Bit Reversal in an 8-bit Value

Code

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

uint8_t C_DEC(char *arr){
    int result = 0;
for(int i = 0 ; i <= 7 ; i++){
result += arr[i] * pow(2, i);
}
return result;
}

uint8_t reverse_bits(uint8_t val) {
    // Your logic here
     
   char ram[8];
    uint8_t mask = 0;
    for(int i = 7; i >= 0; i--){
       mask = (1 << i);
       if((mask & val) > 0){
        ram[7 - i] = 1;
       }
       else{
        ram[7 - i] = 0;
       }
    }
   return C_DEC(ram);

    
   
}

int main() {
    uint8_t val;
    scanf("%hhu", &val);

    uint8_t result = reverse_bits(val);
    printf("%u", result);
    return 0;
}

Solving Approach