Question

I have a function that will overlay images over another image and return as a UIImage. However, a white line-ish thing comes up on it. Here’s what I have

- (UIImage * ) romanticFilterImage: (UIImage *) imageA {

UIImage *vintageOne = [UIImage imageNamed:@"filter_vintage_1.png"];
UIImage *vintageTwo= [UIImage imageNamed:@"filter_vintage_2.png"];
UIImage *vintageThree = [UIImage imageNamed:@"filter_vintage_3.png"];

UIGraphicsBeginImageContextWithOptions(CGSizeMake(imageA.size.width, imageA.size.height), YES, 0.0); 
[imageA drawAtPoint: CGPointMake(0,0)];
[vintageThree drawInRect:CGRectMake(0, 0, imageA.size.width, imageA.size.height)  blendMode: kCGBlendModeMultiply alpha: 1.0];
[vintageTwo drawInRect:CGRectMake(0, 0, imageA.size.width + 2, imageA.size.height + 2)  blendMode: kCGBlendModeColorBurn alpha: 1.0];
[vintageOne drawInRect:CGRectMake(0, 0, imageA.size.width, imageA.size.height)  blendMode: kCGBlendModeMultiply alpha: 1.0];
UIImage *source = UIGraphicsGetImageFromCurrentImageContext(); 
UIGraphicsEndImageContext();

CGSize size = [source size];
UIGraphicsBeginImageContext(size);
CGRect rect = CGRectMake(0, 0, size.width, size.height);
[source drawInRect:rect blendMode:kCGBlendModeNormal alpha:1.0];

CGContextRef context = UIGraphicsGetCurrentContext();
CGContextSetRGBStrokeColor(context, 5.0, 5.0, 5.0, 5.0); 
CGContextStrokeRect(context, rect);
UIImage *testImg =  UIGraphicsGetImageFromCurrentImageContext();
UIGraphicsEndImageContext();
return testImg;

}

Here is a screenshot: (I have added a black background to enhance the slighly transparent border.

Answer 1

Aren’t you stroking the image white in these lines?

CGContextRef context = UIGraphicsGetCurrentContext();
CGContextSetRGBStrokeColor(context, 5.0, 5.0, 5.0, 5.0); 
CGContextStrokeRect(context, rect);
UIImage *testImg =  UIGraphicsGetImageFromCurrentImageContext();
UIGraphicsEndImageContext();

Although I am not sure if you intended GContextSetRGBStrokeColor(context, 5.0, 5.0, 5.0, 5.0); as 1.0 is the maximum value for the rgba values in the function.