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Old 02-07-2011, 12:03 PM   #1
oscaringolilingo
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Registered: Dec 2010
Posts: 34

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


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.
 
Old 02-07-2011, 02:02 PM   #2
ta0kira
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Registered: Sep 2004
Distribution: FreeBSD 9.1, Kubuntu 12.10
Posts: 3,078

Rep: Reputation: Disabled
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

Last edited by ta0kira; 02-07-2011 at 02:15 PM.
 
1 members found this post helpful.
Old 02-07-2011, 02:11 PM   #3
johnsfine
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Distribution: Centos
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Quote:
Originally Posted by oscaringolilingo View Post
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.

Last edited by johnsfine; 02-07-2011 at 02:17 PM.
 
1 members found this post helpful.
Old 02-07-2011, 02:18 PM   #4
ta0kira
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Registered: Sep 2004
Distribution: FreeBSD 9.1, Kubuntu 12.10
Posts: 3,078

Rep: Reputation: Disabled
Quote:
Originally Posted by johnsfine View Post
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
 
1 members found this post helpful.
Old 02-07-2011, 05:37 PM   #5
oscaringolilingo
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Registered: Dec 2010
Posts: 34

Original Poster
Rep: Reputation: 3
Oh wow, that was so simple, thank you all for illustrating me!
 
  


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