CODEWITHSUNDEEP
c++parsingcommand-lineconsole
Ionut Daniel
Ionut Daniel

Reputation: 312

command line parser in c ++

I have the next code : 

#include "CmdLine.h"

void main(int argc, TCHAR **argv)
{

  CCmdLine cmdLine;



  // parse argc,argv 
  if (cmdLine.SplitLine(argc, argv) < 1)
  {
     // no switches were given on the command line, abort
     ASSERT(0);
     exit(-1);
  }

  // test for the 'help' case
  if (cmdLine.HasSwitch("-h"))
  {
     show_help();
     exit(0);
  }

  // get the required arguments
  StringType p1_1, p1_2, p2_1;
  try
  {  
     // if any of these fail, we'll end up in the catch() block
     p1_1 = cmdLine.GetArgument("-p1", 0);
     p1_2 = cmdLine.GetArgument("-p1", 1);
     p2_1 = cmdLine.GetArgument("-p2", 0);

  }
  catch (...)
  {
     // one of the required arguments was missing, abort
     ASSERT(0);
     exit(-1);
  }

  // get the optional parameters

  // convert to an int, default to '100'
  int iOpt1Val =    atoi(cmdLine.GetSafeArgument("-opt1", 0, 100));

  // since opt2 has no arguments, just test for the presence of
  // the '-opt2' switch
  bool bOptVal2 =   cmdLine.HasSwitch("-opt2");

  .... and so on....

}

I have the CCmdLine class implemented and this main is an exemple of how to use it . I am having difficulties understanding how i get input values . I have tried to read them with scanf from the console but the argc won't increment and results faulty reading.

I am a beginner in c++ and i would like to know who to make this code work .

Thanks .

Upvotes: 0

Views: 733

Answers (3)

SingerOfTheFall
SingerOfTheFall

Reputation: 29966

Argc and argv only contain the arguments that were passed when the program started. So if you execute it with myapp.exe option1 option2 option3, than in your argv you will have:

  • myapp.exe //<--argv[0]
  • option1 //<--argv[1]
  • option2 //<--argv[2]
  • option3 //<--argv[3]

In a nutshell, when a program starts, the arguments to main are initialized to meet the following conditions:

  • argc is greater than zero.
  • argv[argc] is a null pointer.
  • argv[0] through to argv[argc-1] are pointers to strings representing the actual arguments.
  • argv[0] will be a string containing the program's name or a null string if that is not available. Remaining elements of argv represent the arguments supplied to the program.

You can find some more information for example here.

All attempts to read input later (either with cin, scanf or whatever else) will not save the inputed values to argv, you will need to handle the input yourself.

Upvotes: 1

Denis Ermolin
Denis Ermolin

Reputation: 5546

This is quite easy: 

void main(int argc, char **argv)
{
    std::string arg1(argv[0]);
    std::string arg2(argv[1]);
}

Upvotes: 0

Ravindra Bagale
Ravindra Bagale

Reputation: 17655

pass the input values from commandline while run the programs e.g

program_name.exe arg1 arg2

Upvotes: 0

Related Questions