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用索引直接访问替换循环(以及循环中的条件)

qt c++

-1

Megidd · 技术社区 · 7 年前

我有一个for循环,循环中有一个条件,可以在 index :

// uint index = ...
// const float *bufferPtr = ...
// uint stride = ...
// uint vertexCount = ...

for (uint i = 0; i < vertexCount; i++) {
    float xVal = *bufferPtr++;
    float yVal = *bufferPtr++;
    float zVal = *bufferPtr++;
    bufferPtr += stride;
    if (i == index) {
        qDebug() << "Vertex coord: " << xVal << " , " << yVal << " , " << zVal;
    }
}

我试图用索引直接访问替换for循环(及其内部条件):

float xVal = *(bufferPtr + index * stride + 0);
float yVal = *(bufferPtr + index * stride + 1);
float zVal = *(bufferPtr + index * stride + 2);
qDebug() << "Vertex coord without loop: " << xVal << " , " << yVal << " , " << zVal;

但是输出日志给了我 不同的 结果:

Vertex coord:  14.574  ,  -8.236  ,  7.644
Vertex coord without loop:  20.67  ,  -19.098  ,  18.536
Vertex coord:  14.552  ,  -8.024  ,  7.842
Vertex coord without loop:  -0.361096  ,  0.109164  ,  0.926117
Vertex coord:  14.722  ,  -8.18  ,  7.842
Vertex coord without loop:  20.648  ,  -19.052  ,  18.522

我不明白为什么结果不同:(

修理

正如@LanceDeGate answer所建议的,通过减少 stride 通过 3 循环之前:


stride = stride - 3; // Three floats per vertex

for (uint i = 0; i < vertexCount; i++) {
    float xVal = *bufferPtr++;
    float yVal = *bufferPtr++;
    float zVal = *bufferPtr++;
    bufferPtr += stride;
    if (i == index) {
        qDebug() << "Vertex coord: " << xVal << " , " << yVal << " , " << zVal;
    }
}

现在日志是一样的:

Vertex coord:  -0.522632  ,  -0.803892  ,  -9.02102
Vertex coord without loop:  -0.522632  ,  -0.803892  ,  -9.02102
Vertex coord:  -0.39095  ,  -2.04955  ,  -8.91668
Vertex coord without loop:  -0.39095  ,  -2.04955  ,  -8.91668
Vertex coord:  -0.259928  ,  -0.804899  ,  -9.03231
Vertex coord without loop:  -0.259928  ,  -0.804899  ,  -9.03231

3 回复 | 直到 7 年前

Lance DeGate 7 年前

也许是因为 大步走 在三个“bufferPtr++”之后,添加到bufferPtr。

也许这就是你的意思:

float xVal = *bufferPtr;
float yVal = *(bufferPtr+1);
float zVal = *(bufferPtr+2);
bufferPtr += stride;

或

float xVal = *bufferPtr++;
float yVal = *bufferPtr++;
float zVal = *bufferPtr++;
bufferPtr += (stride-3);

Klaus 7 年前

第一个提示:

如果可能的话,请提供一个每个人都可以编译的完整示例。这需要一些时间来启动和运行你的代码。。。

好吧,据我所知你的!代码是这样的:

float var[]= { 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
size_t elements = sizeof(var)/sizeof(float);
int stride = 2;
int vertexCount = elements/(3+stride);

void f( float* bufferPtr, int index )
{
    for (uint i = 0; i < vertexCount; i++)
    {
        float xVal = *bufferPtr++;
        float yVal = *bufferPtr++;
        float zVal = *bufferPtr++;
        bufferPtr += stride;
        if (i == index) {
        std::cout << "Vertex coord: " << xVal << " , " << yVal << " , " << zVal << std::endl;
        }
    }
}

可以简化为:

void f2( float* bufferPtr, int index )
{   
    struct Data
    {
        float x;
        float y;
        float z;
        float dummy[2]; // stride
    };

   Data& d = (reinterpret_cast<Data*>(bufferPtr))[index];
   std::cout << "Vertex coord: " << d.x << " " << d.y << " " << d.z << std::endl;
}   


int main()
{   
    f( var, 2 );
    f2( var, 2 );
}

borizzzzz 7 年前

通过以下测试用例,我得到了正确的结果:

#include <iostream>
int main()
{
  float tabla[16] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
  unsigned int index = 0;
  const float *bufferPtr = &tabla[0];
  unsigned int stride = 2;
  unsigned int vertexCount = 2;

  for (uint i = 0; i < vertexCount; i++) {
    float xVal = *bufferPtr++;
    float yVal = *bufferPtr++;
    float zVal = *bufferPtr++;
    bufferPtr += stride;
    if (i == index) {
      std::cout << "Vertex coord: " << xVal << " , " << yVal << " , " << zVal << std::endl;
    }
  }
  const float *bufferPtr2 = &tabla[0];
  float xVal2 = *(bufferPtr2 + index * stride + 0);
  float yVal2 = *(bufferPtr2 + index * stride + 1);
  float zVal2 = *(bufferPtr2 + index * stride + 2);
  std::cout << "Vertex coord without loop: " << xVal2 << " , " << yVal2 << " , " << zVal2 << std::endl;
  return 0;
}

输出:

Vertex coord: 0 , 1 , 2
Vertex coord without loop: 0 , 1 , 2

我根本没有改变你的代码。唯一的区别是,我在一个主函数中添加了两个测试,显然使用了不同的缓冲区指针( bufferPtr2 )我用表格的第一个地址初始化 tabla .您确定在尝试其他方法之前重置了指针吗?很难说,因为您只提供了代码片段。