I'm trying to make a binary to decimal converter and I wanted to use _getch() so the user doesn't have to press enter every number he enters, but…
#include <stdio.h>
#include <conio.h>
int main() {
int intnum[8], i = 0, ris;
char charnum;
for (i = 0; i < 8; i++) {
charnum = _getch();
printf("%c", charnum);
intnum[i] = (charnum - '0');
}
ris = intnum[8] * 2^0;
for (i = 7; i > 0; i--) {
ris += (intnum[i] * 2 ^ j);
j++;
}
printf("%d", ris);
getchar();
}
…but the problem is that when they enter in the loop the user can enter only 4 numbers and not 8; can you explain why?
After debugging of code,following changes you should do with your code..
declare int j = 0.
Also instead of ris = intnum[8] * 2^0 use ris = intnum[7] * 1 since array index start with 0 in c.
Also do same for for(i = 6;i >= 0;i--) and ris += intnum[i] * 2<<j with left shift operator(<<).
Basically use of getch() is unnecessary and makes your program
non-portable,but if you want to use it,then I will suggest you to use
below codes..
For linux,use this code..
#include <stdio.h>
#include <termios.h>
#include <unistd.h>
int getch (void){
int ch;
struct termios oldt, newt;
tcgetattr(STDIN_FILENO, &oldt);
newt = oldt;
newt.c_lflag &= ~(ICANON|ECHO);
tcsetattr(STDIN_FILENO, TCSANOW, &newt);
ch = getchar();
tcsetattr(STDIN_FILENO, TCSANOW, &oldt);
return ch;
}
int main() {
int intnum[8], i = 0, ris,j = 0;
char charnum;
for (i = 0; i < 8; i++) {
charnum = getch();
intnum[i] = (charnum-'0');
}
ris = intnum[7] * 1;
for (i = 6; i >= 0; i--) {
ris += (intnum[i] * 2<<j);
j++;
}
printf("%d", ris);
getchar();
}
For windows,use this code..
#include <stdio.h>
#include <conio.h>
int main() {
int intnum[8], i = 0, ris,j = 0;
char charnum;
for (i = 0; i < 8; i++) {
charnum = getch();
intnum[i] = (charnum-'0');
}
ris = intnum[7] * 1;
for (i = 6; i >= 0; i--) {
ris += (intnum[i] * 2<<j);
j++;
}
printf("%d", ris);
getchar();
}
I keep getting the error message that my I have an undefined reference to the power function, but I'm not really sure where that is occurring or why my code is coming up with that error because I have used to power function before in this way. If anyone could help me figure out why it isn't working now I would really appreciate it.
#include "stdio.h"
#include "string.h" //Needed for strlen()
#include "math.h"
#define MAX_BITS 32
#define MAX_LENGTH 49
#define NUMBER_TWO 2
#define NUMBER_ONE 1
#define TERMINATOR '\0'
//Code to find the index of where the string ends
int last_index_of(char in_str[], char ch) {
for (int i = 0; i < MAX_LENGTH; i++) {
if(in_str[i] == ch) {
last_index_of == i;
}
}
return last_index_of;
}
//Code to find the start of the fractional aspect
void sub_string(char in_str[], char out_str[], int start, int end){
int i = 0;
while (i < 1) {
out_str[i] = in_str[start] + in_str[end-1];
i++;
}
}
int main()
{
//Declaration of variable
char input[MAX_LENGTH +1]; // +1 for '\0'
int number;
double exponent;
char output[MAX_BITS];
int fraction;
sub_string(input, output, 0, TERMINATOR);
//Input from the user
printf("Enter a floating point value in binary: ");
scanf("%s", input);
//Calculates the Decimal Part
for (int i = 0; i < last_index_of(input, TERMINATOR) ; i++) {
number = number + number + input[i];
}
printf("%d", number);
exponent = -1;
//Calculates the Fractional Part
for (int j = 0; j < last_index_of(input, TERMINATOR); j++) {
if (j == last_index_of) {
fraction = NUMBER_ONE/(pow(NUMBER_TWO, exponent));
printf("%d/n", fraction);
}
else {
fraction = NUMBER_ONE/(pow(NUMBER_TWO, exponent));
printf("%d + ", fraction);
exponent--;
}
}
return 0;
}
Some problems:
you need -lm option to linker to tell it where to find pow function
last_index_of is not correctly written, you use the function name as an internal variable, you can correct it this way:
//Code to find the index of where the string ends
int last_index_of(char in_str[], char ch) {
int ret = 0;
for (int i = 0; i < MAX_LENGTH; i++) {
if(in_str[i] == ch) {
ret = i;
}
}
return ret;
}
Note that you can replace your last_index_of() function by strlen()
