CODEWITHSUNDEEP
iosobjective-ccocoa-touchuilabeluislider
Sergey Grishchev
Sergey Grishchev

Reputation: 12051

UISLider change UILabel's value the right way

So I have this code to change the values of UILabel based on the value of UISlider:

- (void)slider03Changed:(UISlider *)slider{

    int value = 10;

    if (slider.value >= 0.1){
        value = 20;
    }
    if (slider.value >= 0.2){
        value = 30;
    }
    if (slider.value >= 0.3){
        value = 40;
    }
    if (slider.value >= 0.4){
        value = 50;
    }
    if (slider.value >= 0.5){
        value = 60;
    }
    if (slider.value >= 0.6){
        value = 70;
    }
    if (slider.value >= 0.7){
        value = 80;
    }
    if (slider.value >= 0.8){
        value = 90;
    }
    if (slider.value >= 0.9){
        value = 100;
    }

    lbl01.text = [NSString stringWithFormat:@"%u hours", value];
}

Everything works fine, but I'm worried about the quality of code here.

This code is quite, well, long. Is there any way to optimise it somehow? Thanks!

Upvotes: 0

Views: 579

Answers (4)

Charlie Price
Charlie Price

Reputation: 1376

Well first off, you might consider changing the UISlider so that the minimum and maximum values matched the output that you want -- i.e. min==0, max=99. Those would be the minimumValue and maximumValue properties if you want to do it in code or the Minimum and Maximum fields in the Attributes Inspector if you're using Interface Builder.

Then the computation of the output value would be:

int value = slider.value;
    value = (value + 10) - value % 10;

Upvotes: 0

Balu
Balu

Reputation: 8460

once try like this i think here no need to use loop every time,

- (void)slider03Changed:(UISlider *)slider{

    float value = slider.value;

        int k=((int)(value*10)*10)+10;
        if(k==110){
            k=100;
        }
   lbl01.text = [NSString stringWithFormat:@"%d hours", value];
}

Upvotes: 1

Thanakron Tandavas
Thanakron Tandavas

Reputation: 5683

Use for loop : 

- (void)slider03Changed:(UISlider *)slider{

    int value = 10;

    for (float i = 0.9; i > 0; i-=0.1)
    {
        if (slider.value >= i)
        {
           value = (i*100)+10; 
           break;
        }
    }

    lbl01.text = [NSString stringWithFormat:@"%u hours", value];
}

The formula used is:

(i*100)+10

For example:

(0.6*100)+10 = 70
(0.7*100)+10 = 80
and so on...

UPDATE:

You can optimize it further by start checking this case slider.value >= 0.9 first.

Why ?

Because if it satisfy the case with higher value, then the rest of the your if statement will be useless. In the code I wrote, whenever the condition with most value of slider.value is satisfied, we set the value of value variable and break out of the loop. This will prevent your code to evaluate unnecessary cases.

For example:

Assume your slider.value = 0.95

This is what will happen in your code:

(slider.value >= 0.1) ----> true ----> then ---> set value = 20
(slider.value >= 0.2) ----> true ----> then ---> set value = 30
(slider.value >= 0.3) ----> true ----> then ---> set value = 40
(slider.value >= 0.4) ----> true ----> then ---> set value = 50
(slider.value >= 0.5) ----> true ----> then ---> set value = 60
(slider.value >= 0.6) ----> true ----> then ---> set value = 70
(slider.value >= 0.7) ----> true ----> then ---> set value = 80
(slider.value >= 0.8) ----> true ----> then ---> set value = 90
(slider.value >= 0.9) ----> true ----> then ---> set value = 100

With my code:

(slider.value >= 0.9) ----> true ----> then ---> set value = 100 and break out of the loop

As you may see, with my code, it evaluates only 1 case.

Upvotes: 3

Anoop Vaidya
Anoop Vaidya

Reputation: 46563

Whenever you need many if's and you know only one is going to be executed, use if-else ladder.

In this case you can use as:(this is not the best way)

if (slider.value >= 0.9){
    value = 100;
}
else if (slider.value >= 0.8){
    value = 90;
}
else ....

Now to automate/making shorter code, this you can use as :

- (void)slider03Changed:(UISlider *)slider{
    lbl01.text = [NSString stringWithFormat:@"%.0f hours", round(value*10-.5)*10+10]);];
}

Upvotes: 1

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