showbits(int)

[Bhushan]

There is a bitwize function showbits(int) in C language which displays what is saved in individual bit of integer. I am trying it from hours but it is not working if any one know which header file have to be used then please tell.

[laserlight]

I believe that showbits() is not a standard C function.

[Bhushan]

I believe that showbits() is not a standard C function.

If showbits() is not a standard function then is there any other alternative for same.

[laserlight]

If showbits() is not a standard function then is there any other alternative for same.

I am guessing that it returns or prints the binary representation of an integer. That should be rather trivial to write, methinks.

[brewbuck]

There is a bitwize function showbits(int) in C language which displays what is saved in individual bit of integer. I am trying it from hours but it is not working if any one know which header file have to be used then please tell.

Hours? You've been trying to call a nonexistent function for HOURS?

[mike_g]

I've never heard of showbits(), but this will print the binary value of an integer to the screen:

void PrintBin(unsigned int x)
{
    int i; 
    for(i=(sizeof(int)*8); i>=0; i--)
        (x&(1<<i))?putchar('1'):putchar('0');
}

[matsp]

void PrintBin(unsigned int x)
{
    int i; 
    for(i=(sizeof(int)*8)-1; i>=0; i--)
        (x&(1<<i))?putchar('1'):putchar('0');
}

Fix: Note -1 on the i initializer... a 32-bit (4 * 8 bit) integer has bits from 31..0, not 32..0...

--
Mats

[robatino]

For portability you might want to compute the number of bits in UINT_MAX by right-shifting until you get 0, since it's not always true that the number of bits in a byte is 8, nor that all the bits in an unsigned int are value bits.

[mike_g]

Fix: Note -1 on the i initializer... a 32-bit (4 * 8 bit) integer has bits from 31..0, not 32..0...

Oh yeah, that was quite dumb of me. I did it right, then edited it to take the -1 out because for some reason I though it was wrong o_0.

For portability you might want to compute the number of bits in UINT_MAX by right-shifting until you get 0, since it's not always true that the number of bits in a byte is 8, nor that all the bits in an unsigned int are value bits.

Yeah that should be better. Just out of curiosity is there any way to get the size in bits of a Byte?

[laserlight]

Yes. CHAR_BIT should have it.

[mike_g]

Ok, thanks

[robatino]

> Yeah that should be better. Just out of curiosity is there any way to get the size in bits of a Byte?
<limits.h> gives you both UINT_MAX (maximum value of unsigned int) and CHAR_BIT (number of bits in a byte). But although sizeof(unsigned int)*CHAR_BIT is always the number of bits in an unsigned int (and you could shorten that to sizeof(int)*CHAR_BIT since the signed and unsigned versions are always the same size) it's not guaranteed that all of those bits are value bits - for example, even if the number of bits is 32, it could happen that only 23 of those are value bits, so UINT_MAX would be equal to 2^23 - 1, not 2^32 - 1 as you would expect. The whole business of "value bits" and "padding bits" is something it's not easy to get information on - all I've found are newsgroup postings of people quoting the Standard at each other. But basically, UINT_MAX is equal to 2^N - 1, where N is equal to the number of value bits, which is <= the number of bits. To find the number of bits in UINT_MAX, you can do something like

unsigned int j = UINT_MAX, n = 0;
while (j >>= 1) ++n;

and then initialize i = n-1.

Edit: Actually, that should be i = n.

[Dave_Sinkula]

As merely a side note on what has already been mentioned, you don't really need to know the exact number of value bits in the int. By using a bitmask as the loop "counter", you can achieve the same effect. My preferred method is along this line:

void showbits(unsigned int value)
{
   unsigned int bit;
   for ( bit = /* msb */(-1U >> 1) + 1; bit > 0; bit >>= 1 )
   {
      putchar(value & bit ? '1' : '0');
   }
   putchar('\n');
}

If anyone is interested in more of me babbling on this, here you go.

[robatino]

That's much more elegant and efficient than mine. I didn't like having a separate loop.

[mike_g]

Hey thats cool I read your article about how this part works: bit=(-1U >> 1)+1

One thing I am unfamiliar with is the U by the number. Here it seems as if it treats an unsigned int as if it was signed. Is that what the U does?