在directx中渲染粗实线11

Question

我想在directx11中实现粗线。 我想，我可以使用实例化呈现技术来呈现“高品质”的几何形状为这样的图像的每一行示出了：

P1和P2表示不等距线verticies WICH在给定的作为“D3D11_PRIMITIVE_TOPOLOGY_LINELIST”。线条粗细存储在常量缓冲区中。每条线具有相同的厚度。 实例几何也有一个索引缓冲区，其中包含有关如何将顶点连接到三角形的信息（图中的顶点是I0-I11）。

我应该将P1和P2位置和SV_VertexID放到一个顶点着色器线程中，我可以计算出I0-I11顶点的每个位置。所以这不是问题。

问题是：是否可以使用实例化渲染技术来实现这一点？

如果是这样的话：像这样使用它是个好主意吗？还是有更多的性能方法来实现粗圆线？例如使用geometryshader，或者只是使用几何图形进行1000次绘图...

Answer 1

我尝试了很多使用实例渲染的想法，但现在我已将想法更改为geometryshader，并且它非常易于实现。 作为输入它变成一条线（2个顶点）并且输出是30个三角形。

这里pixelshader输入结构

struct PS_INPUT 
{ 
    float4 Pos : SV_POSITION; 
}; 

这里几何着色器：

#include <psiPosition.hlsli> 

//#pragma warning (disable:3206) 
//#pragma warning (disable:3554) 

static const float PI = 3.1415926f; 
static const float fRatio = 2.0f; 
static float fThickness = 0.01f; 

void addHalfCircle(inout TriangleStream<PS_INPUT> triangleStream, int nCountTriangles, float4 linePointToConnect, float fPointWComponent, float fAngle) 
{ 
    PS_INPUT output = (PS_INPUT)0; 
    for (int nI = 0; nI < nCountTriangles; ++nI) 
    { 
     output.Pos.x = cos(fAngle + (PI/nCountTriangles * nI)) * fThickness/fRatio; 
     output.Pos.y = sin(fAngle + (PI/nCountTriangles * nI)) * fThickness; 
     output.Pos.z = 0.0f; 
     output.Pos.w = 0.0f; 
     output.Pos += linePointToConnect; 
     output.Pos *= fPointWComponent; 
     triangleStream.Append(output); 

     output.Pos = linePointToConnect * fPointWComponent; 
     triangleStream.Append(output); 

     output.Pos.x = cos(fAngle + (PI/nCountTriangles * (nI + 1))) * fThickness/fRatio; 
     output.Pos.y = sin(fAngle + (PI/nCountTriangles * (nI + 1))) * fThickness; 
     output.Pos.z = 0.0f; 
     output.Pos.w = 0.0f; 
     output.Pos += linePointToConnect; 
     output.Pos *= fPointWComponent; 
     triangleStream.Append(output); 

     triangleStream.RestartStrip(); 
    } 
} 

[maxvertexcount(42)] 
void main(line PS_INPUT input[2], inout TriangleStream<PS_INPUT> triangleStream) 
{ 
    PS_INPUT output= (PS_INPUT)0; 

    int nCountTriangles =6; 

    float4 positionPoint0Transformed = input[0].Pos; 
    float4 positionPoint1Transformed = input[1].Pos; 

    float fPoint0w = positionPoint0Transformed.w; 
    float fPoint1w = positionPoint1Transformed.w; 

    //calculate out the W parameter, because of usage of perspective rendering 
    positionPoint0Transformed.xyz = positionPoint0Transformed.xyz/positionPoint0Transformed.w; 
    positionPoint0Transformed.w = 1.0f; 
    positionPoint1Transformed.xyz = positionPoint1Transformed.xyz/positionPoint1Transformed.w; 
    positionPoint1Transformed.w = 1.0f; 

    //calculate the angle between the 2 points on the screen 
    float3 positionDifference = positionPoint0Transformed.xyz - positionPoint1Transformed.xyz; 
    float3 coordinateSysten = float3(1.0f, 0.0f, 0.0f); 

    positionDifference.z = 0.0f; 
    coordinateSysten.z = 0.0f; 

    float fAngle = acos(dot(positionDifference.xy, coordinateSysten.xy)/(length(positionDifference.xy) * length(coordinateSysten.xy))); 

    if (cross(positionDifference, coordinateSysten).z < 0.0f) 
    { 
     fAngle = 2.0f * PI - fAngle; 
    } 

    fAngle *= -1.0f; 
    fAngle -= PI *0.5f; 

    //first half circle of the line 
    addHalfCircle(triangleStream, nCountTriangles, positionPoint0Transformed, fPoint0w, fAngle); 
    addHalfCircle(triangleStream, nCountTriangles, positionPoint1Transformed, fPoint1w, fAngle + PI); 

    //connection between the two circles 
    //triangle1 
    output.Pos.x = cos(fAngle) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint0Transformed; 
    output.Pos *= fPoint0w; //undo calculate out the W parameter, because of usage of perspective rendering 
    triangleStream.Append(output); 

    output.Pos.x = cos(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint0Transformed; 
    output.Pos *= fPoint0w; 
    triangleStream.Append(output); 

    output.Pos.x = cos(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint1Transformed; 
    output.Pos *= fPoint1w; 
    triangleStream.Append(output); 

    //triangle2 
    output.Pos.x = cos(fAngle) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint0Transformed; 
    output.Pos *= fPoint0w; 
    triangleStream.Append(output); 

    output.Pos.x = cos(fAngle) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint1Transformed; 
    output.Pos *= fPoint1w; 
    triangleStream.Append(output); 

    output.Pos.x = cos(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness/fRatio; 
    output.Pos.y = sin(fAngle + (PI/nCountTriangles * (nCountTriangles))) * fThickness; 
    output.Pos.z = 0.0f; 
    output.Pos.w = 0.0f; 
    output.Pos += positionPoint1Transformed; 
    output.Pos *= fPoint1w; 
    triangleStream.Append(output); 
} 

我知道这是非常硬编码的，但至少它的工作原理;） 现在图片立方体的线条较粗（第一次是透视投影，第二次是正投影）

我希望我可以帮助这个人。如果有人有更好的实现粗线的想法，请发表评论。