Clear the Bit in an 8-bit Register

Code

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

uint8_t clear_bit(uint8_t reg, uint8_t pos) {
    // Your code here
    int bitvalue= pow(2,pos);
    reg &= (0xFF ^ bitvalue);
    return reg;
}

int main() {
    uint8_t reg, pos;
    scanf("%hhu %hhu", &reg, &pos);
    uint8_t result = clear_bit(reg, pos);
    printf("%u", result);
    return 0;
}

Solving Approach

Do a bitwise AND with the one complement of the bit value.

So if bit 3, it is 0x08= 00001000 , and its ones complement is 0xF7=11110111

doing bitwise AND with the register will clear bit 3