c++ - 尝试获取命令行参数的核心转储。 (C++)

Question

我一直在尝试使用 C++ 命令行参数，但遇到了一些问题。最初我试图使用“==”运算符将“argv”与字符串进行比较。我很快了解到比较的是指针而不是值。我修复了那个错误，但现在我在运行时遇到了这个错误。

terminate called after throwing an instance of 'std::logic_error'
  what():  basic_string::_S_construct null not valid
Aborted (core dumped)

程序编译正常，我也没有收到编译器的警告。这是我的源代码，以便您帮助我找到问题。

#include <iostream>
#include <string>

using namespace std;

int main(int argc, char *argv[])
{
    //Deal with arguments and send them to the correct functions
    if (argc >= 2){

        string op = argv[2];
        if (op == "-a" || op == "--automatic"){
            cout << "Test";
        }

        return 0;
    }
    //Or, just write help and info
    cout << "\n";
    cout << "bwc v0.0.1U-(Unstable)\n\n";
    cout << "Usage: bwc <operation> [...]\n";
    cout << "Operations:\n";
    cout << "   bwc {-a --automatic} <file(s)>\n";
    cout << "   bwc {-i --interactive}\n";
    cout << "   bwc {-c --error-codes}\n";
    cout << "\n";
    return 0;
}

Answer 1

您对 argv[] 的索引差了一个。变化:

string op = argv[2];

到:

string op = argv[1];