How come *++argv works and *++myarray doesn't?

oscaringolilingo · 02-07-2011, 11:03 AM

Ok, so I'm reading the C programming language, 2nd ed by Kernighan and Ritchie and I decide to compile this bit of code from part "5.10 Command line arguments":

echo2.c

Code:

#include <stdio.h>
/* echo command-line arguments; 2nd version */
main(int argc, char *argv[])
{
while (--argc > 0)
printf("%s%s", *++argv, (argc > 1) ? " " : "");
printf("\n");
return 0;
}

So it compiles without complaints and runs smoothly as expected:

Code:

$ ./echo2 hello world
hello world

I notice they use "*++argv" to scan through "char *argv[]", so I decide to try my own version:

pointerarray2.c

Code:

#include <stdio.h>
/* test */
main()
{

int mylength = 5;
char *myarray[] = {"abcd","efg","hijklm","nopq","rs","tuvwxyz"};

while (--mylength > 0)
printf("%s%s", *++myarray, (mylength > 1) ? " " : "");
printf("\n");
return 0;
}

And when trying to compile I get this

Code:

$ gcc -o pointerarray2 pointerarray2.c
pointerarray2.c: In function ‘main’:
pointerarray2.c:10: error: lvalue required as increment operand

WHY? Isn't "char *myarray[]" the same as "char *argv[]" ?? Is this just whimsy compiler design or what is it? I mean, I could use argv[n] or something to scan through the array but my aim is not to get it working, but to understand why ANSI C makes this difference.

ta0kira · 02-07-2011, 01:02 PM

When char *something[] is the point of definition of an array, the symbol something is the value indicating the location of the array; the value itself has no location (i.e. the compiler inserts the value where it's needed,) which makes it an rvalue. When you use char *something[] as a function argument, you indicate that you're expecting a pointer to an array of fixed size (it's really taken as char **something); to do this, a pointer to the array must be pushed to the stack, which makes it an lvalue, which can be incremented.
Kevin Barry

johnsfine · 02-07-2011, 01:11 PM

Quote:

Originally Posted by oscaringolilingo

Is this just whimsy compiler design or what is it?

Certainly it is nothing to be blamed on the compiler design. It is a questionable choice in the language design, and one that confuses most beginning C programmers.

To try to clarify Kevin's correct answer, consider this program

Code:

#include <stdio.h>
/* test */
void work(int mylength, char *myarray[])
{
   while (--mylength > 0)
      printf("%s%s", *++myarray, (mylength > 1) ? " " : "");
   printf("\n");
}
main()
{
   int mylength = 5;
   char *myarray[] = {"abcd","efg","hijklm","nopq","rs","tuvwxyz"};
   work( mylength, myarray );
   return 0;
}

That might look like it does the same thing as your code that I copied and split part into a function. But this version compiles and runs.

The key is in the meaning of

Code:

char *myarray[]

which has a very different meaning when declared as an argument of a function than when declared as a variable.

ta0kira · 02-07-2011, 01:18 PM

Quote:

Originally Posted by johnsfine

Code:

#include <stdio.h>
/* test */
void work(int mylength, char*myarray[])
{
   while (--mylength > 0)
      printf("%s%s", *++myarray, (mylength > 1) ? " " : "");
   printf("\n");

}
main()
{
   int mylength = 5;
   char *myarray[] = {"abcd","efg","hijklm","nopq","rs","tuvwxyz"};
   work( mylength, myarray );
   return 0;
}

Here is an additional clarification:

Code:

#include <stdio.h>
/* test */
main()
{

int mylength = 5;
char *mydata[] = {"abcd","efg","hijklm","nopq","rs","tuvwxyz"};
//char *myarray[] = mydata; /*this will not get you anywhere!*/
char **myarray = mydata;

while (--mylength > 0)
printf("%s%s", *++myarray, (mylength > 1) ? " " : "");
printf("\n");
return 0;
}

Kevin Barry

oscaringolilingo · 02-07-2011, 04:37 PM

Oh wow, that was so simple, thank you all for illustrating me!