Question

I have a problem when handling command line arguments in a simple C++ application written in Visual Studio 2008. I have written the following code:

#include <iostream>

using namespace std;

int _tmain(int argc, char **argv)
{
    char* c = *argv;    

    for(int i=0; i< argc ; ++i) 
    {
        cout << argv[i] << endl;
    }

    int a;
    cin >> a;

    return 0;    
}

My problem is that only the first character of each command line argument is written to cout.

I identified the cause to be that the characters in the command line arguments are represented as 2 bytes, making every other 1 byte char contain only zeros, i.e ‘\0’.

My question is, why do this happen? from what i found in samples on the net it should work as i have written it. Also, is there a way to force the characters in the arguments to be of 1 byte representation?

I hope my question is clear enough.

Answer 1

Your _tmain is receiving UTF-16 characters which, when fed Latin text, have a 0 in every other byte. You are interpreting them as single byte characters. You need to interpret them as wide characters.

I would write it something like this:

#include <iostream>

using namespace std;

int wmain(int argc, wchar_t* argv[])
{
  for(int i=0; i<argc; ++i) 
  {
    wcout << argv[i] << endl;
  }

  int a;
  wcin >> a;

  return 0;
}

If you don’t want to use UTF-16 then you can stick with char like this:

int main(int argc, char* argv[])
{
  for(int i=0; i<argc; ++i) 
  {
    cout << argv[i] << endl;
  }

  int a;
  cin >> a;

  return 0;
}

Note the change in the naming of the main function. In MS world, wmain receives wchar_t* and main receives char*.

If you do switch to char* then you should also update your project configuration to target MBCS rather than Unicode.

More information on the main function handling of the MS compiler can be found here: main: Program Startup.