Storing a hexadecimal string without using arrays in C - c

Question

Our school gave us an assignment and in some part of that assignment we are supposed to take a hexadecimal input then convert it to binary. But the problem is that using arrays are not permitted. When using an array to store the hexadecimal string I can easily convert it to binary but is there anyway to store it without using arrays?

Answer 1

Your task seems to be:
Convert input data entered as hexadecimal values to a stream of binary digits.
You can just read the input ne byte at a time and convert any hexadecimal character to the corresponding binary digits.
There are many different ways to approach this problem, here is an implementation with a switch statement.
#include <stdio.h>
int main() {
int c;
while ((c = getchar()) != EOF) {
switch (c) {
case '0': fputs("0000", stdout); break;
case '1': fputs("0001", stdout); break;
case '2': fputs("0010", stdout); break;
case '3': fputs("0011", stdout); break;
case '4': fputs("0100", stdout); break;
case '5': fputs("0101", stdout); break;
case '6': fputs("0110", stdout); break;
case '7': fputs("0111", stdout); break;
case '8': fputs("1000", stdout); break;
case '9': fputs("1001", stdout); break;
case 'A':
case 'a': fputs("1010", stdout); break;
case 'B':
case 'b': fputs("1011", stdout); break;
case 'C':
case 'c': fputs("1100", stdout); break;
case 'D':
case 'd': fputs("1101", stdout); break;
case 'E':
case 'e': fputs("1110", stdout); break;
case 'F':
case 'f': fputs("1111", stdout); break;
case '\n':
case ' ': putchar(c); break;
default: break;
}
}
return 0;
}
If you just need to read a value encoded as a hexadecimal string and store it into a variable, scanf() is a simple solution:
#include <stdio.h>
void print_binary(unsigned int x) {
if (x > 1) {
print_binary(x >> 1);
}
putchar('0' + (x & 1));
}
int main(void) {
unsigned int value;
if (scanf("%x", &value) == 1) {
print_binary(value);
putchar('\n');
}
return 0;
}