I have a constant defined with screen width to 320. This is being used at multiple places in my code. I have to change this to take value from current device and I do not want to touch all these places. So, I want to define a constant for this with value:
[[UIScreen mainScreen] bounds].size.width
#define kScreenWidth [[UIScreen mainScreen] bounds].size.width
But it is giving me lot of compilation errors. Any clue?
[[UIScreen mainScreen] applicationFrame].size.width
Related
Is this possible? I want the number of inches, not the number of pixels. I know it is approximately 160 ppi. But not exactly.
There isn't an API that will give you this. Your best bet is to look at the device's screen size (in points) and from that surmise if it's an iPad or iPhone etc., and then use hard-coded values for the screen sizes.
Here's some code to get the screen size:
CGRect screenRect = [[UIScreen mainScreen] bounds];
CGFloat screenWidth = screenRect.size.width;
CGFloat screenHeight = screenRect.size.height;
Be aware that width and height might be swapped, depending on device orientation.
If it were available it would be in UIScreen or UIDevice but it is not there.
You can infer it from info in Erica's UIDevice-extension and the specs for each device listed here on Wikipedia.
Here's a short method that estimates the device screen size. It's updated as to the latest devices, but may fail on future ones (as all methods of guessing might). It will also get confused if the device is being mirrored (returns the device's screen size, not the mirrored screen size)
#define SCREEN_SIZE_IPHONE_CLASSIC 3.5
#define SCREEN_SIZE_IPHONE_TALL 4.0
#define SCREEN_SIZE_IPAD_CLASSIC 9.7
+ (CGFloat)screenPhysicalSize
{
if(UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
{
CGSize result = [[UIScreen mainScreen] bounds].size;
if (result.height < 500)
return SCREEN_SIZE_IPHONE_CLASSIC; // iPhone 4S / 4th Gen iPod Touch or earlier
else
return SCREEN_SIZE_IPHONE_TALL; // iPhone 5
}
else
{
return SCREEN_SIZE_IPAD_CLASSIC; // iPad
}
}
You might need use [UIScreen mainScreen].scale;
CGFloat scale = [UIScreen mainScreen].scale;
CGRect screenRect = [[UIScreen mainScreen] bounds];
CGFloat physicalWidth = screenRect.size.width * scale;
CGFloat physicalHeight = screenRect.size.height * scale;
Maybe Jeff Hay's code can be adapted to include iPad Mini. The trick is to get the device's model identifier. The most recent non-retina iPad is "iPad2,4" and the first iPad mini is "iPad2,5". Now all you need to check is if the screen scaling is 1.0 (non-retina)
Although this code is not future-proof, you can always add more rules for model identifiers.
#import <sys/utsname.h>
#define SCREEN_SIZE_IPAD_MINI 7.9
struct utsname systemInfo;
uname(&systemInfo);
if(UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad && strcasecmp(systemInfo.machine, "iPad2,5") >= 0 [[UIScreen mainScreen] scale] == 1.0)
return SCREEN_SIZE_IPAD_MINI
The "formula" I use is
#define IS_iPhone5 ( fabs( (double)[ [ UIScreen mainScreen ] bounds ].size.height - (double)568 ) < DBL_EPSILON )
Since this question has been asked, I’ve created an open-source library to handle this problem: IRLSize. It can be used in either direction: to measure the size of a view (or the whole screen) in real-world dimensions, or to set the size of a view to a specific real-world dimension.
note: screen rotation matters here
extension UIScreen {
var physicalSize:CGSize {
return CGSize(width: bounds.width*scale, height: bounds.height*scale)
}
}
using:
print(UIScreen.main.physicalSize)
Here is a Swift way to get screen sizes:
var screenWidth: CGFloat {
if UIInterfaceOrientationIsPortrait(screenOrientation) {
return UIScreen.mainScreen().bounds.size.width
} else {
return UIScreen.mainScreen().bounds.size.height
}
}
var screenHeight: CGFloat {
if UIInterfaceOrientationIsPortrait(screenOrientation) {
return UIScreen.mainScreen().bounds.size.height
} else {
return UIScreen.mainScreen().bounds.size.width
}
}
var screenOrientation: UIInterfaceOrientation {
return UIApplication.sharedApplication().statusBarOrientation
}
These are included as a standard function in:
https://github.com/goktugyil/EZSwiftExtensions
Nobody said about fixedCoordinateSpace. In Swift 3 to get the screen dimensions in a portrait-up orientation you should use: UIScreen.main.fixedCoordinateSpace.bounds
CGFloat scale = [UIScreen mainScreen].scale;
CGRect screenRect = [[UIScreen mainScreen] bounds];
CGFloat screenWidth = screenRect.size.width * (scale/100.0f);
CGFloat screenHeight = screenRect.size.height * (scale/100.0f);
scale is persent!
