创建“如果不是整数”if 语句条件 (C++)

Question

处理用户以某种格式输入一些数据（例如“Ernest Hemingway, 9”）的作业，然后将这些信息放入表格中。

我必须考虑用户输入中的错误。不放逗号，放太多逗号，检查“，”后面的数据是否为有效整数。

例如。 欧内斯特·海明威 9

错误：字符串中没有逗号。

欧内斯特，海明威，9

错误：输入的逗号过多。

欧内斯特·海明威，九岁（输入我遇到问题）

错误：逗号后面没有整数。

欧内斯特·海明威，9 岁

数据字符串：欧内斯特·海明威 数据整数：9

下面是我当前的代码

    int commaCount = 0; // keeps track of commas in users input
    int currentIndex = 0; // creating variable to create elements in the vectors

    do
    {
        cout << "Enter a data point (-1 to stop input) :" << endl;
        getline(cin, userInput);

        if (userInput == "-1") { break; } // meant to break out of loop if the user inputs "-1" on their first attempt

        for (int i = 0; i < userInput.length(); i++) {
            if (userInput.at(i) == ',') {
                commaCount++;
            }
        }

        if (commaCount == 0) {
            cout << "Error: No comma in string." << endl;
        }

        else if (commaCount > 1) {
            cout << "Error: Too many commas in input." << endl;
        }

        //else if ( second half of string != a POSITIVE int) "If entry after the comma is not an integer"

        else {
        breakPoint = userInput.find(',');

        listOfAuthors.push_back(userInput.substr(0, breakPoint));
        novelsPerAuthor.push_back(stoi(userInput.substr(breakPoint + 1, userInput.length() - breakPoint + 1)));


        cout << "Data string: " << listOfAuthors.at(currentIndex) << endl;
        cout << "Data integer: " << novelsPerAuthor.at(currentIndex) << endl;
        currentIndex++;

        }

        commaCount = 0; // resets comma count between entries

        cout << " " << endl;
    } while (userInput != "-1"); //exits user input loop if "-1" is inputed

下面是我如何分隔字符串并将后半部分转换为 int 的代码，如果它有助于为上面的代码提供上下文。

breakPoint = userInput.find(',');

        listOfAuthors.push_back(userInput.substr(0, breakPoint));
        novelsPerAuthor.push_back(stoi(userInput.substr(breakPoint + 1, userInput.length() - breakPoint + 1)));

如果 userInput.substr(breakPoint + 1, userInput.length() - breakPoint + 1)) 不等于正整数，有没有办法编写一个条件来说明某些内容，程序将打印“错误: 逗号后面不跟整数。"

Answer 1

您需要使用从 std::string 到整数的转换器。在 C++ 中有几种方法可以做到这一点，但我认为使用 stoi 函数将涵盖所有情况（请注意，如果您尝试转换不是整数的字符串，stoi 会引发异常）

这是一个如何实现的小例子

#include <iostream>
#include <string>

std::string /* error string */ SplitLine(const std::string& inLine, std::string& outString, int& outNumber)
{
    const std::string separator = ",";
    auto pos = inLine.find(separator);
    auto pos2 = inLine.rfind(separator);

    if (pos == std::string::npos) {
        return "No comma in string.";
    }

    if (pos != pos2 && pos2 != std::string::npos) {
        return "Too many commas in input.";
    }

    outString = inLine.substr(0, pos);
    std::string remain_part = inLine.substr(pos+1);
    try {
        outNumber = stoi(remain_part);
    }
    catch (...) {
        return "Comma not followed by an integer.";
    }

    return "";
}

void ProcessLine(const std::string& inLine)
{
    std::string result;
    int num = 0;
    auto error = SplitLine(inLine, result, num);

    if (!error.empty()) {
        std::cout << "Error: " << error << std::endl;
    }
    else {
        std::cout << "Data string: " << result << " Data integer: " << num << std::endl;
    }

}

int main()
{
    ProcessLine("Ernest Hemingway 9");
    ProcessLine("Ernest, Hemingway, 9");
    ProcessLine("Ernest Hemingway, nine");
    ProcessLine("Ernest Hemingway, 9");

    return 0;
}