Question

我有一个QTableWidget，用户输入各种样式的复数。例如，复数（-15 + 8.14i）可以写成：

-15 + 8.14i

-15 + 8.14j

-15 + j 8,14

-15+ i8,14

我也可以 j ！两个值都可以很大（它们保存为double）也可以是负值。它们可以用＆＃34;，＆＃34;和＆＃34;。＆＃34; （所以3.14和3,14是相同的）。当用户输入错误的数字时，应该会显示错误消息。

CKomplex fromString(QString str) { // ckomplex is my custom class for complex numbers
    double numReal, numImag;

    QString strNew = "";
    // delete all spaces
    for (int i= 0; i< str.length(); i++) {
        if (!str[i].isSpace()) {
            strNew += str[i];
        }
    }


    QString part1 = "";
    int index;
    // get the first number
    for (int i= 0; i < strNew.length(); i++) { // iterate string
        if (strNew[i] != '+' && strNew[i] != '-') {
            part1 += strNew[i];
        } else { // e.g.: 5 + 3j -> the loop is at the "+"
            if (i != 0) {
                index = i; // save index at "+" to start for next number
                break;
            }
        }
    }
    numReal = part1.toDouble();


    QString part2 = "";
    // get the second number
    for (int i= index; i < strNew.length(); i++) {
        if (strNew[i].isDigit() || strNew[i] == '+' || strNew[i] == '-' || strNew[i] == '.' || strNew[i] == ',') { // ignore j or i
            part2 += strNew[i];
        }
    }
    numImag = part2.toDouble();

    return CKomplex(numReal, numImag);
}

这适用于基本输入。但不是非常快或可读或有用。它确实涵盖了很少的输入可能性（例如＆＃34; -3 - 5,14和＃34;不起作用）。有没有更简单的方法将字符串转换为复数而没有太多的循环和变量？

Answer 1

单个regular expression可以解析每一行：

#include <string>
#include <sstream>
#include <vector>
#include <iterator>
#include <regex>
#include <iostream>
#include <iomanip>
#include <exception>

class CKomplex {
public:
    CKomplex(double numReal, double numImag) : numReal{numReal}, numImag{numImag} {}
    double numReal;
    double numImag;
};

auto input_text{
R"(-15 + 8.14i
-15+8.14j
-15 +j 8,14
-15+ i8,14
bad line here
+23.4-j24
-35+42.3j
+24i
+2.342j
+24.523-i 432,52
24.523-i 432,52
23.4-j24
35+42.3j
24i
2.342j)"};

CKomplex fromString(std::string str) {
    double numReal{};
    double numImag{};

    std::regex r{R"(([+-]?) *([ij]?) *(\d+)[.,]?(\d*)([ij])?)"}; // 6 groups
    std::istringstream iss(str);
    auto it = std::sregex_iterator(str.begin(), str.end(), r);
    auto end = std::sregex_iterator();
    if(it == end || it->size() != 6)
        throw std::runtime_error("Could not parse line containing the following text: " + str);
    for(; it != end; ++it) {
        auto match = *it;
        auto sign = match[1].str();
        auto iorj_pre = match[2].str();
        auto decimal = match[3].str();
        auto fraction = match[4].str();
        auto iorj_post = match[5].str();

        double val{sign == "-" ? -1.F : 1.F};
        val *= std::stod(decimal + "." + fraction);
        if(iorj_pre == "i" || iorj_pre == "j" || iorj_post == "i" || iorj_post == "j")
            numImag += val;
        else
            numReal += val;
    }
    return{numReal,numImag};
}

std::ostream& operator<<(std::ostream& os, const CKomplex& complex_number)
{
    os << std::showpos << "(" << complex_number.numReal << " " << complex_number.numImag << "i)";
    return os;
}

int main()
{
    std::istringstream input_stream{input_text};

    for(std::string line{}; std::getline(input_stream, line);) {
        try { std::cout << std::setw(20) << line << ": " << fromString(line) << '\n'; }
        catch(std::exception& e) { std::cout << e.what() << '\n'; }
    }
    return 0;
}

制作（live demo）：

         -15 + 8.14i: (-15 +8.14i)
           -15+8.14j: (-15 +8.14i)
         -15 +j 8,14: (-15 +8.14i)
          -15+ i8,14: (-15 +8.14i)
Could not parse line containing the following text: bad line here
           +23.4-j24: (+23.4 -24i)
           -35+42.3j: (-35 +42.3i)
                +24i: (+0 +24i)
             +2.342j: (+0 +2.342i)
    +24.523-i 432,52: (+24.523 -432.52i)
     24.523-i 432,52: (+24.523 -432.52i)
            23.4-j24: (+23.4 -24i)
            35+42.3j: (+35 +42.3i)
                 24i: (+0 +24i)
              2.342j: (+0 +2.342i)

如何从用户输入中获取数据？

1 个答案: