按自定义度数旋转 UIImage

Question

我正在使用以下代码来实现这一点。问题是当图像旋转时，它会将图像缩放得越接近 45 度。

当图像设置为 90' 及其倍数时，图像大小没有问题。但是当度数接近 45' 及其倍数时，图像尺寸会明显缩小。我不想更改图像大小。我应该如何解决这个问题？

- (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 的方式绘制图像。因为当图像旋转时，它的宽度变为（当指的是完美正方形时）是宽度乘以2 次 sqr(2)。因此，如果您的宽度和高度为 10，那么您的新宽度（以 45 度角）将在 14 到 15 之间......所以您的上下文是在 10x10 区域中绘制 14x14 图像

你可以通过一些复杂的数学运算来使用几何获得完美的尺寸，或者你可以用它说 | double percentRotated = (弧度% M_PI_4)/M_PI_4;双 newWidth = (percentRotated == 0 ? img.size.width : img.size.width * (1 + (percentRotated % 0.25)) );并为高度做同样的事情，#lazyprogrammer