Printing Binary Number Backward - c++

Question

I need to print a binary number backward without explicitly converting to binary or using an array (i.e. if the binary number is 10, it should print as 01). Here is the code I've done for printing the number forward. I'm fairly certain that I just need to tell the code to run through the loop starting at the other end in order to have the number render backward. However, I have no idea how to go about doing that, or if that's even correct.
Bonus question -- can someone walk me through what this code is really doing? It's modified from one we were given in class, and I don't fully understand what it actually does.
NOTE: the test case I have been using is 50.
#include <stdio.h>
char str [sizeof(int)];
const int maxbit = 5;
char* IntToBinary (int n, char * BackwardBinaryString) {
int i;
for(i = 0; i <= maxbit; i++) {
if(n & 1 << i) {
BackwardBinaryString[maxbit - i] = '1';
}
else {
BackwardBinaryString[maxbit - i] = '0';
}
}
BackwardBinaryString[maxbit + 1] = '\0';
return BackwardBinaryString;
}
int main () {
int base10input;
scanf("%d", &base10input);
printf("The backwards binary representation is: %s\n", IntToBinary(base10input, str));
return 0;
}

Answer 1

To your disappointment, your code is wrong in these aspects.
sizeof(int) returns the bytes an int takes, but we need the bit it takes as we store each bit in a char, so we need to multiply it by 8.
Your char array str have a size of 4, which means only str[0] to str[3] are vaild. However, you modified str[4], str[5] and str[6] which are out of bounds and such undefined behavior will result in a disaster.
What you should do first is to create an array holds at least sizeof(int) * 8 + 1 chars. (sizeof(int) * 8 for the binary representation, one for the null-terminator) Then start your convention.
And I also suggest that the str should not be a global variable. It will be better to be a local variable of main function.
Your code should be modified like this. I've explained what it does in the comments.
#include <stdio.h>
#define INTBITS (sizeof(int) * 8) // bits an integer takes
char* IntToBinary(int n, char* backwardBinaryString) {
// convert in reverse order (str[INTBITS - 1] to str[0])
// remember that array subscript starts from 0
for (int i = 0; i < INTBITS; i++) {
// (n & (1 << i)) checks the i th bit of n is 0 or 1
// if it is 1, the value of this expression will be true
if (n & (1 << i)) {
backwardBinaryString[INTBITS - 1 - i] = '1';
}
else {
backwardBinaryString[INTBITS - 1 - i] = '0';
}
// here replacing the if-else with and conditional operator like this
// will make the code shorter and easier to read
// backwardBinaryString[INTBITS - 1 - i] = (n & (1 << i)) ? '1' : '0';
}
// add the null-terminator at the end of str (str[INTBITS + 1 - 1])
backwardBinaryString[INTBITS] = '\0';
return backwardBinaryString;
}
int main() {
char str[INTBITS + 1];
int base10input;
scanf("%d", &base10input);
printf("The backwards binary representation is: %s\n", IntToBinary(base10input, str));
return 0;
}

Answer 2

That code is far more elaborate than it needs to be. Since the requirement is to print the bits, there's no need to store them. Just print each one when it's generated. And that, in turn, means that you don't need to use i to keep track of which bit you're generating:
if (n == 0)
std::cout << '0';
else
while (n != 0) {
std::cout << (n & 1) ? '1' : '0';
n >>= 1;
}
std::cout << '\n';