如何在给定文件路径的情况下获取带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？

#include <iostream>
#include <string>

std::string GetFileName(const std::string& filePath);

size_t lastSlashIndex = filePath.find_last_of("\\/");

std::string fileName = filePath.substr(lastSlashIndex + 1);

size_t lastDotIndex = fileName.find_last_of(".");

std::string fileExtension = fileName.substr(lastDotIndex + 1);

std::cout << "File Extension: " << fileExtension << std::endl;
return fileName;

#include <iostream>
#include <string>

std::string GetFileName(const std::string& filePath) {
    size_t lastSlashIndex = filePath.find_last_of("\\/");
    std::string fileName = filePath.substr(lastSlashIndex + 1);
    size_t lastDotIndex = fileName.find_last_of(".");
    std::string fileExtension = fileName.substr(lastDotIndex + 1);
    std::cout << "File Extension: " << fileExtension << std::endl;
    return fileName;
}

int main() {
    std::string filePath = "C:\\path\\to\\file.txt";
    std::string fileName = GetFileName(filePath);
    std::cout << "File Name: " << fileName << std::endl;
    return 0;
}