To practice my C, I've written a small function in C to display an integer as an 8-bit binary number. Please let me know how to improve this function, and other C-style coding conventions. Thanks!

 * Converts an integer to binary.
 * @param num Integer to convert.
 * @return c-string.
const char* to_binary(int num) {

    static char binary[10];
    binary[0] = '0';
    binary[1] = 'b';

    int value = 128;
    int number = num;

    for (int i = 2; i < 10; i++) {
        if (number - value >= 0) {
            binary[i] = '1';
            number = number - value;
        } else {
            binary[i] = '0';
        value = value / 2;

    return binary;


And here's how you call it:

int main() {

    const char* result = to_binary(146);

    printf("%s\n", result);

  • More specific types are available for an eight bit number, such as uint8_t
  • static keyword is dangerous (when used as an output buffer) and results in weird behaviour if your method is called twice
  • The arithmetic can be replaced with a more simple & (bitwise AND)
void to_binary(uint8_t x, char *output) {
    *output++ = '0';
    *output++ = 'b';
    for (int mask = 1<<CHAR_BIT-1; mask!=0 ; mask>>=1) {
        *output++ = x & mask ? '1':'0'; 
    *output = 0; // null terminate the string
  • 6
    \$\begingroup\$ static isn't weird and dangerous. It means the data is allocated in either .DATA or the .BBS section of the binary, instead of the stack. This means a variable is allocated once, and the value "kept" between calls. It has legitimate uses. \$\endgroup\$
    – AnnoyinC
    Jan 21 at 10:07
  • 9
    \$\begingroup\$ @AnnoyinC, I think Ted's point is that using static storage makes the function dangerous and weird. Users don't expect such a function to be stateful like that. BTW, C standard says nothing about "segments" - that's a platform-specific mechanism. \$\endgroup\$ Jan 21 at 12:05
  • 2
    \$\begingroup\$ 2<<i should be 1<<i and you need to start from the MSB, for (int i = total_bits - 1; i >= 0 ; --i). \$\endgroup\$ Jan 21 at 16:31
  • 2
    \$\begingroup\$ Instead of counting with i and generating the mask on the fly, you can do something like for (mask = 1 << 7; mask != 0; mask >>=1), this is naively more efficient, but the compiler probably does this already at O1 or so. \$\endgroup\$
    – 12345ieee
    Jan 21 at 17:55
  • 2
    \$\begingroup\$ Your placement of spaces in the code is strange. \$\endgroup\$ Jan 22 at 1:07


Code attempts to form a string. Yet it lacks a terminating null character and space for it.

// static char binary[10];
static char binary[11];

As a static, it is initialized to all zeros. With increased size, an explicit setting of the null character is not needed.

OP's code undefined behavior perhaps "works" as a zero may exist just past binary[10].

Loop Simplification

Iterate right (most significant) to left (least significant).

for (int i = 10; --i >= 2; ) {
  binary[i] = '0' + (num & 1);
  num /= 2;

Unclear functionality

When num > 255 code prints all '1'.
When num < 0 code prints all '0'.

Code should 1) not use an int argument or 2) state functionality when num is outside the [0-255] range.


I'd avoid the use of static and provide a buffer to the function. Example.

  • \$\begingroup\$ Just a thought, but it might work with 10 because there could be unused bytes after for alignment. \$\endgroup\$
    – Ben
    Jan 21 at 22:02
  • \$\begingroup\$ @ben Agree. See "OP's code undefined behavior perhaps "works" as a zero may exist just past binary[10]." \$\endgroup\$ Jan 21 at 23:11
  • \$\begingroup\$ @Ben even if there are "unused bytes for alignment", there is no guarantee that these bytes are initialized to 0. Still undefined behavior. \$\endgroup\$ Jan 22 at 1:05
  • \$\begingroup\$ @RolandIllig, hence the might. \$\endgroup\$
    – Ben
    Jan 22 at 13:05
  • \$\begingroup\$ @L.F. Agree static char are zero-initialized. Yet the bytes near, though not part of, binary[10] are the issue. They are not specified to be zero nor even accessible. \$\endgroup\$ Jan 22 at 14:15

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