as pointed in comment, sub_string() is not functionnal. A corrected version could be:
//Code to find the start of the fractional aspect
void sub_string(char in_str[], char out_str[], int start, int end){
int i = 0;
while (start != end) {
/* warning, bounds are still not tested...*/
out_str[i++] = in_str[start++];
}
out_str[i] = '\0'
}
Instead of calling last_index_of() in your exist for loop condition, you should take its value to re-use it:
for (int j = 0; j < last_index_of(input, TERMINATOR); j++) {
/* Error here: will never be TRUE */
if (j == last_index_of) {
/* ... */
}
else {
/* ... */
}
}
would become:
int last_index = last_index_of(input, TERMINATOR);
for (int j = 0; j < last_index; j++) {
if (j == last_index) {
/* ... */
}
else {
/* ... */
}
}
Another problem, you use number variable without initializing it, you should write int number = 0 instead of int number;
After that, there is also a problem with your logic.
You have some idea of what you want to do, but it is not clear in your code.
It seems that you want
the user to input some string in the form 10010.100111
to split this string into two parts 10010 and 100111
to convert the first part into integer part 10010 -> 18
to convert the second part into fractional part 100111 -> 0.609...
This decomposition may lead you to write this kind of code:
#include "stdio.h"
#include "string.h"
#define MAX_BITS 32
#define MAX_LENGTH 49
//Code to find the index of where the string ends
int last_index_of(char in_str[], char ch)
{
int ret = 0;
for (int i = 0; i < MAX_LENGTH; i++) {
if (in_str[i] == ch) {
ret = i;
}
}
return ret;
}
void sub_string(char in_str[], char out_str[], int start, int end)
{
int i = 0;
while (start != end) {
/* warning, bounds are still not tested... */
out_str[i++] = in_str[start++];
}
out_str[i] = '\0';
}
void split(char *input, char *first, char *second)
{
int idx = last_index_of(input, '.');
sub_string(input, first, 0, idx);
sub_string(input, second, idx + 1, strlen(input));
}
int main()
{
//Declaration of variable
char input[MAX_LENGTH + 1]; // +1 for '\0'
char first[MAX_BITS];
char second[MAX_BITS];
/* Input from the user */
printf("Enter a floating point value in binary: ");
scanf("%s", input);
/* split integer and fractionnal parts */
split(input, first, second);
/* decode integer part */
printf("integer part:\n");
for (int i = strlen(first) - 1, j = 1; i > -1; --i, j <<= 1) {
if (first[i] == '1') {
printf("%d ", j);
}
}
/* decode frac part */
printf("\nfractionnal part:\n");
for (int i = 0; i < strlen(second); ++i) {
if (second[i] == '1') {
printf("1/%d ", 2 << i);
}
}
return 0;
}
I'm trying to convert an ascii string to a binary string in C. I found this example Converting Ascii to binary in C but I rather not use a recursive function. I tried to write an iterative function as opposed to a recursive function, but the binary string is missing the leading digit. I'm using itoa to convert the string, however itoa is a non standard function so I used the implementation from What is the proper way of implementing a good "itoa()" function? , the one provided by Minh Nguyen.
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
int32_t ascii_to_binary(char *input, char **out, uint64_t len)
{
uint32_t i;
uint32_t str_len = len * 8;
if(len == 0)
{
printf("Length argument is zero\n");
return (-1);
}
(*out) = malloc(str_len + 1);
if((*out) == NULL)
{
printf("Can't allocate binary string: %s\n", strerror(errno));
return (-1);
}
if(memset((*out), 0, (str_len)) == NULL)
{
printf("Can't initialize memory to zero: %s\n", strerror(errno));
return (-1);
}
for(i = 0; i < len; i++)
itoa((int32_t)input[i], &(*out)[(i * 8)], 2);
(*out)[str_len] = '\0';
return (str_len);
}
int main(void)
{
int32_t rtrn = 0;
char *buffer = NULL;
rtrn = ascii_to_binary("a", &buffer, 1);
if(rtrn < 0)
{
printf("Can't convert string\n");
return (-1);
}
printf("str: %s\n", buffer);
return (0);
}
I get 1100001 for ascii character a, but I should get 01100001, so how do I convert the ascii string to the whole binary string?
