Urgent C Question ===> need fast replies

Hi Experts,
               Iam writing a command line program. My commnad looks like below

c:\commandname -f "c:\temp.txt" systemname:port GetStatus

The command is exactly same as shown above

I am able to get the whole command inside a string say dummy. Now I want is when a -f is encountered in the command then take

c:\temp.txt in string test1.
systemname:port in string test2.
GetStatus in string test3.

As Iam new to C language , iam not so versed in string manipulation function. So please provide me code for checking the above. and getting the substrings in the respetive strings.

Early responses are greatly appreciated. Code responses too.


Warm Regards..............

LK
local_kiduAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

cookreCommented:
Which flavor C?

If you have a main with argc and argv, just parse the parms one at a time:

char CmdParms[10][256];
if (argc>10) say too many parms
for (argidx=0; argidx<argc; argidx++)
     {
     strcpy(CmdParms[argidx],argv[argidx])
     }
That will leave the exe reference in CmdParms[0] and parm n in CmdParms[n].

Of course, if your command line has a rigid structure, you can just work straight from argv.
andyinlondonCommented:
There are several ways to achieve this.

Either do this...

int main(int argc, char *argv[])
{
  int i;  // declare a loop counter
  char testStrings[3][256];  // declare an array of strings
  // test the number of arguments
  if (argc < 4)
  { // not enough arguments
    return 1;
  }
  if (argv[1][0] != '-' || argv[1][1] != 'f')
  { // -f not found
    return 1;
  }
  // -f found, copy the arguments into your strings
  for (i = 2; i < argc; i++)
  {
    strcpy(testStrings[i - 2], argv[i]);
  }
  return 0;
}

or this...

int main(int argc, char *argv[])
{
  char *test1, *test2, *test3;  // declare char pointers
  // test the number of arguments
  if (argc < 4)
  { // not enough arguments
    return 1;
  }
  if (argv[1][0] != '-' || argv[1][1] != 'f')
  { // -f not found
    return 1;
  }
  // -f found
  test1 = argv[2];
  test2 = argv[3];
  test3 = argv[4];
  return 0;
}
andyinlondonCommented:
OOPS!
Lines reading "if (argc < 4)" should read "if (argc < 5)"
Big Business Goals? Which KPIs Will Help You

The most successful MSPs rely on metrics – known as key performance indicators (KPIs) – for making informed decisions that help their businesses thrive, rather than just survive. This eBook provides an overview of the most important KPIs used by top MSPs.

local_kiduAuthor Commented:
Dear andyinlondon,
                           I am not using argc or agrv , instead Iam getting the whole commnad inside a string(dummy), as I specified in the question. So I want the answer based on that.

LK
cookreCommented:
In that case, short of parsing the string yourself either character by charactrer or with strtok(), you can use CommandLineToArgvW():
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dllproc/base/commandlinetoargvw.asp

Note that the input string is Unocode.  If you used GetCommandLine(), it's already in Unicode, otherwise, you'll have to convert it.
andyinlondonCommented:
Ok LK
Try this code instead.


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

#define NUM_ARGS  3
char yourDUMMYstring[] = "c:\\commandname -f \"c:\\temp.txt\" systemname:port GetStatus";

char *getNextArg(char **str);

void main(void)
{ int i;
  char *cp, *ptr;
  char test[NUM_ARGS][256];
  // test for -f
  if ((cp = strstr(yourDUMMYstring, "-f")) != NULL)
  { // -f found
    cp += 2;  // move pointer past the -f
    for (i = 0; i < NUM_ARGS; i++)
    { if (!(ptr = getNextArg(&cp)))
      { printf("\nWHOOPS!  arg %d missing\n", i);
        return;
      }
      // copy the arg into test array
      strcpy(test[i], ptr);
      // display it
      printf("test[%i] = %s\n", i, test[i]);
    }
  }
  getchar(); // wait for user to press <CR>
}

/* This function is interesting in that the parameter 'str' is a pointer to a pointer. 'str' is used as
 * both an 'IN' and an 'OUT' parameter; supplying the function with the string to parse, and returning
 * the remains of the string after the argument has been extracted.
 * Because it is necessary (and convenient) for this function to update the input string (otherwise the
 * same string would be found every time), you have two choices: either create a buffer, fill it with
 * the argument string and shift the remains of the string to the beginning of 'str', or employ the
 * pointer to pointer method which can return the updated pointer position itself.
 * The problem with the first method is that it involves a string moving operation which is costly in
 * terms of processing time, whereas the pointer to pointer method doesn't - but it's more complicated.
 * The function returns the argument (if found), otherwise it returns the NULL pointer.
 */
char *getNextArg(char **str)
{ char *cp, *strStart = *str;
  // ignore whitespace before arg
  for (; *strStart && *strStart <= ' '; strStart++);
  if (*strStart == '"')
  { // arg is enclosed in quotes
    strStart++; // not interested in the quote character
    // find the end of the arg
    if ((cp = strchr(strStart, '"')) == NULL)
    { // WHOOPS!  no closing quote found
      return NULL;
    }
  }
  else
  { // arg is not enclosed in quotes
    // find the end of the arg
    for (cp = strStart; *cp && *cp > ' '; cp++);
  }
  // replace quote or space with the string terminator
  if (*cp)
  { //
    *cp++ = '\0';
  }
  *str = cp;
  return strStart;
}

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
local_kiduAuthor Commented:
Thanks andyinlondon, You have done a great work gor me. Thanks alot

LK
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Programming

From novice to tech pro — start learning today.