Invalid Operands to binary / (have 'int *' and 'int')?

Question

Every time I try this:

long crypt(int *integer)
{
printf("Enter five digit integer:\n");  
scanf("%i",integer);

int digit1=integer/10000;
int digit2=(integer%10000)/1000;
int digit3=(integer%1000)/100;
int digit4=(integer%100)/10;
int digit5=(integer%10)/1;

const char *digit1c[10];
const char *digit2c[10];
const char *digit3c[10];
const char *digit4c[10];
const char *digit5c[10];

(There's more but this seems to be the problem, I'll add the rest by request.)

then it return this error:

math2.h:44:20: error: invalid operands to binary / (have ‘int *’ and ‘int’)
math2.h:45:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:46:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:47:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)
math2.h:48:21: error: invalid operands to binary % (have ‘int *’ and ‘int’)

I know it has something to do with the operators I used to initialize the digits and I did try changing their type to "int *" but that didn't work. So what's happening here exactly?

Answer 1

integer is a pointer to int (int*), so when you want to use the int it points to, you need to dereference it:

int digit1=(*integer)/10000; // and so on...