如何在给定文件路径的情况下获取具有文件扩展名的文件名并将其存储在 C++ 控制台应用程序中的字符串中？

Question

I'm using Visual Studio 2019. The file path must be taken as an User input. I'm new to C++ and will appreciate any help i can get.

Answer 1

#include <iostream>
#include <string>

int main()
{
    std::string filename;
    std::cout << "Enter file path: "; 
    std::cin >> filename; // "Example: C:/test/file1.txt"

    // Remove directory if present.
    // Do this before extension removal incase directory has a period character.
    const size_t last_slash_idx = filename.find_last_of("\\/");
    if (std::string::npos != last_slash_idx)
    {
        filename.erase(0, last_slash_idx + 1);
    }

    std::cout << filename;

    return 0;
}