>     why doesn't the following work.  I know that I can get it to work
 > by allocating the memory that str points to before calling printstr(); but
 > I want to know why?  Is it because the allocated memory in
 > printstr(); goes out of scope when you leave the function?


No, the variables in each case are different.  If you assign the memory
prior passing it to the function, the _value_ contained in the original
pointer is passed. If you later change it or assign to that variable, that
change won't be reflected in the variable declared in main().

This is a modified version that passes a pointer to the original pointer
(not the value of the pointer).


 #include
 #include
 #include
 #define _MAX_SIZE_      255

 printstr(char *argv[], char **s );
                           /* ^ */


 main( int argc, char *argv[] )
 {
 char *str;
 printstr( argv, &str );
              /* ^ */
 printf("The name of this file is : %s\n", str );
 free(str);
 }

 printstr(char *argv[], char **s )
                           /* ^ */
 {
      *s = malloc( sizeof(char) * _MAX_SIZE_ );
   /* ^ */
         strcpy(*s,argv[0]);
             /* ^ */
         printf("In (printstr), The name of this file is : %s\n", *s);
                                                               /* ^ */
 }


This should work in the way you want.


  cheers,
  david

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