Printing 1D array using pointer in C

Question

I want to print the data of array by using pointers so I try to save the address of array in the pointer. But the pointer doesn't print the data. I will print a second array as well later on so there are some extra variables declared.

Output: Output

Code

//print 1D array and 2D array
#include<stdio.h>
#include<stdlib.h>
int Arr1[10];
int Arr2[10][10];
int i, j, n1, n2;
int (*p1)[10];
int (*p2)[10][10];

int main()
{
    printf("For the 1D Array: \n");
    printf("Enter the number of elements you want to add: ");
    scanf("%d", &n1);

    printf("Enter the data for the elements:\n");
    for(i=0;i<n1;i++)
    {
        scanf("%d", &Arr1[i]);
    }
    printf("Displaying Array:\n");
    for(i=0;i<n1;i++)
    {
        printf("%d\t", Arr1[i]);
    }
    
    printf("\nDisplaying using pointer: \n");
    p1=Arr1;
    printf("1D Array is: \n");
    for(i=0;i<n1;i++)
    {
        printf("Arr[%d] is %d\t", i, *(p1[i]));
        printf("\nAddress of %d th array is %u\n", i, p1[i]);
    }

}

Arr1 IS the address of the array! printf("Arr[%d] is %d\t", i, *(Arr1+i)) works — chrslg
– chrslg, Commented Nov 26, 2022 at 11:11
You need the address p1=&Arr1; also to print use printf("Arr[%d] is %d\t", i, (*p1)[i]); and printf("\nAddress of %d th array is %p\n", i, *p1+i); — QuentinUK
– QuentinUK, Commented Nov 26, 2022 at 11:46

Vlad from Moscow · Accepted Answer · 2022-11-26 11:24:07Z

0

The pointer p1 is declared like

int (*p1)[10];

In this assignment statement

p1=Arr1;

the pointer is assigned with an expression of the type int * due to the implicit conversion of the array designator to pointer to its first element of the type int *.

The pointer types of the operands are not compatible. So the compiler should issue a message.

You could either write

int (*p1)[10];

//...

p1 = &Arr1;
printf("1D Array is: \n");
for(i=0;i<n1;i++)
{
    printf("Arr[%d] is %d\t", i, ( *p1 )[i] );
    printf("\nAddress of %d th array is %p\n", i,( void * ) ( *p1 + i ) );
}

The expression with the subscript operator

( *p1 )[i]

is equivalent to

*( *p1 + i )

Or you could write

int *p1;

//...

p1 = Arr1;
printf("1D Array is: \n");
for(i=0;i<n1;i++)
{
    printf("Arr[%d] is %d\t", i, p1[i]);
    printf("\nAddress of %d th array is %p\n", i,( void * ) ( p1 + i));
}

I suppose that in the second call of printf you want to output the address of each element of the array..

edited Nov 26, 2022 at 11:24

answered Nov 26, 2022 at 11:11

Vlad from Moscow

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2 Comments

L127Bangtanned Over a year ago

Okay, I see the mistake in the incompatibility of pointer types. Yes, I wanted to print the address of each element in the array. What is the purpose of using ( void * ) ?

Vlad from Moscow Over a year ago

@L127Bangtanned The argument of the conversion specifier p shall be a pointer of the type void *.

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