Question

我使用以下代码来实现此目的。问题是当图像旋转时，它将图像缩放到更接近度数的45'。

当图像设置为90'及其倍数时，图像尺寸没有问题。但是当度数接近45'及其倍数时，图像尺寸会明显缩小。我不想改变图像大小。我该如何处理这个问题？

- (UIImage *)imageRotatedByRadians:(CGFloat)radians img:(UIImage *)img
{
    // calculate the size of the rotated view's containing box for our drawing space
    UIView *rotatedViewBox = [[UIView alloc] initWithFrame:CGRectMake(0,0,img.size.width, img.size.height)];
    UIView *nonrotatedViewBox = rotatedViewBox;
    CGAffineTransform t = CGAffineTransformMakeRotation(radians);
    rotatedViewBox.transform = t;
    CGSize rotatedSize = rotatedViewBox.frame.size;

    // Create the bitmap context
    UIGraphicsBeginImageContext(nonrotatedViewBox.frame.size);
    CGContextRef bitmap = UIGraphicsGetCurrentContext();

    // Move the origin to the middle of the image so we will rotate and scale around the center.
    CGContextTranslateCTM(bitmap, rotatedSize.width/2, rotatedSize.height/2);

    //Rotate the image context
    CGContextRotateCTM(bitmap, radians);

    // Now, draw the rotated/scaled image into the context
    CGContextScaleCTM(bitmap, 1.0, -1.0);
    CGContextDrawImage(bitmap, CGRectMake(-img.size.width / 2, -img.size.height / 2, img.size.width, img.size.height), [img CGImage]);

    UIImage *newImage = UIGraphicsGetImageFromCurrentImageContext();
    UIGraphicsEndImageContext();
    return newImage;
}

Answer 1

您的视图尺寸错误，您不想以宽度和高度除以2绘制图像。因为当图像旋转时，它的宽度变为（指的是完美的square）是宽度乘以2倍sqr（2）。所以如果你的宽度和高度都是10，那么你的新宽度（45度角）将在14到15之间...所以你的上下文在10x10区域绘制一个14x14图像

你可以通过一些复杂的数学运算来获得使用几何体的完美尺寸，或者你可以将其展开并说出double percentRotated =（弧度％M_PI_4）/ M_PI_4; double newWidth =（percentRotated == 0？img.size.width：img.size.width *（1 +（percentRotated％0.25）））;并为高度做同样的事情，＃lazyprogrammer

通过自定义程度旋转UIImage

1 个答案: