Using Pointers In Loops - C Programming

For starters the header <stdlib.h> is redundant in your program because neither declaration from the header is used in your program.

It would be better to declare the array x without specifying the number of its elements. In this case your program will be more flexible. It will be enough just to change the number of its initializers if you will want to add or remove some elements from the array in its declaration.

And do not use magic numbers as 10 used in your for loop. Using magic numbers makes programs error prone.

As for your this for loop

for(int* ptr=x;ptr < 10;ptr++){

then in the condition of the loop you are trying to compare the address stored in the pointer ptr with the magic integer value 10 that does not make sense.

And in general declare pointers with qualifier const if you are not going to change data pointed to by pointers.

The program can look the following way.

#include <stdio.h>

int main( void )
{
    int x[] = { 1, 3, 2, 5, 43, 67, 8, 64, 22, 123 };
    const size_t N = sizeof( x ) / sizeof( *x );

    for ( const int *first = x, *last = x + N; first != last; ++first )
    {
        printf( "%d ", *first );
    }
    putchar( '\n' );

    return 0;
}

Inside the for loop the pointer last points to the memory after the last element of the array due to the pointer arithmetic used in the expression x + N. So when the pointer first will traverse all elements of the array it will be equal to the pointer last and the loop will be ended.

Try this version.

#include <stdio.h>

int main() {
  int x[10] = { 1,3,2,5,43,67,8,64,22,123 };

  for (int* ptr = x; ptr < x+10; ptr++) {
    printf("The value at %p is %d\n", (void*) ptr, *ptr);
  }
  return 0;
}

ptr stops when it's 10 elements further than when it started (starts at x, ends at x+10).

Each time we print what ptr itself is, the address (the void* cast is there to adhere to C standard -- doesn't affect the logic of the program) and also print the element it will produce.

Here's my output (your pointers will differ, but not the values):

The value at 000000BBA258F928 is 1
The value at 000000BBA258F92C is 3
The value at 000000BBA258F930 is 2
...

We can't use x[i] because i doesn't exist, but that's fine, because ptr does and get us the values we want.

hmmmmmmmm....

    for(int* ptr=x;ptr < 10;ptr++){
        printf("%d\n",x[i]);
        printf("%d\n",*ptr);
    }

wrong!!!! yoou can do the following, instead:

    for(int* ptr=x;ptr < x + 10;ptr++){
        /* printf("%d\n",x[i]); in this case you don't have i */
        printf("%d\n",*ptr);
    }

Don't worry, as x + 10 is a constant pointer in the context of the loop and will not have to be calculated on each iteration. Indeed, that code is equivalent to this:

    const int * const x_end = x + 10; /* get x_end before entering the loop */
    for(int* ptr=x;ptr < x_end;ptr++){
        printf("%d\n",*ptr);
    }

The code you are intending to will not work

you have initialized the variable ptr as a pointer inside the for loop and in the ptr you stored the x ,here (in c) x is an array and it will contain the starting address of the array which is also a pointer so

but when you start to compare the ptr with 10 which will create an error logically since the value a pointer will have value of 64000,6784579,.....etc the program will consider this pointer as an integer perform comparison . due to this conidering the value of ptr exceeded the 10 the for loop will be exited

and the i you are using is not defined

the program i given below uses your idea to print the array's elements using a pointer the base value is incremented by the 1 integer each time which will give you each element sequentially similary to using a pointer loop

#include <stdio.h>
#include <stdlib.h>

int main(){
    int x[10]={1,3,2,5,43,67,8,64,22,123};

    for( int ptr=0;ptr < 10;ptr++){
        printf("%d\n",*(x+ptr));
    }
    return 0;
}