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# I know my problem is inside the for in loop, what can I do to improve the code?

I f*ck up with running the average, x float value, through runningTotal. It's does miner errors that trip me up, but this is just my second attempt so it's not too bad.

variable_assignment.mm
```NSArray *temps= @[@75.5, @83.3, @96, @99.7];
float average;
float runningTotal;
for(NSNumber *temps in tempsArray ){
float runningTotal( xfloatValue && flat avarage);
}
```

Hi Luis,

You've nearly done that - very close!

A few points to make. Here:

```for(NSNumber *temps in tempsArray)
```

The loop will iterate over an array holding each value in the array, in turn, within the local variable declared there. Your array is called `temps` so you'll need to call your local variable something different. Just use `x`; it's easy! And `tempsArray` doesn't exist; change that to `temps`. That gives us:

```for(NSNumber *x in temps){
// add up all the x's
}
```

We want to add the value held by `x` into `runningTotal` at each iteration. You've got:

```float runningTotal( xfloatValue && flat avarage);
```

First, don't redeclare `runningTotal` as a `float`; omit the `float` word as the compiler knows `runningTotal` is a `float` already. Next, ignore `average` for now - we come to that later. You can't create the average inside the loop. What we're doing is adding up all the elements of `temps` with the loop, we'll then divide by the length of `temps` after the loop and store the result of that in `average`.

Inside the loop, you want to add to `runningTotal` with `+=` just like you have done. But we need to convert the `NSNumber` into `float`. Use square brackets and `floatValue` for that:

```runningTotal += [x floatValue];
```

That completes the loop. It looks like:

```for(NSNumber *x in temps){
runningTotal += [x floatValue];
}
```

Now, find the number of elements in the `temps` array. Use the `count` property and square brackets again `[temps count]`, and divide `runningTotal` by that. Store the result in `average`.

Let me know how you get on.

Steve.