iPhone图像拉伸（倾斜）

Question

如何歪斜图像？例如，每个角都有一个CGPoint，坐标为p1，p2，p3，p4。然后，我需要设置-p4.x + = 50，p4.y + = 30。因此，应在2D透视图中拉伸此角（p4），并使图像变形。替代文字（来源：polar-b.com）我尝试使用CATransform3D，但是似乎不能以这种方式完成，因为这只是改变视角（旋转，使一侧更近/更远）。也许CGAffineTransform有用吗？如果您知道答案，请编写示例代码。

Answer 1

我在Swift中尝试了一个很棒的答案@KennyTM，并收到一个错误“表达式太复杂，无法在合理的时间内解决”。

因此，这是Swift的简化版本：

let y21 = y2a - y1a

let y32 = y3a - y2a

let y43 = y4a - y3a

let y14 = y1a - y4a

let y31 = y3a - y1a

let y42 = y4a - y2a

let a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)

let b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)

let c0 = -H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43)

let cx = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)

let cy = -W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)

let c = c0 + cx + cy

let d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a)

let e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)

let f0 = -W*H*(x4a*y1a*y32 - x3a*y1a*y42 + x2a*y1a*y43)

let fx = H*X*(x4a*y21*y3a - x2a*y1a*y43 - x3a*y21*y4a + x1a*y2a*y43)

let fy = -W*Y*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)

let f = f0 + fx + fy;

let g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43)

let h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42)

let i0 = H*W*(x3a*y42 - x4a*y32 - x2a*y43)

let ix = H*X*(x4a*y21 - x3a*y21 + x1a*y43 - x2a*y43)

let iy = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42)

var i = i0 + ix + iy

let epsilon = CGFloat(0.0001);

if fabs(i) < epsilon {

    i = epsilon * (i > 0 ? 1 : -1);

}

return CATransform3D(m11: a/i, m12: d/i, m13: 0, m14: g/i, m21: b/i, m22: e/i, m23: 0, m24: h/i, m31: 0, m32: 0, m33: 1, m34: 0, m41: c/i, m42: f/i, m43: 0, m44: 1.0)