Decimal to binary in C language - c

Question

I wrote simple code in C language (without libraries), but the result is good only when you read it from right to left, how to reverse it? I want the most simplified code as it is possible. My code below:
#include <stdio.h>
int main() {
int number;
printf("Changes from decimal to binary\n");
printf("Enter the number: ");
scanf("%d",&number);
do{
if(number % 2 == 1){
printf("1");
}else{
printf("0");
}
number = number / 2;
}
while(number>0);
return 0;
}

Answer 1

You could store the binary number in an array and then print it backwards. So you would have for example int binaryDigits[1024]; and then instead of printing you will store the number in the array with a counter. Before the loop: int i = 0; and after that instead of printf("1"); and printf("0");, binaryDigits[i++] = 1 and binaryDigits[i++] = 0 respectively. Finally you could print the number in order with a for loop:
for (i = i - 1; i >= 0; i--)
printf("%d", binaryDigits[i])

Answer 2

You could use recursion and print the digits as you fall off the stack. I changed number from int to unsigned int.
void print_dec_to_bin_hlp(unsigned int number)
{
if(number > 0)
{
print_dec_to_bin_hlp(number/2);
printf("%d", (number%2));
}
}
void print_dec_to_bin(unsigned int number)
{
if (number == 0) printf("%d", number);
else print_dec_to_bin_hlp(number);
}

Answer 3

A recursive solution using bitwise operations, for unsigned integers:
void print_binary(unsigned int num) {
if(num >> 1) print_binary(num >> 1);
putchar(num & 1 ? '1' : '0');
}
The bitwise operators >>, &, etc. access the bits in the value. The bits form a binary representation of the number. >> shifts the bits to the right, where the bits are ordered from MSB (most significant bit) to LSB, like the normal way numbers are written.
num & 1 returns non-zero only if the LSB (rightmost) bit is set. (AND mask with ..0001). So (num >> 1) & 1 does the same on the second-rightmost bit. So this recursive function prints the bits in MSB to LSB order.
The if(num >> 1) check makes sure prefix zeroes are not printed: If after the rightshift, num is all-zeroes, it is not printed any further. The check is made outside the function call, so that print_binary(0) (initial call) will still always print a single 0.
For signed integers:
void print_signed_binary(int num) {
if(num < 0) {
putchar('-');
print_binary(-num);
} else {
print_binary(num);
}
}