The showbits() function

Question

While reading a book called "Let us C" I read that a function showbit() exists which can show you the bits of the number. There wasn't any special header file mentioned for it. Searched for it on the internet and didn't found anything useful. Is there such a function? I want this to print the binary of decimal numbers. Else please give me a replacement function. Thanks

Answer 1

All integers are actually in binary in your computer. Its just that it is turned into a string that is the decimal representation of that value when you try to print it using printf and "%d". If you want to know what it looks like in some other base (e.g. base 2 or binary), you either have to provide the proper printf format string if it exists (e.g. "%x" for hex) or just build that string yourself and print it out.

Here is some code that can build the string representation of an integer in any base in [2,36].

#include <stdio.h>
#include <string.h>

char digits[]="01234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ";

void reverse(char* start, char* end)
{
    for(end--;start<end;start++,end--)
    {
        char t=*start;
        *start=*end;
        *end=t;
    }
}


int base_change(int n, int base,char* buffer)
{
    int pos=0;
    if (base>strlen(digits)) 
        return -1;
    while(n)
    {
        buffer[pos]=digits[n%base];
        n/=base;
        pos++;
    }
    buffer[pos]='\0';
    reverse(buffer,buffer+pos);
    return 0;
}

int main()
{
    char buffer[32];
    int conv=base_change(1024,2,buffer);
    if (conv==0) printf("%s\n",buffer);
    return 0;
}

Answer 2

You can also try this snippet which uses bit-shifting:

EDIT: (I've made it more portable)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define BITS_IN_BYTE 8
#define INTEGRAL_TYPE unsigned int

void showBits(INTEGRAL_TYPE x) {
    int i;
    static int intSizeInBits = sizeof(INTEGRAL_TYPE) * BITS_IN_BYTE;
    static char symbol[2] = {'0','1'};
    char * binary = (char*) malloc(intSizeInBits + 1);

    memset(binary, 0, intSizeInBits + 1);

    for (i=0; i< intSizeInBits; i++) {
        binary[intSizeInBits-i-1] = symbol[(x>>i) & 0x01];
    }

    printf("%s\n", binary);
    free(binary);
}

int main() {
    showBits(8698513);
    return 0;
}

HTH!

Answer 3

This is a very simple solution for printing the bits of an integer

int value = 14;
int n;
for (n=8*sizeof(int)-1;n>=0;n--) {
    printf("%d",(value >>n)&1);
}

Answer 4

The book "Let Us C" doesn't define it as a built-in function. The showbits() function is defined in later part of the book as a complete user defined function as I personally went through it. This answer is to anyone who haven't reached that later part of the book and are stuck at the first appearance of the function in the book.

Answer 5

this is the header file for showbits

 void showbits(unsigned int x)
    {
        int i;
        for(i=(sizeof(int)*5)-1; i>=0; i--)
            (x&(1u<<i))?putchar('1'):putchar('0');
        printf("\n");
    }

Answer 6

you need to look here

@downvoter It works fine in c also. You just need to reformat your code in c-style.

#include <stdlib.h>
#include <stdio.h>
int main()
{
char buffer[20];
int  i = 3445;
_itoa( i, buffer, 2  );
printf("String of integer %d (radix 2): %s",i,buffer);
return 0;
}

*you need to save your file as .c in MSVC++ for _itoa() to work.*

Answer 7

No there is no pre built function and you do not need to include any specific header file. You will have to provide implementation for function showbits(int).

#include <stdio.h>
void showbits(int);
int main()
{
    int j,k;
    for(j=0;j<=11;j++)
    {
        printf("\nBinary value of decimal %d is :",j);
        showbits(j);

    }
    return 0;
}
showbits(int n){
    int i,k,andmask;
    for(i = 15;i>=0;i--){
        andmask = 1<<i;
        k = n & andmask;
        k==0 ? printf("0"):printf("1");
    }
}

Answer 8

If you want to print out the bits of a float, for example you could do something like:

float myFloat = 45.2;
for (int n=0;n<8*sizeof(float);n++) {
    printf("%d",(myFloat>>n)&1);
}