I'm trying to make certain elements resize according to the screen dimensions when the interface orientation changes. I've read in various places (including a few answers on here) that the way to get the bounds of the screen is to use [[UIScreen] mainScreen] bounds]. So I've added this code to see how it behaves:
- (void) willRotateToInterfaceOrientation:(UIInterfaceOrientation)toInterfaceOrientation duration:(NSTimeInterval)duration
{
NSLog(#"view dimensions: (%#, %#)", [[UIScreen mainScreen] bounds].size.width, [[UIScreen mainScreen] bounds].size.height);
}
But when running it, as soon as I rotate the screen I get an EXC_BAD_ACCESS message. Is there something about the scope of the willRotateToInterfaceOrientation function that doesn't allow me to call for the screen bounds? If so, how would I go about it?
%# format specifier expects format parameter to be objective-c object, while what you provide is plain float. To print float numbers use %f specifier:
NSLog(#"view dimensions: (%f, %f)", [[UIScreen mainScreen] bounds].size.width, [[UIScreen mainScreen] bounds].size.height);
I thought I was doing something relatively simple, but I guess not.
Running:
NSLog(#"%f",[[UIScreen mainScreen] scale];
returns 0.000000
The problem is that I am trying to check for a retina display and the:
if([UIScreen respondsToSelector:#selector(scale)] &&
[[UIScreen mainScreen] scale] == 2.0) {
// does not get called on an iPhone 4
}
Doesnt get called. I have tried this both in the simulator as well as on the device.
I guess the answer is How to call [[UIScreen mainScreen] scale] in a universal app for the iphone and ipad
Hi I'm wondering if there's a way to get the width programmatically.
I'm looking for something general enough to accomodate iphone 3gs, iphone 4, ipad. Also, the width should change based on if the device is portrait or landscape (for ipad).
Anybody know how to do this?? I've been looking for a while... thanks!
Take a look at UIScreen.
eg.
CGFloat width = [UIScreen mainScreen].bounds.size.width;
Take a look at the applicationFrame property if you don't want the status bar included (won't affect the width).
UPDATE: It turns out UIScreen (-bounds or -applicationFrame) doesn't take into account the current interface orientation. A more correct approach would be to ask your UIView for its bounds -- assuming this UIView has been auto-rotated by it's View controller.
- (void)didRotateFromInterfaceOrientation:(UIInterfaceOrientation)fromInterfaceOrientation
{
CGFloat width = CGRectGetWidth(self.view.bounds);
}
If the view is not being auto-rotated by the View Controller then you will need to check the interface orientation to determine which part of the view bounds represents the 'width' and the 'height'. Note that the frame property will give you the rect of the view in the UIWindow's coordinate space which (by default) won't be taking the interface orientation into account.
CGRect screen = [[UIScreen mainScreen] bounds];
CGFloat width = CGRectGetWidth(screen);
//Bonus height.
CGFloat height = CGRectGetHeight(screen);
This can be done in in 3 lines of code:
// grab the window frame and adjust it for orientation
UIView *rootView = [[[UIApplication sharedApplication] keyWindow]
rootViewController].view;
CGRect originalFrame = [[UIScreen mainScreen] bounds];
CGRect adjustedFrame = [rootView convertRect:originalFrame fromView:nil];
use:
NSLog(#"%f",[[UIScreen mainScreen] bounds].size.width) ;
As of iOS 9.0 there's no way to get the orientation reliably. This is the code I used for an app I design for only portrait mode, so if the app is opened in landscape mode it will still be accurate:
screenHeight = [[UIScreen mainScreen] bounds].size.height;
screenWidth = [[UIScreen mainScreen] bounds].size.width;
if (screenWidth > screenHeight) {
float tempHeight = screenWidth;
screenWidth = screenHeight;
screenHeight = tempHeight;
}
Use this code it will help
[[UIScreen mainScreen] bounds].size.height
[[UIScreen mainScreen] bounds].size.width
Here is a Swift way to get screen sizes, this also takes current interface orientation into account:
var screenWidth: CGFloat {
if UIInterfaceOrientationIsPortrait(screenOrientation) {
return UIScreen.mainScreen().bounds.size.width
} else {
return UIScreen.mainScreen().bounds.size.height
}
}
var screenHeight: CGFloat {
if UIInterfaceOrientationIsPortrait(screenOrientation) {
return UIScreen.mainScreen().bounds.size.height
} else {
return UIScreen.mainScreen().bounds.size.width
}
}
var screenOrientation: UIInterfaceOrientation {
return UIApplication.sharedApplication().statusBarOrientation
}
These are included as a standard function in:
https://github.com/goktugyil/EZSwiftExtensions
how can i simplify that code in one line?
CGRect screen = [[UIScreen mainScreen] bounds];
NSLog(#"%#", screen.size.width);
Thanks for your time.
This statement will cause an exception, or should:
NSLog(#"%#", screen.size.width);
The width property returns a CGFloat. You would need to change your log statement to:
NSLog(#"%f", screen.size.width);
If you want everything on one line:
NSLog(#"%f", [[[[UIScreen mainScreen] bounds] size] width]);
I would suggest:
NSLog(#"%1.0f", [UIScreen mainScreen].bounds.size.width);
To get both height and width you can use NSStringFromCGSize:
NSLog(#"%#", NSStringFromCGSize([UIScreen mainScreen].bounds.size));