Converting a decimal to a binary in C (output missing digits) - arrays

Question

Why is it returning 7 digits instead of 8 digits once binary array's length is 8 (meaning 8 bits)?
#include <cs50.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
int number = get_int("Number: ");
char binary[8]; // 8 bits in a byte
for (int i = 0; number != 0 ; i++)
{
if (number % 2 == 0)
{
binary[i] = '0';
}
else
{
binary[i] = '1';
}
number /= 2;
}
printf("%s\n", binary);
}
I'm getting
Number: 72
Binary: 0001001
I know it's reversed from the correct answer for this decimal, just want to correct the missing digit first.

Answer 1

For your program, I only had to make one small correction:
https://ideone.com/87v18W
int main(void)
{
int number = 145;
char binary[8+1] = {}; // 8 bits in a byte, + 1 for string-terminator
// If you're going to print it out as a string, you have to make sure there is space for a nil-terminator ('\0') at the end of the string.
// Hence the +1 for space, and the ={}; to set the whole thing to 0 before filling in the bits.
for (int i = 0; number != 0 ; i++)
{
if (number % 2 == 0)
{
binary[i] = '0';
}
else
{
binary[i] = '1';
}
number /= 2;
}
printf("%s\n", binary);
}
When I run this program, with input of 145, I get
Output
10001001
Here's how I do ints to binary, with a . separator every 8 bits:
#include <stdio.h>
#include <limits.h>
#include <stdint.h>
int main(void) {
uint32_t num = 93935;
for(int i=(sizeof(uint32_t)*CHAR_BIT)-1; i>=0; --i)
{
printf("%c%s", "01"[!!(num & (1<<i))], i%8?"":i?".":"");
}
return 0;
}
Example Output:
Success #stdin #stdout 0s 5404KB
00000000.00000001.01101110.11101111