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:

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#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;
}

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

Code:

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

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

Thanks for any help.

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:

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

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You can do this:

Code:

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

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

Consider subscript notation:

Code:

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

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

I have two further questions about the following code:

Code:

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#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;
}

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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) 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.

Quote:

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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;
}

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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.

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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:

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#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
}

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2. It's fine I guess.