Some help with point adress

Code:

---

#include <stdio.h>

int main()
{
        int i; int a[10];

        for(i = 0; i < 10; i++)
                a[i] = i;


/* what I wanna to do the same thing like this */

       

Code:

       

---

      
    for(i = 0; i < 10; i++)
        printf("%d = %d\n",a+i, *(a+i));
    printf("\n\n");

---

/* But in little difernet style. That is that I do not declare i. something like */

       

Code:

       

---

    for(; *a < 10; /* a += sizeof(int)   this is what I dont know how to increment adress */)
        printf("%d = %d\n",a+i, *(a+i));
    printf("\n\n");

---

        return 0;
}

---

Code:

---

int * b=a;
for(; *b < 10; ++b )
printf("%d = %d\n",b, *(b));
printf("\n\n");

---

The most important aspect of declaring the type of the pointer is when you want to do pointer arithmatic. So, if you delcare "a" as an int* , and you say a++ , the compiler knows how to increment a , i.e. by an amount equal to the size of int

Hope this helps you.

RoshanX
all I got is

Code:

---

      wrong type argument to increment

---

Code:

---

#include <stdio.h>

int main(void)
{
  int a[] = {1,2,3,4,5,6,7,8,9,10};
  int *p;
  for ( p = a; p < &a[sizeof a / sizeof *a - 1]; ++p )
  {
      printf("%p:%d\n", (void*)p, *p);
  }
  return 0;
}

---

In my previous reply, "a" should be substituted by "b".

Anyway,

Code:

---

for (;*b <10 ;++b)

---

is not a good way to impose the termination condition. memory beyond the 10th element will have values, and you are not guranteed an exit. Why you can't increment a is because a is an array which is different from just a pointer to an integer.

Actually, you're guarinteed the ability to test for the location of one past the end of the array. You can't use it, but you can test for it.

Code:

---

char array[ BUFSIZ ];
char *p = NULL;

for( p = array; p != array + BUFSIZ; p++ )
    ...do something...

---

Quzah.