You could change the for loop to something like this:
for(i = 0; i < len; i++) {
unsigned char ch = input[i];
char *o = *out + 8 * i;
int b;
for (b = 7; b >= 0; b--)
*o++ = (ch & (1 << b)) ? '1' : '0';
}
or similar:
for(i = 0; i < len; i++) {
unsigned char ch = input[i];
char *o = &(*out)[8 * i];
unsigned char b;
for (b = 0x80; b; b >>= 1)
*o++ = ch & b ? '1' : '0';
}
This program gets and integer ( which contains 32 bits ) and converts it to binary, Work on it to get it work for ascii strings :
#include <stdio.h>
int main()
{
int n, c, k;
printf("Enter an integer in decimal number system\n");
scanf("%d", &n);
printf("%d in binary number system is:\n", n);
for (c = 31; c >= 0; c--)
{
k = n >> c;
if (k & 1)
printf("1");
else
printf("0");
}
printf("\n");
return 0;
}
Best just write a simple function to do this using bitwise operators...
#define ON_BIT = 0x01
char *strToBin(char c) {
static char strOutput[10];
int bit;
/*Shifting bits to the right, but don't want the output to be in reverse
* so indexing bytes with this...
*/
int byte;
/* Add a nul at byte 9 to terminate. */
strOutput[8] = '\0';
for (bit = 0, byte = 7; bit < 8; bit++, byte--) {
/* Shifting the bits in c to the right, each time and'ing it with
* 0x01 (00000001).
*/
if ((c >> bit) & BIT_ON)
/* We know this is a 1. */
strOutput[byte] = '1';
else
strOutput[byte] = '0';
}
return strOutput;
}
Something like that should work, there's loads of ways you can do it. Hope this helps.
this is my first time posting, so please be patient with
me.. I need help with my Code.. i am very unexperienced. I want to write
the CS_UID (the unique ID of my microcontroller) to UART, using this
function:
void appWriteDataToUart(uint8_t* aData, uint8_t aLength);
Therefore I think I need to convert the hexadecimal CS_UID (64bit long)
which is defined as follows for example:
#define CS_UID 0x1234567890abcdff
into something that makes sense and is possible to be written to UART.
Someone told me this would help:
#include <stdio.h>
#include <string.h>
int hex_to_int(char c) {
int first = c / 16 - 3;
int second = c % 16;
int result = first*10 + second;
if(result > 9) result--;
return result;
}
int hex_to_ascii(char c, char d) {
int high = hex_to_int(c) * 16;
int low = hex_to_int(d);
return high+low;
}
int main() {
const char* st = "48656C6C6F3B";
int length = strlen(st);
int i;
char buf = 0;
for(i = 0; i < length; i++){
if(i % 2 != 0){
printf("%c", hex_to_ascii(buf, st[i]));
}else{
buf = st[i];
}
}
}
But I don't know how to use that. I'm clueless. These functions return
int and use printf and %c.
I tried it like this but it doesn't work:
#include <stdio.h>
#include <string.h>
#define CS_UID 0x1234567890abcdff
int hex_to_int(char c) {
int first = c / 16 - 3;
int second = c % 16;
int result = first*10 + second;
if(result > 9) result--;
return result;
}
int hex_to_ascii(char c, char d) {
int high = hex_to_int(c) * 16;
int low = hex_to_int(d);
return high+low;
}
int main (void) {
char st[16] = CS_UID;
char csuid_array[8]; //this is the array i want to write to UART later? right or wrong?
int length = strlen(st);
int j;
char buf = 0;
for (j=0; j < length; j++){
if(j % 2 != 0){
csuid_array[j] = hex_to_ascii(buf,st[j]);
}
else{
buf = st[j];
}
}
return 0;
}
i'd appreciate any help or other solution!
Thanks in advance.
I'll assume that you want to send the ASCII representation of the value so that it is human readable on a terminal program. printf() (and sprintf()) can do the conversion for you, so you don't need those other "hex" routines.
