Best way to programmatically detect iPad/iPhone hardware

Question

The reason I need to find out is that on an iPad, a UIPickerView has the same height in landscape orientation as it does in portrait. On an iPhone it is different. The iPad programming guide introduces an "idiom" value to UIDevice:

    UIDevice* thisDevice = [UIDevice currentDevice];
    if(thisDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad)
    {
        // iPad
    }
    else
    {
        // iPhone
    }

which works OK while you're in iPad (3.2) but not iPhone (3.1.3) - so it looks like there also needs to be an ifdef to conditionally compile that check, like:

#if __IPHONE_OS_VERSION_MIN_REQUIRED >= 30200
        UIDevice* thisDevice = [UIDevice currentDevice];
        if(thisDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad)
        {
            // etc.
        }
#endif

To me that's starting to look very clumsy. What's a better way?

Answer 1

I like my isPad() function. Same code but keep it out of sight and in only one place.

Answer 2

Checking at runtime (your first way) is completely different from #if at compile time. The preprocessor directives won't give you a universal app.

The preferred way is to use Apple's Macro:

if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
{
     // The device is an iPad running iPhone 3.2 or later.
}
else
{
     // The device is an iPhone or iPod touch.
}

Use 3.2 as the base SDK (because the macro is not defined pre 3.2), you can target prior OS versions to get it running on the iPhone.

Answer 3

Put this method in your App Delegate so that you can call it anywhere using [[[UIApplication sharedApplication] delegate] isPad]

-(BOOL)isPad
{
    BOOL isPad;
    NSRange range = [[[UIDevice currentDevice] model] rangeOfString:@"iPad"];
    if(range.location==NSNotFound)
    {
        isPad=NO;


    }
    else {
        isPad=YES;
    }

    return isPad;
}