Beginner's question concerning pointers and arrays

rcomj · 08-09-2009, 12:14 AM

I am having trouble understanding why I can't get my array of strings to print using a pointer.

Here is my code so far:

Code:

#include <stdio.h>

int main() {

  int a[] = {1, 10, 100, 1000};
  int *b = a;
  char *c[] = {"hello", "how", "are", "you?"};

  printf("%d\n", *b);
  printf("%d\n", *++b);
  printf("%d\n", *++b);

  printf("%s\n", *c);
  printf("%s\n", *++c);

  return 0;
}

When I run try to compile this code, I receive this error:

Code:

pointer-test.c: In function ‘main’:
pointer-test.c:14: error: lvalue required as increment operand

Why does the same pointer syntax not work for an int array and a string array?

Thanks for any help.

BEN10 · 08-09-2009, 12:34 AM

c is an array of char pointers, which means c is actually an array, and that's why you can't use ++ operator with it. Whereas b is a poniter(not an array), that's why ++b is legal. Take this as an example.

Code:

int a[]={1,2,3,4};
printf("%p",(void *)++a); // again the same error coz a is an array

MK27 · 08-09-2009, 12:57 AM

You can do this:

Code:

printf("%s\n", ++(*c));

but that ++ == sizeof(char), which is not what you want.

Consider subscript notation:

Code:

printf("%s\n", c[1]);

rcomj · 08-10-2009, 07:55 PM

Thanks for the help. I appreciate both of your replies.

I have two further questions about the following code:

Code:

#include <stdio.h>

int main(int argc, char *argv[]) {

  int a[] = {1, 10, 100, 1000};
  int *b = a;
  char *c[] = {"hello", "how", "are", "you?"};
  char **d = c;

  printf("%d\n", *b);
  printf("%d\n", *++b);
  printf("%d\n", *++b);
  printf("%d\n", *++b);

  printf("%s\n", *d);
  printf("%s\n", *++d);
  printf("%s\n", *++d);
  printf("%s\n", *++d);

  while (--argc > 0)
    printf("%s\n", *++argv);

  return 0;
}

1) Why is it that I can manipulate argv, which is an array of char pointers, using ++ but couldn't do the same to c (in the original post)?

2) Is my way of using char **d sloppy at all? After reading BEN10's answer, I had the idea of using a pointer to a pointer to a char, but I don't know if that is a convoluted way of doing things.

My goal with this little program is to understand pointers better and not use indexing, since I already understand how indexing works.

Thanks again for any help.

**Salem** · 08-10-2009, 11:31 PM

> 1) Why is it that I can manipulate argv, which is an array of char pointers, using ++ but couldn't do the same to c (in the original post)?
Because it's a pointer.
char *argv[] could equally be written as char **argv

Using [] in the context of a function parameter doesn't add any "true array" properties to it.

BEN10 · 08-11-2009, 09:09 AM

Quote:

Originally Posted by rcomj

Thanks for the help. I appreciate both of your replies.

I have two further questions about the following code:

Code:

#include <stdio.h>

int main(int argc, char *argv[]) {

  int a[] = {1, 10, 100, 1000};
  int *b = a;
  char *c[] = {"hello", "how", "are", "you?"};
  char **d = c;

  printf("%d\n", *b);
  printf("%d\n", *++b);
  printf("%d\n", *++b);
  printf("%d\n", *++b);

  printf("%s\n", *d);
  printf("%s\n", *++d);
  printf("%s\n", *++d);
  printf("%s\n", *++d);

  while (--argc > 0)
    printf("%s\n", *++argv);

  return 0;
}

1) Why is it that I can manipulate argv, which is an array of char pointers, using ++ but couldn't do the same to c (in the original post)?

2) Is my way of using char **d sloppy at all? After reading BEN10's answer, I had the idea of using a pointer to a pointer to a char, but I don't know if that is a convoluted way of doing things.

My goal with this little program is to understand pointers better and not use indexing, since I already understand how indexing works.

Thanks again for any help.

1. When you use any array as an argument to function it decomposes into pointer and thus you can do ++ with argv. Here's an example.

Code:

#include<stdio.h>
void f(int []);
int main(void)
{
	int a[]={1,2,3};
	printf("%p",(void *)a);
	f(a);
	getch();
}
void f(int b[])
{
	printf("\n%p",(void *)++b);//++ is legal here
}

2. It's fine I guess.

08-09-2009, 12:14 AM	#1
rcomj Registered User Join Date: Aug 2009 Posts: 2	Beginner's question concerning pointers and arrays I am having trouble understanding why I can't get my array of strings to print using a pointer. Here is my code so far: Code: #include <stdio.h> int main() { int a[] = {1, 10, 100, 1000}; int b = a; char c[] = {"hello", "how", "are", "you?"}; printf("%d\n", b); printf("%d\n", ++b); printf("%d\n", ++b); printf("%s\n", c); printf("%s\n", *++c); return 0; } When I run try to compile this code, I receive this error: Code: pointer-test.c: In function ‘main’: pointer-test.c:14: error: lvalue required as increment operand Why does the same pointer syntax not work for an int array and a string array? Thanks for any help.
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08-09-2009, 12:34 AM	#2
BEN10 DESTINY Join Date: Jul 2008 Location: in front of my computer Posts: 656	c is an array of char pointers, which means c is actually an array, and that's why you can't use ++ operator with it. Whereas b is a poniter(not an array), that's why ++b is legal. Take this as an example. Code: int a[]={1,2,3,4}; printf("%p",(void *)++a); // again the same error coz a is an array __________________ HOPE YOU UNDERSTAND....... for( ; ; ) printf("If you can't make it good, at least make it look good"); PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D. IDE- Microsoft Visual Studio 2008 Express Edition
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08-09-2009, 12:57 AM	#3
MK27 subminimalist Join Date: Jul 2008 Location: NYC Posts: 3,944	You can do this: Code: printf("%s\n", ++(c)); but that ++ == sizeof(char), which is not what you want. Consider subscript notation: Code: printf("%s\n", c[1]); __________________ Accuracy and integrity mean nothing if you don't make it past the censors...*PYTHAGORAS
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08-10-2009, 07:55 PM	#4
rcomj Registered User Join Date: Aug 2009 Posts: 2	Thanks for the help. I appreciate both of your replies. I have two further questions about the following code: Code: #include <stdio.h> int main(int argc, char argv[]) { int a[] = {1, 10, 100, 1000}; int b = a; char c[] = {"hello", "how", "are", "you?"}; char d = c; printf("%d\n", b); printf("%d\n", ++b); printf("%d\n", ++b); printf("%d\n", ++b); printf("%s\n", d); printf("%s\n", ++d); printf("%s\n", ++d); printf("%s\n", ++d); while (--argc > 0) printf("%s\n", ++argv); return 0; } 1) Why is it that I can manipulate argv, which is an array of char pointers, using ++ but couldn't do the same to c (in the original post)? 2) Is my way of using char d sloppy at all? After reading BEN10**'s answer, I had the idea of using a pointer to a pointer to a char, but I don't know if that is a convoluted way of doing things. My goal with this little program is to understand pointers better and not use indexing, since I already understand how indexing works. Thanks again for any help.
rcomj is offline

08-10-2009, 11:31 PM	#5
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	> 1) Why is it that I can manipulate argv, which is an array of char pointers, using ++ but couldn't do the same to c (in the original post)? Because it's a pointer. char argv[] could equally be written as char *argv Using [] in the context of a function parameter doesn't add any "true array" properties to it. __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
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