#define CS_UID 0x1234567890abcdffU
uint8_t csuid_array[19]; // extra bytes for "0x" and NULL terminator
sprintf((char*)csuid_array, "0x%016llX", CSUID); // that's two letter 'l' between '%016' and 'X'
appWriteDataToUart(csuid_array, 18); // extra bytes for "0x"
I'm looking for a function to allow me to print the binary representation of an int. What I have so far is;
char *int2bin(int a)
{
char *str,*tmp;
int cnt = 31;
str = (char *) malloc(33); /*32 + 1 , because its a 32 bit bin number*/
tmp = str;
while ( cnt > -1 ){
str[cnt]= '0';
cnt --;
}
cnt = 31;
while (a > 0){
if (a%2==1){
str[cnt] = '1';
}
cnt--;
a = a/2 ;
}
return tmp;
}
But when I call
printf("a %s",int2bin(aMask)) // aMask = 0xFF000000
I get output like;
0000000000000000000000000000000000xtpYy (And a bunch of unknown characters.
Is it a flaw in the function or am I printing the address of the character array or something? Sorry, I just can't see where I'm going wrong.
NB The code is from here
EDIT: It's not homework FYI, I'm trying to debug someone else's image manipulation routines in an unfamiliar language. If however it's been tagged as homework because it's an elementary concept then fair play.
Here's another option that is more optimized where you pass in your allocated buffer. Make sure it's the correct size.
// buffer must have length >= sizeof(int) + 1
// Write to the buffer backwards so that the binary representation
// is in the correct order i.e. the LSB is on the far right
// instead of the far left of the printed string
char *int2bin(int a, char *buffer, int buf_size) {
buffer += (buf_size - 1);
for (int i = 31; i >= 0; i--) {
*buffer-- = (a & 1) + '0';
a >>= 1;
}
return buffer;
}
#define BUF_SIZE 33
int main() {
char buffer[BUF_SIZE];
buffer[BUF_SIZE - 1] = '\0';
int2bin(0xFF000000, buffer, BUF_SIZE - 1);
printf("a = %s", buffer);
}
A few suggestions:
null-terminate your string
don't use magic numbers
check the return value of malloc()
don't cast the return value of malloc()
use binary operations instead of arithmetic ones as you're interested in the binary representation
there's no need for looping twice
Here's the code:
#include <stdlib.h>
#include <limits.h>
char * int2bin(int i)
{
size_t bits = sizeof(int) * CHAR_BIT;
char * str = malloc(bits + 1);
if(!str) return NULL;
str[bits] = 0;
// type punning because signed shift is implementation-defined
unsigned u = *(unsigned *)&i;
for(; bits--; u >>= 1)
str[bits] = u & 1 ? '1' : '0';
return str;
}
Your string isn't null-terminated. Make sure you add a '\0' character at the end of the string; or, you could allocate it with calloc instead of malloc, which will zero the memory that is returned to you.
By the way, there are other problems with this code:
As used, it allocates memory when you call it, leaving the caller responsible for free()ing the allocated string. You'll leak memory if you just call it in a printf call.
It makes two passes over the number, which is unnecessary. You can do everything in one loop.
Here's an alternative implementation you could use.
#include <stdlib.h>
#include <limits.h>
char *int2bin(unsigned n, char *buf)
{
#define BITS (sizeof(n) * CHAR_BIT)
static char static_buf[BITS + 1];
int i;
if (buf == NULL)
buf = static_buf;
for (i = BITS - 1; i >= 0; --i) {
buf[i] = (n & 1) ? '1' : '0';
n >>= 1;
}
buf[BITS] = '\0';
return buf;
#undef BITS
}
Usage:
printf("%s\n", int2bin(0xFF00000000, NULL));
The second parameter is a pointer to a buffer you want to store the result string in. If you don't have a buffer you can pass NULL and int2bin will write to a static buffer and return that to you. The advantage of this over the original implementation is that the caller doesn't have to worry about free()ing the string that gets returned.
