Mat i1 = imread("1.jpg", 0); //read as a gray scale image
Mat i2 = imread("2.jpg", 0); //reas as a gray scale image
Mat flowMat;
vector <Point2f> i1_corner, i2_corner;
vector <uchar> status;
vector <float> err;
goodFeaturesToTrack(i1,
下面是在Couldn't match type ‘a’ with ‘a1’错误上失败的Haskell代码:
bar :: [Int] -> (a -> Int -> a) -> a -> a
bar ns fp ap = snd $ foldl fn (fp, ap) ns
where fn :: ((a -> Int -> a), a) -> Int -> ((a -> Int -> a), a)
fn (f, x) i = (fp, (f x (i + length(ns))))
下面是详细的错误
我有一个简单的函数,可以一次读取一个字节的二进制文件。它会得到一个编译时错误,如下。问题似乎是bs2,BSSC.length的结果ByteString,有一个未知的类型。我是否遗漏了r上的类型约束 import qualified Data.ByteString.Streaming.Char8 as BSSC
main :: IO ()
main = runResourceT $ dump $ BSSC.readFile "filename"
dump :: (MonadIO m) => BSSC.ByteString m r -> m ()
我有以下类型和函数定义:
data Tree a = Tree a [Tree a]
list t = list' [t]
where list' [] = []
list' (Tree a hs:ts) = a : list' (ts ++ hs)
我有这样的表达:
list (Tree (-1) [Tree 0 [ Tree 4 [], Tree 7 []],Tree 8 [Tree 5 []]])
这给了我:
[-1,0,8,4,7,5]
问题是我不明白为什么!我认为我的问题是,我不明白为什么li
我正在尝试获取对象具有的原始材质,并添加场景中的平行光:
protected function onMeshComplete(event:AssetEvent):void {
if (event.asset.assetType == AssetType.MESH) {
myMesh = event.asset as Mesh;
for each (var m:SubMesh in myMesh.subMeshes){
var mat:MaterialBase = m.material;
mat.light