How to convert a 16-bit hexadecimal number to binary number and then count number of zeros and one in it ? Using simplest C program

Question

我想到的一个简单程序是将十六进制数转换为二进制数，然后使用计数器来计算零和一的数量。

Answer 1

#include <stdio.h>
#define MAX 1000

int main()
{
    char binarynum[MAX], hexa[MAX];
    long int i = 0,number_of_zeros=0,number_of_ones=0;

    printf("Enter the value for hexadecimal ");
    scanf("%s", hexa);
    printf("\n Equivalent binary value: ");

    while(hexa[i])
    {
        switch (hexa[i])
        {
        case '0':
            printf("0000");
            number_of_zeros +=4;
            break;
        case '1':
            printf("0001");
            number_of_zeros +=3;
            number_of_ones +=1;
            break;
        case '2':
            printf("0010");
            number_of_zeros +=3;
            number_of_ones +=1;
            break;
        case '3':
            printf("0011");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case '4':
            printf("0100");
            number_of_zeros +=3;
            number_of_ones +=1;
            break;
        case '5':
            printf("0101");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case '6':
            printf("0110");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case '7':
            printf("0111");
            number_of_zeros +=1;
            number_of_ones +=3;
            break;
        case '8':
            printf("1000");
            number_of_zeros +=3;
            number_of_ones +=1;
            break;
        case '9':
            printf("1001");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case 'A': case 'a':
            printf("1010");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case 'B': case 'b':
            printf("1011");
            number_of_zeros +=1;
            number_of_ones +=3;
            break;
        case 'C': case 'c':
            printf("1100");
            number_of_zeros +=2;
            number_of_ones +=2;
            break;
        case 'D': case 'd':
            printf("1101");
            number_of_zeros +=1;
            number_of_ones +=3;
            break;
        case 'E': case 'e':
            printf("1110");
            number_of_zeros +=1;
            number_of_ones +=3;
            break;
        case 'F': case 'f':
            printf("1111");
            number_of_ones +=4;
            break;

        default:
            printf("\n Invalid hexa digit %c ", hexa[i]);
            return 0;
        }
        i++;
    }
    printf("\nNumber of zero\'s is %d\n",number_of_zeros);
    printf("Number of One\'s is %d\n",number_of_ones);

    return 0;
}