ios - iOS:将UIImage中的黑色更改为另一种颜色

Question

我有一个白色边框和黑色填充的图像。我只想在运行时将黑色更改为另一种颜色。用户将在运行时以HSB格式选择颜色。我怎样才能做到这一点？我通过尝试尝试 CGImageCreateWithMaskingColors
const float colorMasking [4] = {255，255，255，255};
但我每次都会得到nil CGImageRef 。请帮忙。

- (UIImage*) maskBlackInImage :(UIImage*) image color:(UIColor*)color
{
    const CGFloat colorMasking[4] = { 222, 255, 222, 255 };
    CGImageRef imageRef = CGImageCreateWithMaskingColors(image.CGImage, colorMasking);
    UIImage* imageB = [UIImage imageWithCGImage:imageRef];
    CGImageRelease(imageRef);
    return imageB;

}

我要附加灯泡的图像-灯泡具有黑色填充，白色边框和透明背景

更新:

在接受的答案中使用代码后，我能够用另一种颜色填充黑色。但是，我可以在白色边框上看到一点点颜色。图像看起来不那么清晰。附加输出:

Answer 1

创建一个UIImage类的类别并添加以下方法

- (UIImage *)imageTintedWithColor:(UIColor *)color
{
     UIImage *image;
     if (color) {
        // Construct new image the same size as this one.
        UIGraphicsBeginImageContextWithOptions([self size], NO, 0.0); // 0.0 for scale means "scale for device's main screen".
        CGRect rect = CGRectZero;
        rect.size = [self size];

        // tint the image
        [self drawInRect:rect];
        [color set];
        UIRectFillUsingBlendMode(rect, kCGBlendModeScreen);

        // restore alpha channel
        [self drawInRect:rect blendMode:kCGBlendModeDestinationIn alpha:1.0f];

        image = UIGraphicsGetImageFromCurrentImageContext();
        UIGraphicsEndImageContext();
    }
    return image;
}