C++如何使用argv[1]读取第一个用户参数并将其存储在字符串中以读取/写入文本文件

#include <iostream>
#include <fstream>

int main(int argc, char* argv[]) {
    if (argc < 2) {
        std::cout << "Please provide a filename as a command line argument." << std::endl;
        return 1;
    }

    std::string filename = argv[1];

    // 读取文本文件
    std::ifstream inputFile(filename);
    if (!inputFile) {
        std::cout << "Failed to open file: " << filename << std::endl;
        return 1;
    }

    std::string line;
    while (std::getline(inputFile, line)) {
        // 处理每一行数据
        // ...

        // 输出每一行数据
        std::cout << line << std::endl;
    }

    inputFile.close();

    // 写入文本文件
    std::ofstream outputFile(filename, std::ios::app); // 使用std::ios::app以追加模式打开文件
    if (!outputFile) {
        std::cout << "Failed to open file: " << filename << std::endl;
        return 1;
    }

    std::string data = "Hello, World!";
    outputFile << data << std::endl;

    outputFile.close();

    return 0;
}