A downside is that there's only one static buffer so subsequent calls will overwrite the results from previous calls. You couldn't save the results from multiple calls for later use. Also, it is not threadsafe, meaning if you call the function this way from different threads they could clobber each other's strings. If that's a possibility you'll need to pass in your own buffer instead of passing NULL, like so:
char str[33];
int2bin(0xDEADBEEF, str);
puts(str);
Here is a simple algorithm.
void decimalToBinary (int num) {
//Initialize mask
unsigned int mask = 0x80000000;
size_t bits = sizeof(num) * CHAR_BIT;
for (int count = 0 ;count < bits; count++) {
//print
(mask & num ) ? cout <<"1" : cout <<"0";
//shift one to the right
mask = mask >> 1;
}
}
this is what i made to display an interger as a binairy code it is separated per 4 bits:
int getal = 32; /** To determain the value of a bit 2^i , intergers are 32bits long**/
int binairy[getal]; /** A interger array to put the bits in **/
int i; /** Used in the for loop **/
for(i = 0; i < 32; i++)
{
binairy[i] = (integer >> (getal - i) - 1) & 1;
}
int a , counter = 0;
for(a = 0;a<32;a++)
{
if (counter == 4)
{
counter = 0;
printf(" ");
}
printf("%i", binairy[a]);
teller++;
}
it could be a bit big but i always write it in a way (i hope) that everyone can understand what is going on. hope this helped.
#include<stdio.h>
//#include<conio.h> // use this if you are running your code in visual c++, linux don't
// have this library. i have used it for getch() to hold the screen for input char.
void showbits(int);
int main()
{
int no;
printf("\nEnter number to convert in binary\n");
scanf("%d",&no);
showbits(no);
// getch(); // used to hold screen...
// keep code as it is if using gcc. if using windows uncomment #include & getch()
return 0;
}
void 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");
}
}
Just a enhance of the answer from #Adam Markowitz
To let the function support uint8 uint16 uint32 and uint64:
#include <inttypes.h>
#include <stdint.h>
#include <stdio.h>
#include <string.h>
// Convert integer number to binary representation.
// The buffer must have bits bytes length.
void int2bin(uint64_t number, uint8_t *buffer, int bits) {
memset(buffer, '0', bits);
buffer += bits - 1;
for (int i = bits - 1; i >= 0; i--) {
*buffer-- = (number & 1) + '0';
number >>= 1;
}
}
int main(int argc, char *argv[]) {
char buffer[65];
buffer[8] = '\0';
int2bin(1234567890123, buffer, 8);
printf("1234567890123 in 8 bits: %s\n", buffer);
buffer[16] = '\0';
int2bin(1234567890123, buffer, 16);
printf("1234567890123 in 16 bits: %s\n", buffer);
buffer[32] = '\0';
int2bin(1234567890123, buffer, 32);
printf("1234567890123 in 32 bits: %s\n", buffer);
buffer[64] = '\0';
int2bin(1234567890123, buffer, 64);
printf("1234567890123 in 64 bits: %s\n", buffer);
return 0;
}
The output:
1234567890123 in 8 bits: 11001011
1234567890123 in 16 bits: 0000010011001011
1234567890123 in 32 bits: 01110001111110110000010011001011
1234567890123 in 64 bits: 0000000000000000000000010001111101110001111110110000010011001011
Two things:
Where do you put the NUL character? I can't see a place where '\0' is set.
Int is signed, and 0xFF000000 would be interpreted as a negative value. So while (a > 0) will be false immediately.
Aside: The malloc function inside is ugly. What about providing a buffer to int2bin?
A couple of things:
int f = 32;
int i = 1;
do{
str[--f] = i^a?'1':'0';
}while(i<<1);
It's highly platform dependent, but
maybe this idea above gets you started.
Why not use memset(str, 0, 33) to set
the whole char array to 0?
Don't forget to free()!!! the char*
array after your function call!
Two simple versions coded here (reproduced with mild reformatting).
