用rust语言写一个实现透视变形的函数

好的，以下是一个用 Rust 语言实现透视变形的函数的示例代码：

use image::{ImageBuffer, Rgb};

fn perspective_transform(src: &ImageBuffer<Rgb<u8>, Vec<u8>>, 
                         tl: (f32, f32), 
                         tr: (f32, f32), 
                         bl: (f32, f32), 
                         br: (f32, f32)) -> ImageBuffer<Rgb<u8>, Vec<u8>> {
    let (width, height) = src.dimensions();
    let mut output = ImageBuffer::new(width, height);

    let a = tl.0 * (tr.1 - br.1) + tr.0 * (br.1 - tl.1) + br.0 * (tl.1 - tr.1);
    let b = tl.1 * (tr.0 - br.0) + tr.1 * (br.0 - tl.0) + br.1 * (tl.0 - tr.0);
    let c = tl.0 * (tr.1 * br.0 - br.1 * tr.0) + tr.0 * (br.1 * tl.0 - tl.1 * br.0) + br.0 * (tl.1 * tr.0 - tr.1 * tl.0);
    let d = bl.0 * (tr.1 - br.1) + br.0 * (bl.1 - tr.1) + tr.0 * (br.1 - bl.1);
    let e = bl.1 * (tr.0 - br.0) + br.1 * (bl.0 - tr.0) + tr.1 * (br.0 - bl.0);
    let f = bl.0 * (tr.1 * br.0 - br.1 * tr.0) + br.0 * (bl.1 * tr.0 - tr.1 * bl.0) + tr.0 * (br.1 * bl.1 - bl.1 * br.1);
    let g = tl.0 * (bl.1 - br.1) + bl.0 * (tl.1 - br.1) + br.0 * (bl.1 - tl.1);
    let h = tl.1 * (bl.0 - br.0) + bl.1 * (tl.0 - br.0) + br.1 * (bl.0 - tl.0);
    let i = tl.0 * (bl.1 * br.0 - br.1 * bl.0) + bl.0 * (tl.1 * br.0 - br.1 * tl.0) + br.0 * (bl.1 * tl.0 - tl.1 * bl.0);

    for y in 0..height {
        for x in 0..width {
            let w = a * x as f32 + b * y as f32 + c;
            let xp = ((d * x as f32) + (e * y as f32) + f) / w;
            let yp = ((g * x as f32) + (h * y as f32) + i) / w;

            if xp >= 0.0 && xp < width as f32 && yp >= 0.0 && yp < height as f32 {
                let pixel = src.get_pixel(xp as u32, yp as u32);
                output.put_pixel(x, y, *pixel);
            }
        }
    }

    output
}

这个函数接受一个 RGB 图像和四个点的坐标，返回一个透视变形后的图像。具体实现中，使用了透视变换的矩阵表达式，并遍历图像的每个像素，计算出它在透视变形后的位置，然后将原图像对应位置的像素值复制到输出图像中。注意，我们需要对计算出的像素位置进行边界检查，以确保它们在图像范围内。