.exe程序的C ++命令行参数

Question

I have program.exe that is being created on my Windows machine. For some reason I am not able to pass in command line arguments properly.

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

  ///////testing
  cout << "\n\n(int)ARGV[1]: " << (int)argv[1] << "\n\n";

  return 0;
} 

In terminal I run:

program.exe 4 

I see (int)ARGV[1]: 15333464 printed to my console.

Any idea why this is or how I can modify the code? I should get the number 4 printed out. Thanks.

Answer 1

When you cast from char* to int you get the pointer value bitpattern interpreted as an integer.

A good way to instead interpret the character string pointed to, as a specification of an integer, is to use std::stoi from the <string> header.

Thus:

#include <iostream>
#include <stdlib.h>    // EXIT_FAILURE
#include <string>      // std::stoi
using namespace std;

int main(int argc, char *argv[])
{
    if( argc != 2 )
    {
        return EXIT_FAILURE;
    }
    cout << "ARGV[1]: " << stoi( argv[1] ) << "\n";
} 

For the case where the argument isn't a valid specification of an integer, stoi will throw an exception, and in the code above that will cause the program to terminate with some message displayed – a crash. That's generally preferable to producing incorrect results. But if you want to handle it, read up on try and catch in your C++ textbook.

---

If you want filenames as command line arguments, and more generally filesystem paths, then do note that the Windows convention for encoding of char based strings is Windows ANSI, which has a very limited set of characters. Some filenames and paths on my Norwegian computer can't be represented in your Windows ANSI (yes, Windows ANSI is locale-specific). So for this the C and C++ main argument mechanism, is ungood.

The Windows API provides a pair of wchar_t based functions that can be used as a (working) alternative, namely GetCommandLine (retrieves the raw UTF-16-encoded command line) and CommandLineToArgvW (standard parsing to produce individual arguments).

Some Windows compilers also provide an alternative to standard main called wmain , where argv is declared as wchar_t* argv[] . These compilers include Visual C++ and the MinGW64 variant of g++.

Answer 2

Here argv[1] is a character pointer ( char* ) so you cant convert a character pointer to an integer by type casting. So to convert the character pointer to an integer use the atoi() function, which is available in the c standard library (include the cstdlib library).

#include <iostream>
#include <cstdlib>
using std::cout;
using std::endl;
using std::atoi;
int main(int argc, char *argv[])
{
    cout << atoi(argv[1]) << " " << endl;
    return 0;
}

Note that atoi() returns zero if the argument passed to it cannot be converted to an integer. So you could check the returned value and if it is zero then print the relevant message.

Thus:

#include <iostream>
#include <cstdlib>
#include <cstring>
using std::atoi;
using std::cout;
using std::endl;
using std::strcmp;

int main(int argc, char *argv[])
{
    if(argc == 2)
    {
        if(atoi(argv[1]) == 0 && strcmp(argv[1], "0") != 0)
            cout << "The argument supplied is not an integer" << endl;
        else
            cout << atoi(argv[1]) << " " << endl;
    }
    else if( argc > 2)
        cout << "Too many arguements" << endl;
    else
        cout << "Insufficient arguements" << endl;
    return 0;
}