Array not showing up on console

Question

I have no clue why this code doesn't work.

var tipCalculator = [1.2,1.15,1.10];
var restaurantBills = [124,48,738,10,300,150];
var finalBills = [];

function calculateBills(){
    for (var i = 0 ; i < restaurantBills.length ;  i++){

        switch(true){
            case restaurantBills[i]<50:
                finalBills.push(restaurantBills[i]*tipCalculator[0]);
                break;
            case restaurantBills[i]>50 && restaurantBills[i]<200 :
                finalBills.push(restaurantBills[i]*tipCalculator[1]);
                break;
            case restaurantBills[i]>200:
                finalBills.push(restaurantBills[i]*tipCalculator[2]);
                break;
            default:
                break;
        };
    };

return finalBills;};

console.log(calculateBills);

You never *call* the function. Then the function gets its input from global state, not parameters, and doesn't deal with bills of exactly 50 or 200 dollars. — jonrsharpe, Jul 11 '19 at 18:40
calculateBills is a function and you are not executing it.... your console should clearly show the function reference — epascarello, Jul 11 '19 at 18:40
https://stackoverflow.com/questions/988363/how-can-i-debug-my-javascript-code — Heretic Monkey, Jul 11 '19 at 18:41
to add to the above -- you need to add `()` to call a function for example: `calculateBills()` — Mark, Jul 11 '19 at 18:42
yes i totally missed the () at the end. thank you epascarello — chris, Jul 11 '19 at 18:43
@AlonEitan it is a bad practice (in most people's eyes) way of using it like if statements. — epascarello, Jul 11 '19 at 18:44
@AlonEitan `switch (true) { case condition: ...` will run the first `case` where `condition` evaluates to true. — Heretic Monkey, Jul 11 '19 at 18:45
@AlonEitan you tell me , that's what was shown in the course im taking and its working after i added the () i forgot in the last line — chris, Jul 11 '19 at 18:46
@epascarello is it considered a bad practice because of runtime issues? — chris, Jul 11 '19 at 18:49
@chris I don't think it causes any runtime issues, but usually switch statements contain the expression to evaluate in the brackets, and then each case is a possible value. Using it in reverse like you are can be a little confusing to read at first. — IceMetalPunk, Jul 11 '19 at 19:04

score 1 · Answer 1 · answered Jul 11 '19 at 18:44

1

calculate bills is a function and so must be called

calculateBills();

then your code should be fine

answered Jul 11 '19 at 18:44

stephen leigh

54
1

You should use code notation in your answer like ```caculateBills()``` to indicate your piece of code. – Lalit Goswami Jul 23 '19 at 09:10

Ehtisham Hassan · Answer 2 · 2019-07-12T17:11:14.910

As pointed by jonrsharpe you never called the function that is returning the array. To make a function execute you need to call it.

change the line from

console.log(calculateBills);

to

console.log(calculateBills());

for debugging

Browsers like Chrome and Firefox comes with javascript debugger. Goto dev tools -> debugger where you would be able to set break points and trace the execution of your code.

Get Started with Debugging JavaScript in Chrome DevTools

How can I debug my JavaScript code?

Array not showing up on console

2 Answers2