Multiplying integer with elements of pointer array in C

Question

#include <stdio.h>

void main()
{
    int array[5]={1,2,3,4,5};
    int *parray;
    *parray=&array;
    int i;
    for(i=0;i<5;i++)
    {
        printf("%d%d",*(parray[i]*2));// [Error] invalid type argument of unary '*' (have 'int')
    }
    getch();
}

I don't know why this error is coming. How do I solve it?

score 3 · Answer 1 · edited May 23 '17 at 12:30

This line
```
*parray=&array;
```
is undefined behavior because parray hasn't been initialized. Dereferencing it will point to an indeterminate memory location.
Use
```
parray = array;
```
instead. Read up on the difference between array and &array also.
This expression
```
*(parray[i] * 2)
```
causes your compilation error. Applying the subscript operator to an int* (parray) yields an int. Multiplying that by 2 yields an int again. Now you attempt to dereference that int. int is no pointer, so it cannot be dereferenced.

The format specifier %d%d expects two ints. You only provide one, so this is undefined behavior again. A simple %d suffices.

After all, change
```
printf("%d%d",*(parray[i]*2));
```
to
```
printf("%d", parray[i] * 2);
```

ameyCU · Accepted Answer · 2015-12-06T11:40:46.120

int *parray;
*parray=&array;

1. parray is of type int * and in next line when you dereference it (undefined behaviour as it is uninitialized) becomes type int to which you try to assign address of array . Just do -

parray=array;

2. And in this line -

  printf("%d%d",*(parray[i]*2));  //gave 2 specifiers but pass one argument ??

You use 2 format specifiers but you pass only one argument and that also of wrong type.

When you write parray[i] it means *(parray+i) ,you don't need to further dereference it using * , as it already of type int (and you try to dereference an int variable).

Just writing this would work-

printf("%d",parray[i]*2);

Multiplying integer with elements of pointer array in C

2 Answers2