieve the desired functions, access the length property of arguments and loop through < arguments.length, returning an object with the specified functions.

Keep in mind that using a variable named result retains its value from previous calculations. In this case, the initial result is 15 and later becomes 7, rather than the expected value of 2.

function calculator(firstNumber) {

    function sum() {
        for (let i = 0; i < arguments.length; i++) {
            result += arguments[i];
        }
        return result;
    }

    function dif() {
        for (let i = 0; i < arguments.length; i++) {
            result -= arguments[i];
        }
        return result;
    }

    var result = firstNumber;
    return { sum: sum, dif: dif };
}

var myCalculator = calculator(10);
console.log(myCalculator.sum(2, 3));
console.log(myCalculator.dif(1, 2, 5));

If you want the result to always include the original firstNumber, initialize the result variable with firstNumber.

function calculator(firstNumber) {

    function sum() {
        var result = firstNumber;
        for (let i = 0; i < arguments.length; i++) {
            result += arguments[i];
        }
        return result;
    }

    function dif() {
        var result = firstNumber;
        for (let i = 0; i < arguments.length; i++) {
            result -= arguments[i];
        }
        return result;
    }

    return { sum: sum, dif: dif };
}

var myCalculator = calculator(10);
console.log(myCalculator.sum(2, 3));
console.log(myCalculator.dif(1, 2, 5));

I am attempting to invoke a JavaScript function from within another function, but unfortunately, it does not seem to be functioning

Answer №1

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