使用argv的for循环内的C ++ std :: logic_error

Question

I am new to C++ and I am trying to write a program to take in command line arguments and produce a .desktop file. I am trying to implement identification of the argv values but I keep getting a std::logic_error

My code is:

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string>

using namespace std;

int main(int argc, char* argv[]) {

    string name;
    string comment;

    for(int i = 1; i <= argc; i++) {
        char* tmp[] = {argv[i]};
        string param = *tmp;
        string paramVal = argv[i+1];
        if(param == "-h") {
            cout << "-h        Display this help dialogue" << endl;
            cout << "-n        Set entry name" << endl;
            cout << "-c        Set entry comment" << endl;
            cout << "-e        Set entry executable path" << endl;
            cout << "-i        Set entry icon" << endl;
            break;
        }
        else if(param == "-n") {
            name = paramVal;
            i++;
            continue;
        }
        else if(param == "-c") {
            comment = paramVal;
            i++;
            continue;
        }
        else if(param == "-e") {

        }
        else if(param == "-i") {

        }
        else {
            cout << "ERROR >>> Unrecognised parameter %s" << param << endl;
        }
    }

    cout << "Name: %s\nComment: %s" << name << comment << endl;
    return(0);
}

The program compiles fine (using g++) but when I try to run ./createDesktopIcon -na -cb I get the following error

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

Please help as it is very frustrating

Answer 1

Here are the problems I see:

i <= argc

You want to compare i < argc because the argv[argc] element in the array is actually one past the last element in the argv array.

Also, here:

string paramVal = argv[i+1];

This will access the array out of bounds as well.

You might want to look at getopt to do all of this for you.