为什么我的文件不能用 ifstream C++ 打开？

Question

尽管这是一个相当简单的问题，但我无法通过 inputFileStream(inputFilePath) 打开文件。有人能指引我正确的方向吗（我不是来这里接受学校作业的答案的）？

#include <cmath>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <sstream>

using namespace std;

const string ID_LINE = "James McMillan - CS 1336 050 - Assignment 26";

int main(int argc, char *argv[]) {
    cout << ID_LINE << endl << endl;

    // guard clause - invalid number of arguments
    if (argc != 3) {
        cout << "Usage: " << argv[0] << "<input file path> <output file path>" << endl;
        return 1;
    }

    //extract arguments
    string programPath = argv[0];
    string inputFilePath = argv[1];
    string outputFilePath = argv[2];

    cout << "Program path: " << programPath << endl;
    cout << "Input file path: " << inputFilePath << endl;
    cout << "Output file path: " << outputFilePath << endl << endl;


    cout << "Creating input file stream..." << endl;
    ifstream inputFileStream;
    cout << "Created input file stream." << endl;

    cout << "Opening input file stream: " << inputFilePath << endl;
    inputFileStream.open(inputFilePath);

    if (!inputFileStream.is_open()) {
        cout << "Unable to open input file stream: " << inputFilePath << endl;
        return 1;
    }
}

Answer 1

您的解决方案可能来自检查文件是否存在，尝试

//Add this at the top
#include <experimental/filesystem>
//somewhere in the body
const auto has_the_file = std::experimental::filesystem::exists(inputFilePath);
if (!has_the_file)
{
    std::cout << "Oh No! Can't find" << inputFilePath << std::endl;
}

此外，我不会解析参数，而是手动设置路径以确认您对如何指定文件路径的期望。

为什么我的文件不能用 ifstream C++ 打开？

Why is my file is not opening with ifstream C++?

c++

arguments

ifstream