CODEWITHSUNDEEP
c
Imdad
Imdad

Reputation: 769

Invalid type argument of unary '*' (have 'int') Error in C

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

/*
 * 
 */
int main() {

    int a[] = {5, 15, 34, 54, 14, 2, 52, 72};
    int p = &a[1];
    int q = &a[5]; 

    printf(*(p+3));
    printf(*(q-3));
    printf(*(q-p));
    printf(*p<*q);

    return (EXIT_SUCCESS);
}

Errors: "initialization makes integer from pointer without a cast [-Wint-conversion]" and "invalid type argument of unary '*' (have 'int')". First error is shown twice for the initialisation of variables above. Second error is shown for each print statement.

I'm not sure what is going wrong, anyone know how I can fix this?

Upvotes: 9

Views: 131672

Answers (3)

VHS
VHS

Reputation: 10184

You forgot to make p and q int pointers. Also, you forgot to use the format specifier in the printf statements. Try the following:

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

/*
* 
*/
int main() {
  int a[] = {5, 15, 34, 54, 14, 2, 52, 72};
  int *p = &a[1];
  int *q = &a[5];   

  printf("%d\n", *(p+3));
  printf("%d\n", *(q-3));
  printf("%d\n", *q-*p);
  printf("%d\n", *p<*q);
  return (EXIT_SUCCESS);
}

Upvotes: 16

lost_in_the_source
lost_in_the_source

Reputation: 11237

The unary operator & yields the address of its operand. The type is of T *, not T. Therefore you cannot assign a int * to an int without a cast. The expression

&a[1]

yields the address of a[1].

I think you mean to define the variables as pointers to int, not just ints.

int *p = &a[1];
int *q = &a[5];

Upvotes: 2

Mark Segal
Mark Segal

Reputation: 5550

&a[3] (or &a[5]) is a pointer type, i.e. int *.

p is defined as int.

So you need to define p and q as int *, like this:

int * p = &a[1];
int * q = &a[5];

Upvotes: 3

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