Clear Specific Bits in a 32-bit Register

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

void printBinary(uint32_t x);

uint32_t clear_bits(uint32_t reg, uint8_t pos, uint8_t len) {
    // Your code here
    //1. Create a mask of 1's of len
    //1.1 Put a 1 in the len position (e.g 4)
    //mask =  1<< len (0001 0000)
    //1.2 Subtract 1 to get mask of 1s of length
    //mask = ((1<<len)-1) (0000 1111)
    //1.3 Move mask to correct position pos
    //mask = ((1<<len)-1) << pos
    uint32_t mask = ((1<<len)-1) << pos; 
    //1.4 Apply method to clear reg bits at mask position (invert bits of mask 0000 and and)
    //reg &= ~((1<<len)-1) << pos
    
    /*PRINT TESTS
    printf("Mask\n");
    printBinary(mask);
    printf("Reg\n");
    printBinary(reg);
    printf("~Mask\n");
    printBinary(~mask);
    printf("reg & ~Mask\n");
    printBinary(reg & ~mask);
    */
    
    reg &= ~mask;
    return reg;
}

int main() {
    uint32_t reg;
    uint8_t pos, len;
    scanf("%u %hhu %hhu", &reg, &pos, &len);
    printf("%u", clear_bits(reg, pos, len));
    return 0;
}

//FUNCTION TO PRINT BINARY 8 bit
void printBinary(uint32_t x){
    //1. Find size of passed int
    int num_bits = sizeof(x)*8;
    //2. loop through every bit and perform checkbit
    //printf("Int value is: %hhu, Num of bits is: %d\n",x, num_bits);
    for(int i=num_bits-1;i>=0;i--){
    //Prints a space every 4 bits
    if ((i+1)%4==0 && i+1 != sizeof(x)*8){
        printf(" ");
    }
    if (x & (1 << i)){
        printf("1");
    } 
    else{
        printf("0");
    }
    }
printf("\n");
}