Convert from base 10 to base 2 using bitwise operations

Question

As in title i'm trying to convert decimal number to binary. I set the lowest int as i coudl

char * toBinary(int num)
{

    int i = 1 << 31;
    while(i!=0)
    {
        if((num & i) == 0 )
            printf("0");
        else
            printf("1");
        i >>= 1;
        printf("%d", i);
        getchar();
    }
}

but it does not work, after every shift i is still negative number, what shoudl i change ?

I found same program explained here on YouTube - https://youtu.be/1MJ1o56x-g8 hope this will help you more than anything — Amit Upadhyay, Feb 08 '16 at 16:30

score 6 · Accepted Answer · answered Oct 08 '13 at 17:18

6

There are several errors both with your code and with your description of what it does. Here is a corrected version:

void print_binary(unsigned x)
{
    for (unsigned bit = 1u << 31; bit != 0; bit >>= 1)
        putchar((x & bit) ? '1' : '0');
}

Errors:

The function should not return char *, since it doesn't return anything. If you want to make a function that returns a string, that is different.
You cannot compute 1 << 31, because that is an overflow. You must use unsigned numbers: 1u << 31 is okay (assuming int is 32 bits).
You are not converting from decimal to binary. The input number is already in binary, you are simply printing it out in binary.

answered Oct 08 '13 at 17:18

Dietrich Epp

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i thought only difference between int and unsigned int is that last bit is sign bit, why i cant shift signed int like i did ? – whd Oct 08 '13 at 17:30
The problem is that when you shift a signed number to the right it "sign fills". An unsigned number is zero-filled. – Hot Licks Oct 08 '13 at 17:43
@whd Before coming to SO I was also unaware of this, similarly one more interesting thing is `~0` is undefined (or may be implementation) where as `~0u` is well defined!! – Grijesh Chauhan Oct 08 '13 at 17:59

score 3 · Answer 2 · edited May 23 '17 at 11:45

3

Use unsigned i.

int i = 1 << 31;

correct as:

unsigned int i = 1u << 31;

Additionally, instead of using 31 I would like to suggest you to write size independent code (size of int can be different in different machine).

Do like:

unsigned int mask = 1u << ((sizeof(unsigned int) * CHAR_BIT) - 1);

I have named mask instead i, write like this code.

//conversion 
unsigned int mask = 1u << ((sizeof(unsigned int) * CHAR_BIT) - 1);   
while(mask > 0){
  if((num & mask) == 0 )
       printf("0");
  else
       printf("1")
   mask = mask >> 1 ;  // Right Shift 
}

From @Carl Norum's comment: To correctly write a size platform independent code use CHAR_BIT macro defined in limits.h header file. Note CHAR_BIT is the number of bits in char and it is possible that in some implementation of C a byte may not be equals to 8 bits.

Read: What is CHAR_BIT?

edited May 23 '17 at 11:45

Community

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answered Oct 08 '13 at 17:18

Grijesh Chauhan

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`1 << 31` is undefined behavior anyway. – Dietrich Epp Oct 08 '13 at 17:18
@DietrichEpp Because of bit overflow ? – Grijesh Chauhan Oct 08 '13 at 17:20
@DietrichEpp Corrected now. – Grijesh Chauhan Oct 08 '13 at 17:25
2

If you want platform independent code you need to use `CHAR_BIT` instead of `8`, too. – Carl Norum Oct 08 '13 at 17:26
@CarlNorum I know in some implementation a byte may not be equals to 8 bit but I don't know `CHAR_BIT`. Which header file I need to include? – Grijesh Chauhan Oct 08 '13 at 17:29
@GrijeshChauhan - it's in `limits.h`. – Carl Norum Oct 08 '13 at 17:31
@CarlNorum Thanks Carl I should improve my answer. I was unaware of this Thanks. – Grijesh Chauhan Oct 08 '13 at 17:32
Is i>>1 when i=1 undefined behaviour, too?, or just zero? What if i=0? – zubergu Oct 08 '13 at 17:34
@zubergu No, also note `1 << 31` is undefined because `1` was singed but I corrected this as `1u << 31` from Dietrich Epp's answer. for unsigned number it is not undefined behavior. – Grijesh Chauhan Oct 08 '13 at 17:38
So right-shift guarantees that variable will be zero at some point(when variable is unsigned)? And going one step further also is guaranteed to result in 0, correct? – zubergu Oct 08 '13 at 17:40
@zubergu Yes `<<` is implementation dependent/undefined because of trap set. – Grijesh Chauhan Oct 08 '13 at 17:45
@zubergu I read sometime from [here](https://is.muni.cz/www/408176/38744863/International_Standard_for_C.txt) last time when I was answering. – Grijesh Chauhan Oct 08 '13 at 17:49

Convert from base 10 to base 2 using bitwise operations

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