Changing the value of a pointer but keeping the address

Question

I want to write a program which changes the value in the address a pointer points to but keep the address the same. I wrote a small program to test it but I'm having some problems.

#include <stdio.h>

void swap(int *entry1, int *entry2, int length)

{
  int i,temp;

  for(i=0; i<length; i++)
    {
      temp = *entry1[i];
      *entry1[i] = *entry2[i];
      *entry2[i] = temp;
    }
}

int main()

{
  int length=5, i;

  int *entry1 = malloc(length * sizeof(int));
  int *entry2 = malloc(length * sizeof(int));

  for(i=0;i<length;i++)
    {
      entry1[i] = i + 1;
      entry2[i] = length - i;

    }
swap(entry1, entry2, length);

return 0;
}

I was under the impression that I could use an asterisk to access the value at stored in the address the pointer points at. However, when I run this code I get the error:

error: invalid type argument of unary ‘*’ (have ‘int’)

I believe if I do:

temp = entry1[i];
entry1[i] = entry2[i];
entry2[i] = temp;

This would effectively swap the values, however, it would also swap the address' which I don't want.

Answer 1

Well yes you did wrong. Correct one would be

  temp = entry1[i];
  entry1[i] = entry2[i];
  entry2[i] = temp;

This would be enough given that array decays into pointer to first element - enabling you to change the value of the array.

invalid type argument of unary ‘*’ (have ‘int’) 

The error itself says that you are trying to dereference an int not an address or some pointer value.

And for your information, entry1[i] is *(entry+i) . But you did this **(entry+i) . You passed a single level pointer so it is expected that you won't dereference it twice without any error.

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Also check the return value of malloc and free dynamically allocated memory when you are done working with it.