Set the Bit in an 8-bit Register

Code

#include <stdio.h>
#include <stdint.h>

uint8_t set_bit(uint8_t reg, uint8_t pos) {
    // Your code here
    reg |= (1<<pos);
    return reg;
}

int main() {
    uint8_t reg, pos;
    scanf("%hhu %hhu", &reg, &pos);  // Accept register value and position
    uint8_t result = set_bit(reg, pos);
    printf("%u", result);         // Output the result as an integer
    return 0;
}

Solving Approach
I performed OR operation with the register value by left shifting 1 by position value. 
So if the position if 5 and the reg = 10001111, 1<<5 means 00100000.
now, reg | (1<<5) = 10001111 | 00100000 = 10101111
Hence bit is set!