Printing binary char in c

Question

I cannot, for the love of myself figure out what the logic is behind this C code.

Supposedly, the following should print out the binary representation of an unsigned char x, and we are only allowed to fill in the blanks.

void print_binary(unsigned char x) {
    int b = 128;

    while (__________________) {

        if (b <= x) {
            x -= b;
            printf("1");
        } else
            printf("0");

        ______________________;
    }
}

Of course I could game the program by simply ignoring the lines above. However I'm under the impression that this is not really the way to do things (it's more of a hack).

I mean really now, the first condition checks whether 128 is <= the char, but isn't an unsigned char, what, 255 bytes? So why is it only printing '1' in it.

Perhaps I'm missing something quite obvious (not really a c programmer) but the logic just doesn't sink into me this time.

Can anyone point me in the right direction? And if you can give me a clue without completely saying the answer, that would be heavenly.

The second blank is probably `b /= 2`; the first is presumably `b != 0`. — Jonathan Leffler, Sep 02 '13 at 06:18
I invite you to look to the list on the right/left of Related questions. and pick [this one](http://stackoverflow.com/questions/111928/is-there-a-printf-converter-to-print-in-binary-format?rq=1). Further, "but isn't an unsigned char, what, 255 bytes" ?? I dunno about your platform, but on mine an unsigned char is *one* byte. — WhozCraig, Sep 02 '13 at 06:18
@Joni My exact feeling - which is weird since this is supposed to be an "Are you familiar enough with C?" test. — Secret, Sep 02 '13 at 06:24

Thanushan · Answer 1 · 2013-09-02T06:40:59.423

3

void print_binary(unsigned char x) {
    int b = 128;

    while (b != 0) {

        if (b <= x) {
            x -= b;
            printf("1");
        } else
            printf("0");

        b = b >> 1;
    }
}

The binary representaion for b is 10000000. By doing b >> 1 and checking b <= x we can check each bit on x is 1 or 0.

edited Sep 02 '13 at 06:40

answered Sep 02 '13 at 06:34

Thanushan

532
2
8

1

did you miss a "`b = `"?? "`b = b >> 1;`" – raj raj Sep 02 '13 at 06:38
@raj raj. yes, Thanks. was a typo. – Thanushan Sep 02 '13 at 06:40
he asked for a hint: "And if you can give me a clue without completely saying the answer", but you gave him the answer – AndersK Sep 02 '13 at 07:21

user694733 · Answer 2 · 2013-09-02T06:55:25.463

1

You wanted only a clue: Value of the current bit is always bigger, than the combination of less significant bits after it. Thus code tries to test only the most significant '1'-bit on each iteration of loop.

edited Sep 02 '13 at 06:55

answered Sep 02 '13 at 06:50

user694733

15,208
2
42
68

score 0 · Answer 3 · answered Sep 02 '13 at 06:28

0

If we disregard the original code, the most intuitive way to do this would be:

void print_binary (uint8_t x) 
{
  for(uint8_t mask=0x80; mask!=0; mask>>=1)
  {
    if(x & mask)
    {
      printf("1");
    }
    else
    {
      printf("0");
    }
  }
  printf("\n");
}

answered Sep 02 '13 at 06:28

Lundin

195,001
40
254
396

There are advantages to not including the newline in the output from the function. – Jonathan Leffler Sep 02 '13 at 06:30
@JonathanLeffler Without a specification of what the function should do, who can tell. – Lundin Sep 02 '13 at 06:31
Do we really need an `if else`? What about this? `for(int i=7; i>=0; i--) printf("%d", (x>>i)&1);` – raj raj Sep 02 '13 at 06:43
The original code was a form of specification; it does not include a newline at the end of the output, and for the sake of being generally useful, it should not include the newline. – Jonathan Leffler Sep 02 '13 at 13:55

Printing binary char in c

3 Answers3