Checking if device is iPad

Question

What is wrong with this code? I want to be able to check if the current device the user is using is an iPad but it keeps giving me errors.

if (UIUserInterfaceIdiom == UIUserInterfaceIdiomPad)
{
    //do stuff
}

Solution 1

You can use

if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)

Solution 2

In Swift you can use:

if UIDevice.current.userInterfaceIdiom == .pad {
    //do stuff
}

Solution 3

you must cheack the condition wheather device is iphone/ipad

if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
    {
//do ur  ipad logic
}else
{
//do ur  iphone logic
}

Solution 4

If you are using swift, use this

    if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiom.Pad)
    {
      // device is ipad
    }
    else
    {
        //  device is iPhone
    }

Solution 5

/* The UI_USER_INTERFACE_IDIOM() function is provided for use when deploying to a version of the iOS less than 3.2. If the earliest version of iPhone/iOS that you will be deploying for is 3.2 or greater, you may use -[UIDevice userInterfaceIdiom] directly. */

Better use UIDevice.current.userInterfaceIdiom unless you are supporting super old iOS.

Possible cases are as below:

public enum UIUserInterfaceIdiom : Int {
    case unspecified

    @available(iOS 3.2, *)
    case phone // iPhone and iPod touch style UI

    @available(iOS 3.2, *)
    case pad // iPad style UI

    @available(iOS 9.0, *)
    case tv // Apple TV style UI

    @available(iOS 9.0, *)
    case carPlay // CarPlay style UI
}

View more solutions

Comments

[epascarello]

You probably want to tag this with more than just ipad.

[Daniel Monteiro]

Maybe you're calling it too late? Are you loading your XIB manually? I remember doing this and succeeding. But I guess there are better solutions.

[magma]

Note that an UIUserInterfaceIdiomPhone value does not necessarily mean that you're on an iPhone device - rather, that the interface idiom matches the iPhone one. You might be on an iPad, running an iPhone-only app in 1x/2x mode.

[rmaddy]

Actually it's UI_USER_INTERFACE_IDIOM().

[Makleesh]

That's an interesting point magma thanks for bringing it to my attention. I had no idea that was the case