# Card count - code passes but doesn't seem to work in all cases

Other forum threads on the card counting challenge were really helpful, I feel like I got to a good understanding of how to arrive at my solution, but it doesn’t seem to give the correct outputs for all the examples given.

Here is my solution:

``````let count = 0;

function cc(card) {
// Only change code below this line
switch (card) {
case 2:
case 3:
case 4:
case 5:
case 6:
count += 1;
break;
case 7:
case 8:
case 9:
count += 0;
break;
case 10:
case 'J':
case 'Q':
case 'K':
case 'A':
count -= 1;
}

if (count > 0) {
return count + " Bet";
}
return count + " Hold";
// Only change code above this line
}

console.log(cc(10, 'J', 'Q', 'K', 'A'));
cc(2); cc(3); cc(7); cc('K'); cc('A');
``````

This solution is currently passing the lesson. However…

In the example I’ve tried to log, the lesson says “Cards Sequence 10, J, Q, K, A should return the string `-5 Hold`”, but my log displays “-1 Hold” instead. In fact, none of the negative count values (10, J, Q, K, A) submitted seem to be returning values below -1, which makes me think something is either wrong with the way I’m testing it in the console or with the function count -= 1?

The calls to the function only take one argument at a time

1 Like

got it, I thought it might be something like that! So is there a way to test a sequence like in the example given?

hold on, I think I figured it out?

``````console.log(cc(10));
console.log(cc('J'));
console.log(cc('Q'));
console.log(cc('K'));
console.log(cc('A'));
``````

this does arrive at -5 Hold. Not as straightforward as I imagined, but I guess it works!

1 Like

You could definitely write the function in such a way to take one or more arguments and be able to write: `cc(2, 3, 7, 'K', 'A');` instead of `cc(2); cc(3); cc(7); cc('K'); cc('A');` Once you get a bit more knowledge of JavaScript, you should come back and try to implement such a function for this challenge.

1 Like

ooh thank you for the idea! I think that was where I was initially coming from, which made my first attempts at coding the solution a mess. It makes sense there are other ways to write the function that are a bit more advanced - I’ll put a bookmark in this challenge then.