C++:strod() 和 atof() 没有按预期返回 double

C++: strod() and atof() are not returning double as expected

我正在尝试使用从 .txt 文件中获得的值将字符串转换为双精度值。

我得到的双打没有小数点。我相信这是因为在 .txt 中,数字的小数点由逗号而不是点分隔。但是我不知道怎么解决。

这是我的代码的简化版:

#include <iostream>
#include <fstream> // read text file
#include <stdlib.h> // strtod, atof

int main() {

std::ifstream readWindData("winddata.txt");

// check that document has been opened correctly
if (!readWindData.is_open()) {
    std::cout << "Wind data could not be opened." << std::endl;
    return 1;
}

// skip headers of the table
std::string firstLine;
std::getline(readWindData, firstLine);

int numberOfRows = 0; // variable to count rows of the table

// initialise strings that separate each value
std::string time, string u10st, u40st, u60st, u80st, u100st, 
            u116st, u160st, dir10st, dir60st, dir100st, timeDecst;

// initialise doubles
double u10, u40, u60, u80, u100, u116, u160, dir10, dir60, dir100, timeDec;

std::string nextLine;

// Read strings and turn it into doubles line by line until end
while (readWindData >> time >> u10st >> u40st >> u60st >> u80st >> u100st
        >> u116st >> u160st >> dir10st >> dir60st >> dir100st >> timeDecst) {

    // try two different functions to turn strings into doubles:
    u10 = strtod(u10st.c_str(), NULL);
    u40 = atof(u40st.c_str());

    // ensure numbers are displaying all their decimals
    std::cout.precision(std::numeric_limits<double>::digits10 + 1);

    // see what I am getting
    std::cout << time << " " << u10st << " " << u10 << " " << u40 << "\n";

    std::getline(readWindData, nextLine); // this line skips some crap on the side of some rows

    numberOfRows++; // counts rows
}

std::cout << "Number of rows = " << numberOfRows << "\n";
readWindData.close();

return 0;
}

这是文件的三行:

time (hour) u10(m/s)u40(m/s)u60 (m/s)u80(m/s)u100(m/s)u116(Um/s)u160(m/s)dir10 dir60 dir100         time decimal hours      
00:00       4,25636 7,18414 8,56345 9,75567 10,9667 12,1298 13,8083 110,616 131,652 141,809         0       midnight
00:10       4,54607 7,40763 8,62832 9,91782 11,2024 12,2694 14,1229 114,551 133,624 142,565         0,166666667 

这些是用上面的代码输出的那些行: (提醒,我std::cout时间(字符串),u10st(字符串),u10(双),u40(双)。

00:00 4,25636 4 7
00:10 4,54607 4 7

关于如何将 4,25636 字符串读入双精度 4.25636 有什么想法吗?文件太长无法修改。

浮点数使用逗号小数点分隔符,而您需要句点。

这表明数据是使用与您的locale 不同的序列化的。

要解决此问题,请在解析数字之前设置您的语言环境(然后再恢复)。

例如,这里有一些代码使用默认句点分隔符解析数字“12,34”,然后我们将区域设置设置为丹麦并重试:

const char* number = "12,34";
double parsed_number = 0;
parsed_number = std::strtod(number, nullptr);
std::cout << "parsed number in default locale: " << parsed_number << std::endl;
std::cout << "Setting locale to Denmark (comma decimal delimiter)" << std::endl;
std::locale::global(std::locale("en_DK.utf8"));
parsed_number = std::strtod(number, nullptr);
std::cout << "parsed number in Denmark locale: " << parsed_number << std::endl;
//restore default locale
std::locale::global(std::locale(""));

输出:

parsed number in default locale: 12
Setting locale to Denmark (comma decimal delimiter)
parsed number in Denmark locale: 12.34

Live Demo

四处询问,看看是谁序列化了这些数据,并获得正确的语言环境。 您可以使用 locale -a

在 *nix 系统中找到可用的语言环境

在 Windows 上似乎是 a little more difficult

为了避免弄乱全局语言环境,您可以在调用 strtod()

之前将 , 替换为 .
    std::replace(u10st.begin(), u10st.end(), ',', '.');
    u10 = strtod(u10st.c_str(), nullptr);

但是在上面显示的程序中,如果您使用 comma 是小数的语言环境,您也可以使用运算符 >> 直接从 ifstream 读取到 double分隔符。

    readWindData.imbue(std::locale("de_DE.UTF-8")); // or fr, nl, ru, etc.

    double u10, u40, u60, u80, u100, u116, u160, dir10, dir60, dir100, timeDec;

    while (readWindData >> time >> u10 >> u40 >> u60 >> u80 >> u100
            >> u116 >> u160 >> dir10 >> dir60 >> dir100 >> timeDec) {

Live demo