Rounded Corners on UIImage

I'm gonna go ahead here and actually answer the question in the title.

Try this category.

UIImage+additions.h

#import <UIKit/UIKit.h>

@interface UIImage (additions)
-(UIImage*)makeRoundCornersWithRadius:(const CGFloat)RADIUS;
@end

UIImage+additions.m

#import "UIImage+additions.h"

@implementation UIImage (additions)
-(UIImage*)makeRoundCornersWithRadius:(const CGFloat)RADIUS {
    UIImage *image = self;

    // Begin a new image that will be the new image with the rounded corners
    // (here with the size of an UIImageView)
    UIGraphicsBeginImageContextWithOptions(image.size, NO, image.scale);

    const CGRect RECT = CGRectMake(0, 0, image.size.width, image.size.height);
    // Add a clip before drawing anything, in the shape of an rounded rect
    [[UIBezierPath bezierPathWithRoundedRect:RECT cornerRadius:RADIUS] addClip];
    // Draw your image
    [image drawInRect:RECT];

    // Get the image, here setting the UIImageView image
    //imageView.image
    UIImage* imageNew = UIGraphicsGetImageFromCurrentImageContext();

    // Lets forget about that we were drawing
    UIGraphicsEndImageContext();

    return imageNew;
}
@end

If appIconImage is an UIImageView, then:

appIconImage.image = [UIImage imageWithContentsOfFile:@"image.png"]; 
appIconImage.layer.masksToBounds = YES;
appIconImage.layer.cornerRadius = 10.0;
appIconImage.layer.borderWidth = 1.0;
appIconImage.layer.borderColor = [[UIColor grayColor] CGColor];

And also remember:

#import <QuartzCore/QuartzCore.h>

Here is an even easier method that is available in iPhone 3.0 and up. Every View-based object has an associated layer. Each layer can have a corner radius set, this will give you just what you want:

UIImageView * roundedView = [[UIImageView alloc] initWithImage: [UIImage imageNamed:@"wood.jpg"]];
// Get the Layer of any view
CALayer * l = [roundedView layer];
[l setMasksToBounds:YES];
[l setCornerRadius:10.0];

// You can even add a border
[l setBorderWidth:4.0];
[l setBorderColor:[[UIColor blueColor] CGColor]];