Setting A CGContext Transparent Background

Question

I am still struggling with drawing a line with CGContext. I have actually go to line to draw, but now I need the background of the Rect to be transparent so the existing background shows thru. Here's my test code:

(void)drawRect:(CGRect)rect
{
    CGContextRef context = UIGraphicsGetCurrentContext();
    CGContextSetFillColorWithColor(context, [UIColor clearColor].CGColor);
    CGContextSetAlpha(context,0.0);
    CGContextFillRect(context, rect);

    CGContextSetStrokeColorWithColor(context, [UIColor whiteColor].CGColor);
    CGContextSetLineWidth(context, 5.0);
    CGContextMoveToPoint(context, 100.0,0.0);
    CGContextAddLineToPoint(context,100.0, 100.0);
    CGContextStrokePath(context);
}

Any ideas?

Answer 1

Init a context with opaque == false, Swift 3

UIGraphicsBeginImageContextWithOptions(size, false, UIScreen.main.scale) 

opaque

A Boolean flag indicating whether the bitmap is opaque. If you know the bitmap is fully opaque, specify true to ignore the alpha channel and optimize the bitmap’s storage. Specifying false means that the bitmap must include an alpha channel to handle any partially transparent pixels.

Answer 2

This is what worked for me with a UIImage which had been manually added using InterfaceBuilder.

- (id)initWithCoder:(NSCoder *)aDecoder {

    if(self = [super initWithCoder:aDecoder]) {
        self.backgroundColor = [UIColor clearColor];
    }

    return self;
}


-(void)drawRect:(CGRect)rect
{
    CGContextRef context = UIGraphicsGetCurrentContext();    
    CGContextSetStrokeColorWithColor(context, [UIColor whiteColor].CGColor);
    CGContextSetLineWidth(context, 5.0);
    CGContextMoveToPoint(context, 100.0,0.0);
    CGContextAddLineToPoint(context,100.0, 100.0);
    CGContextStrokePath(context);
}

David Kanarek's answer only works when you're manually creating your own UIImageView. If you've created a UIView and manually added it via Interface Builder then you will need a different approach like this calling the initWithCoder method instead.