Thread: Pointers and parameters in the command line

Good afternoon.

Code:

/* 
 * **************************** how to run *******************************
 * in the command line: 
 * 
 * 1) [Executable] -x [pattern] 
 * - Only words which don't match the pattern are going to be printed   
 * -x is the "except" parameter
 * 
 * 2) [Executable] -x -n [pattern]
 * Words that don't match the pattern are going to be printed numerically
 * 
 * 3) [Executable] -n [pattern]
 * The words that match the pattern are going to be printed numerically
 * ***********************************************************************
 */

/* I didn't understand how those two while work 
 
 * argv[0] is the name of the executable, right? 
 * is (*++argv)[0] incrementing the pointer to argv[1] (-n or -x) 
 * and accessing the first value pointed ? (-) 
 * 
 * Due to that, I thought *++argv[0] was doing the same of (*++argv)[0] because that 
 * increments the pointer to point to the first value of the parameter argv[1] which 
 * is '-'. How does it point to 'n' or 'x' ? 
 */ 

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

#define MAXLINE 1000

int getLine(char s[], int lim) 
{ 
   int i,c; 
   for(i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i) 
   { 
      s[i] = c;   
   }
   if(c == '\n') 
   { 
     s[i] = c;  
     ++i; 
   }
   s[i] = '\0'; 
   return i;                      
}

int main(int argc, char* argv[]) 
{ 
  char line[MAXLINE];
  long lineno = 0;
  int c, except = 0, number = 0, found = 0;
  
  while(--argc > 0 && (*++argv)[0] == '-')
    while( (c = *++argv[0]) ) 
     switch (c) 
     { 
        case 'x': 
          except = 1; 
          break;
        case 'n': 
          number = 1;
          break;
        default: 
          printf("find: illegal option %c\n", c);
          argc = 0; 
          found = -1; 
          break;
     }
  if(argc != 1) 
    printf("Usage: find -x -n pattern\n");             
  else 
    while(getLine(line, MAXLINE) > 0)
    { 
      lineno++;
      if((strstr(line, *argv) != NULL) != except)
      { 
        if(number) 
          printf("%ld:", lineno);
        printf("%s", line);
        found++;    
      }               
    }       
  return found;    
}

Originally Posted by thames

Good afternoon.

I didn't understand how those two while work

argv[0] is the name of the executable, right?
is (*++argv)[0] incrementing the pointer to argv[1] (-n or -x)
and accessing the first value pointed ? (-)

Due to that, I thought *++argv[0] was doing the same of (*++argv)[0] because that
increments the pointer to point to the first value of the parameter argv[1] which
is '-'. How does it point to 'n' or 'x' ?

The purpose of the two while loops is to loop through command-line switches. Consider you have the program foo.c and you run it like so:

Code:

Foo.exe -xyz -a -b -c bar.c

What will be the contents of argc and argv as you enter main?

Code:

//argc: 6
//argv: {"Foo.exe", "-xyz", "-a", "-b", "-c", "bar.c"}

Now consider the two while loops:

Code:

    while (--argc > 0 && (*++argv)[0] == '-')
        while ((c = *++argv[0]))
            printf("got c: %c\n", c);

The outer loop says to loop through all arguments (starting with argv[1]) as long as the argv begins with '-'. The inner loop says to loop through all characters in that argument (for example, "-xyz") and process each one.

Notice in the inner while loop, the [0] part takes precedence over the ++, so it first dereferences argv into a single argument and THEN increments to the next character with ++ (i.e. skipping the '-'). The inner while loop may also be written like this:

Code:

        while ((c = *++(*argv)))

As you first enter main, your argc and argv look like this:

Code:

//argc: 6 
//argv: {"Foo.exe", "-xyz", "-a", "-b", "-c", "bar.c"}

Then after the first while loop, --argc and ++argv modify them, so now they look like this:

Code:

//argc: 5 
//argv: {"-xyz", "-a", "-b", "-c", "bar.c"}

So argv[0] (or equivalently, *argv) always refers to the "current" argument being processed.