Question

我有一个二进制数据文件，按顺序包含2d和3d坐标：

uint32 numberOfUVvectors;
2Dvec uv[numberOfUVvectors];
uint32 numberOfPositionVectors;
3Dvec position[numberOfPositionVectors];
uint32 numberOfNormalVectors;
3Dvec normal[numberOfNormalVectors];

2Dvec和3Dvec是分别由2个和3个浮点组成的结构。

首先，我使用“通常”的方式阅读所有这些值：

in.read(reinterpret_cast<char *>(&num2d), sizeof(uint32));
2Dvectors.reserve(num2d); // It's for an std::vector<2DVec> 2Dvectors();
for (int i = 0; i < num2d; i++){
    2Dvec 2Dvector;
    in.read(reinterpret_cast<char *>(&2Dvector), sizeof(2DVec));
    2Dvectors.push_back(2Dvector);
}

它运行良好，但速度很慢（文件中可能有超过200k条目，并且有如此多的读取调用，hdd访问成为瓶颈）。我决定立刻将整个文件读入缓冲区：

in.seekg (0, in.end);
int length = in.tellg();
in.seekg (0, in.beg);

char * buffer = new char [length];

is.read (buffer,length);

现在阅读速度更快，但问题是：如何将char缓冲区解析为整数和结构？

Answer 1

回答您的具体问题：

unsigned char * pbuffer = (unsigned char *)buffer;

uint32 num2d = *((uint32 *)pbuffer);
pbuffer += sizeof(uint32);
if(num2d)
{
    2Dvec * p2Dvec = (2Dvec *)pbuffer;
    2Dvectors.assign(p2Dvec, p2Dvec + num2d);
    pbuffer += (num2d * sizeof(2Dvec));
}

uint32 numpos = *((uint32 *)pbuffer);
pbuffer += sizeof(uint32);
if(numpos)
{
        3Dvec * p3Dvec = (3Dvec *)pbuffer;
    Pos3Dvectors.assign(p3Dvec, p3Dvec + numpos);
    pbuffer += (numpos * sizeof(3Dvec));
}

uint32 numnorm = *((uint32 *)pbuffer);
pbuffer += sizeof(uint32);
if(numnorm)
{
    3Dvec * p3Dvec = (3Dvec *)pbuffer;
    Normal3Dvectors.assign(p3Dvec, p3Dvec + numnorm);
    pbuffer += (numnorm * sizeof(3Dvec));
}
// do not forget to release the allocated buffer

更快的方法是：

in.read(reinterpret_cast<char *>(&num2d), sizeof(uint32));
if(num2d)
{
    2Dvectors.resize(num2d);
    2Dvec * p2Dvec = &2Dvectors[0];
    in.read(reinterpret_cast<char *>(&p2Dvec), num2d * sizeof(2Dvec));
}
//repeat for position & normal vectors

Answer 2

使用具有适当大小和起始值的memcpy

或转换值（示例）：

#include <iostream>

void copy_array(void *a, void const *b, std::size_t size, int amount)
{
    std::size_t bytes = size * amount;
    for (int i = 0; i < bytes; ++i)
        reinterpret_cast<char *>(a)[i] = static_cast<char const *>(b)[i];
}

int main()
{
    int a[10], b[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

    copy_array(a, b, sizeof(b[0]), 10);

    for (int i = 0; i < 10; ++i)
        std::cout << a[i] << ' ';
}

从已读入内存的文件中提取数据

2 个答案: