# FAQ: Learn TDD With Mocha - Edge Case

This community-built FAQ covers the “Edge Case” exercise from the lesson “Learn TDD With Mocha”.

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## FAQs on the exercise Edge Case

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The task here is redundant. we do not need to have if condition to check whether the array is empty.

2 Likes

Interesting,
from what I understand: you set the initial value of the reduce method to zero with that second parameter? Therefore even if the array is empty, the value of zero still comes out of the calculation?

I tried throwing an error if the data type of the array was not numbers, but I got stuck along the way. I only managed to make it work using this code in index_test.js:

``````    it('raises an error if the array passed in is not made of numbers', () => {
// Setup
const inputArray = ['lorem', 'ipsum'];
// Exercise
const result = () => {
Calculate.sum(inputArray)
};
// Verification
assert.throws(result, /only use numbers/);
});
``````

And writing this block in index.js:

``````    if (typeof inputArray.reduce((sum, value) => {
return sum + value;
}) !== 'number') {
throw new Error('ERROR: only use numbers');
}
``````

Seems like it could be shorter, but I failed to find a more concise way that passed all tests. Do any of you know a better way?

1 Like

Hey, your challenge seems interesting, i tried to and implement this solution… Not a lot shorter but maybe a bit more readable to know we’re throwing an error if it’s not a number. I used the .some() methods

``````    if (inputArray.some(el => typeof el !== 'number')) {
throw new Error('ERROR: only use numbers');
}

``````

The sort() method will return a boolean true if one of the element in the array is not a number

1 Like

Thanks @thibault_deverge! That’s a good solution using a method I did not know about.
For anyone interested, here is the MDN documentation for the .some() method:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some

1 Like

Hey Sebastien,
I know this discussion is old but I just came across it today.
this is a shorter solution:

``````if (inputArray.some(isNaN) {
throw new Error('ERROR: only use numbers');
}
``````

The `isNaN()` function determines whether a value is “Not a Number” or not

1 Like

The given solution has edge case bugs.

``````const Calculate = {
sum(inputArray) {

if(inputArray.length === 0){
return 0
}

return inputArray.reduce((sum, value) => {
return sum + value;
})
}
}
``````
• this doesn’t take into account if the passed in value is not an array
• the if statement is also unnecessary when reduce can take an inital value of 0
• there is no handling if a string is mixed in with the array

My solution

``````const Calculate = {
sum(inputArray = []) {
if (!Array.isArray(inputArray)) return 0;
return inputArray
.map(el => Number(el))
.filter(el => !Number.isNaN(el))
.reduce((acc, curr) => acc + curr, 0);
}
}
``````
• convert any non-numbers, then filter out any that still aren’t number values and sum with initial value of 0

Test case

``````  it('returns the sum of numbers in array', () => {
const expectedResult = 5;
const inputArray = ["A", 3, "C", 1, "1"];

const result = Calculate.sum(inputArray);

assert.strictEqual(result, expectedResult);
});
``````