如何将include "e+“字符串转换为long？

#include <iostream>
#include <string>

int main() {
    std::string str = "e+";
    try {
        long num = std::stol(str);
        std::cout << "Converted number: " << num << std::endl;
    } catch (const std::exception& e) {
        std::cout << "Conversion error: " << e.what() << std::endl;
    }
    return 0;
}

str = "e+"
try:
    num = int(str)
    print("Converted number:", num)
except ValueError as e:
    print("Conversion error:", str(e))