Working in Simple Cesar Cipher in C

Question

I'm working in a program for university and they use a problem like the typical Cesar Cipher. It's more like a functional program and needs to be the most basic possible.

The program will receive a number from the user from 65 to 90 and for example when the user inserts 65 will show 68. Will add 3 numbers but when the user gives 90 will give 67. 90+3 ---->90,65,66,67. It's a cycle from 65 to 90.

#include <stdio.h>

int cesar_encrypted(int x)
{
  return (x+3);
}


void test_cesar_encrypted(void)
{
    int x;
    scanf("%d", &x);
    int z = cesar_encrypted(x);
    printf("%s\n", z);
}

int main(){
    test_cesar_basic();

}

I did this sample code, but we can only go further and if you give 90 he will give 93 and I want 67.

Can anyone help me to wrap it around 90?

Answer 1

You can use the modulo operator, it gives the remainder of a division:

int cesar_encrypted(int x)
{
  return (x - 65 + 3)%(91 - 65) + 65;
}

---

Implementing the suggestions of Sulthan (see comment), it would look like this:

int cesar_encrypted(int x)
{
  const int n_chars = 'Z' - 'A' + 1;
  const int shift = 3;
  return (x - 'A' + shift)%n_chars + 'A';
}