c++ - 读取文件c++的最快方法

Question

我想读入这样的文件:

13.3027 29.2191 2.39999
13.3606 29.1612 2.39999
13.3586 29.0953 2.46377
13.4192 29.106 2.37817

它有超过 1mio 行。

我当前的 cpp 代码是:

loadCloud(const string &filename, PointCloud<PointXYZ> &cloud)
{
    print_info("\nLoad the Cloud .... (this takes some time!!!) \n");
    ifstream fs;
    fs.open(filename.c_str(), ios::binary);
    if (!fs.is_open() || fs.fail())
    {
        PCL_ERROR(" Could not open file '%s'! Error : %s\n", filename.c_str(), strerror(errno));
        fs.close();
        return (false);
    }

    string line;
    vector<string> st;

    while (!fs.eof())
    {
        getline(fs, line);
        // Ignore empty lines
        if (line == "") 
        {
            std::cout << "  this line is empty...." << std::endl;
            continue;
        }

        // Tokenize the line
        boost::trim(line);
        boost::split(st, line, boost::is_any_of("\t\r "), boost::token_compress_on);

        cloud.push_back(PointXYZ(float(atof(st[0].c_str())), float(atof(st[1].c_str())), float(atof(st[2].c_str()))));
    }
    fs.close();
    std::cout<<"    Size of loaded cloud:   " << cloud.size()<<" points" << std::endl;
    cloud.width = uint32_t(cloud.size()); cloud.height = 1; cloud.is_dense = true;
    return (true);
}

目前读取这个文件需要很长时间。我想加快这个速度，有什么想法该怎么做吗？

Answer 1

只要数字始终以三个为一组出现，您就可以只读取数字而不是整行加上解析。

void readFile(const std::string& fileName)
{
    std::ifstream infile(fileName);

    float vertex[3];
    int coordinateCounter = 0;

    while (infile >> vertex[coordinateCounter])
    {
        coordinateCounter++;
        if (coordinateCounter == 3)
        {
            cloud.push_back(PointXYZ(vertex[0], vertex[1], vertex[2]));
            coordinateCounter = 0;
        }
    }
}