#include <stdio.h>
/* Print n as a binary number */
void printbitssimple(int n)
{
unsigned int i;
i = 1<<(sizeof(n) * 8 - 1);
while (i > 0)
{
if (n & i)
printf("1");
else
printf("0");
i >>= 1;
}
}
/* Print n as a binary number */
void printbits(int n)
{
unsigned int i, step;
if (0 == n) /* For simplicity's sake, I treat 0 as a special case*/
{
printf("0000");
return;
}
i = 1<<(sizeof(n) * 8 - 1);
step = -1; /* Only print the relevant digits */
step >>= 4; /* In groups of 4 */
while (step >= n)
{
i >>= 4;
step >>= 4;
}
/* At this point, i is the smallest power of two larger or equal to n */
while (i > 0)
{
if (n & i)
printf("1");
else
printf("0");
i >>= 1;
}
}
int main(int argc, char *argv[])
{
int i;
for (i = 0; i < 32; ++i)
{
printf("%d = ", i);
//printbitssimple(i);
printbits(i);
printf("\n");
}
return 0;
}
//This is what i did when our teacher asked us to do this
int main (int argc, char *argv[]) {
int number, i, size, mask; // our input,the counter,sizeofint,out mask
size = sizeof(int);
mask = 1<<(size*8-1);
printf("Enter integer: ");
scanf("%d", &number);
printf("Integer is :\t%d 0x%X\n", number, number);
printf("Bin format :\t");
for(i=0 ; i<size*8 ;++i ) {
if ((i % 4 == 0) && (i != 0)) {
printf(" ");
}
printf("%u",number&mask ? 1 : 0);
number = number<<1;
}
printf("\n");
return (0);
}
the simplest way for me doing this (for a 8bit representation):
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
char *intToBinary(int z, int bit_length){
int div;
int counter = 0;
int counter_length = (int)pow(2, bit_length);
char *bin_str = calloc(bit_length, sizeof(char));
for (int i=counter_length; i > 1; i=i/2, counter++) {
div = z % i;
div = div / (i / 2);
sprintf(&bin_str[counter], "%i", div);
}
return bin_str;
}
int main(int argc, const char * argv[]) {
for (int i = 0; i < 256; i++) {
printf("%s\n", intToBinary(i, 8)); //8bit but you could do 16 bit as well
}
return 0;
}
Here is another solution that does not require a char *.
#include <stdio.h>
#include <stdlib.h>
void print_int(int i)
{
int j = -1;
while (++j < 32)
putchar(i & (1 << j) ? '1' : '0');
putchar('\n');
}
int main(void)
{
int i = -1;
while (i < 6)
print_int(i++);
return (0);
}
Or here for more readability:
#define GRN "\x1B[32;1m"
#define NRM "\x1B[0m"
void print_int(int i)
{
int j = -1;
while (++j < 32)
{
if (i & (1 << j))
printf(GRN "1");
else
printf(NRM "0");
}
putchar('\n');
}
And here is the output:
11111111111111111111111111111111
00000000000000000000000000000000
10000000000000000000000000000000
01000000000000000000000000000000
11000000000000000000000000000000
00100000000000000000000000000000
10100000000000000000000000000000
#include <stdio.h>
#define BITS_SIZE 8
void
int2Bin ( int a )
{
int i = BITS_SIZE - 1;
/*
* Tests each bit and prints; starts with
* the MSB
*/
for ( i; i >= 0; i-- )
{
( a & 1 << i ) ? printf ( "1" ) : printf ( "0" );
}
return;
}
int
main ()
{
int d = 5;
printf ( "Decinal: %d\n", d );
printf ( "Binary: " );
int2Bin ( d );
printf ( "\n" );
return 0;
}
Not so elegant, but accomplishes your goal and it is very easy to understand:
#include<stdio.h>
int binario(int x, int bits)
{
int matriz[bits];
int resto=0,i=0;
float rest =0.0 ;
for(int i=0;i<8;i++)
{
resto = x/2;
rest = x%2;
x = resto;
if (rest>0)
{
matriz[i]=1;
}
else matriz[i]=0;
}
for(int j=bits-1;j>=0;j--)
{
printf("%d",matriz[j]);
}
printf("\n");
}
int main()
{
int num,bits;
bits = 8;
for (int i = 0; i < 256; i++)
{
num = binario(i,bits);
}
return 0;
}
#include <stdio.h>
int main(void) {
int a,i,k=1;
int arr[32]; \\ taken an array of size 32
for(i=0;i <32;i++)
{
arr[i] = 0; \\initialised array elements to zero
}
printf("enter a number\n");
scanf("%d",&a); \\get input from the user
for(i = 0;i < 32 ;i++)
{
if(a&k) \\bit wise and operation
{
arr[i]=1;
}
else
{
arr[i]=0;
}
k = k<<1; \\left shift by one place evry time
}
for(i = 31 ;i >= 0;i--)
{
printf("%d",arr[i]); \\print the array in reverse
}
return 0;
}
void print_binary(int n) {
if (n == 0 || n ==1)
cout << n;
else {
print_binary(n >> 1);
cout << (n & 0x1);
}
}