Related
I am working on code to reverse an unsigned int whose bits are the same but in reverse order. My code wont stop running while taking user input. What am I doing wrong?
#include <stdio.h>
unsigned int reverse_bits(unsigned int n);
int main(void) {
unsigned int n;
printf("Enter an unsigned integer: ");
scanf("%u",&n);
printf("%u\n",reverse_bits(n));
return 0;
}
unsigned int reverse_bits(unsigned int n) {
unsigned int reverse = 0;
while(n>0) {
reverse <<= 1;
if((n & 1) == 1) {
reverse = reverse^1;
}
}
return reverse;
}
In your program, the while condition is checking whether n is greater than zero. However, inside the loop, there was no operation that modifies the n. So please try using this. It will work.
unsigned int reverse_bits(unsigned int n) {
unsigned int reverse = 0;
while (n>0) {
reverse <<=1;
if((n & 1) == 1) {
reverse = reverse | 1;
}
n=n>>1;
}
return reverse;
}
as #Mark Benningfield said you forget to modify n inside the loop and the
best way to reverse a string is:-
It only runs upto when the leftmost bit is 1.
unsigned int reverse_bits(unsigned int n)
{
unsigned int reverse = 0;
int x = 1;
while ( n > 0 )
{
reverse = reverse << 1;
if ( n & x )
reverese = reverse | 1;
n = n >> 1;
}
return reverse;
Usually bit operations are done in smaller data width such as int, unsigned int or wchar_t. Assuming we want to use the bit strings in a longer format, how to shift, get and set bits for bits in char string?
One way may be to divide and conquer using the conventional method, but how do we ensure the bit carry over?
Given
#define numberOfState 2000 // number of bits
#define numberOfBitsIn1Byte 8
char* record;
int numberOfCharRequiredToRepresentBits =
ceil(((float)numberOfState/(float)numberOfBitsIn1Byte));
record = (char*) malloc(sizeof(char)*numberOfCharRequiredToRepresentBits);
// record = "NAXHDKAN552ajdasdadNDfadsEBEAfA8gda5214S";
// optional : initialization by doing the set bit according to
// input from files. After which, do free(record);
How may we conduct bit operations such as to
i. shift the *record
ii. get bits from a specific bit position in *record
iii. set bits from a specific bit position in *record
Please have a try with following code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int isLittleEndian = 1;
void checkEndian(void)
{
union
{
short inum;
char c[sizeof(short)];
} un;
un.inum=0x0102;
if(un.c[0]==1 && un.c[1]==2)
{
printf("big_endian.\n");
isLittleEndian = 0;
}
else if(un.c[0]==2 && un.c[1]==1)
{
printf("little_endian.\n");
isLittleEndian = 1;
}
}
void shift_L(char *src, char * dst, int len, int n)
{
int shiftBytes = n/8;
int shiftBits = n%8;
memset(dst, 0, len);
memcpy(dst, src + shiftBytes, len - shiftBytes);
if (shiftBits)
{
int i = 0;
unsigned short tmp = 0;
for ( i = 0; i < len; i++)
{
if (isLittleEndian)
{
tmp = *(dst+i) << 8 | *(dst+i+1);
tmp <<= shiftBits;
*(dst+i) = *((char *)&tmp + 1);
}
else
{
tmp = *(short *)(dst+i);
tmp <<= shiftBits;
*(dst+i) = *((char *)&tmp);
}
}
}
}
void shift_R(char *src, char * dst, int len, int n)
{
int shiftBytes = n/8;
int shiftBits = n%8;
memset(dst, 0, len);
memcpy(dst + shiftBytes, src, len - shiftBytes);
if (shiftBits)
{
int i = 0;
unsigned short tmp = 0;
for ( i = len -1; i >= 0; i--)
{
if (isLittleEndian)
{
tmp = *(dst+i-1) << 8 | *(dst+i);
tmp >>= shiftBits;
*(dst+i) = *((char *)&tmp);
}
else
{
tmp = *(short *)(dst+i-1);
tmp >>= shiftBits;
*(dst+i) = *((char *)&tmp+1);
}
}
}
}
int getBit(char *src, int n)
{
unsigned char tmp = *(src + n/8);
unsigned char mask = (0x1 << (8 - n%8 - 1));
int bit = 0;
bit = (tmp & mask) > 0;
printf("%d", bit);
}
void setBit(char *src, int n, int bit)
{
unsigned char * pTmp = src + n/8;
unsigned char mask = (0x1 << (8 - n%8 - 1));
if (bit)
{
*pTmp |= mask;
}
else
{
*pTmp &= ~mask;
}
}
void dumpBin(unsigned char *src, int len)
{
int i = 0;
int j = 0;
unsigned char mask = 0;
for ( i = 0; i < len; i++)
{
for ( j = 0; j < 8; j++)
{
mask = 0x1 << 8 - j - 1;
printf("%d",(*(src + i) & mask) > 0);
}
}
}
void main()
{
char *record = "NAXHDKAN552ajdasdadNDfadsEBEAfA8gda5214S";
//char *record = "NAXHDKA";
int recordLen = strlen(record);
char * buffer = NULL;
int i = 0;
checkEndian();
recordLen = recordLen + recordLen%2;
buffer = malloc(recordLen);
memcpy(buffer, record, recordLen);
printf("\n input bit stream:\n");
dumpBin(buffer, recordLen);
printf("\n bit stream from getBit:\n");
for ( i = 0; i < recordLen*8; i++)
{
getBit(buffer, i);
}
setBit(buffer, 8, 1);
setBit(buffer, 9, 0);
setBit(buffer, 10, 1);
setBit(buffer, 11, 1);
printf("\n bit stream after setBit:\n");
dumpBin(buffer, recordLen);
shift_L(record, buffer, recordLen, 1);
printf("\n bit stream after shift_L:\n");
dumpBin(buffer, recordLen);
shift_R(record, buffer, recordLen, 9);
printf("\n bit stream after shift_R:\n");
dumpBin(buffer, recordLen);
printf("\n");
free(buffer);
}
Your bitstream is essentially an array of char. So, to perform these operations you work on these char elements.
i. The shifting operation depends on the number of bits you want to shift.
If the number is a multiple of 8, it is pretty straightforward, you just copy the elements right of left as many bytes as the number is a multiple of 8.
If the number is less than 8, you perform the operation on every element of the array, but you need to or the overflowing bits of the previous element. For example, in left shift, element i must incorporate the overflowing bits of element i+1, and on right shift, the overflowing bits of element i-1.
Any other number of bits you want to shift can be achieved by a combination of these two actions. For example, a left shift by 18 is a shift by 16 followed by a shift by 2.
In any case, you need to be careful on which side of the bitstring you start, so that you do not lose data.
ii. In order to get the n-th bit of the bitstream, you access the element with index n/8 (integer division) and get the n%8 bit from it.
iii. Pretty much the same as ii.
I'm trying to convert an integer 10 into the binary number 1010.
This code attempts it, but I get a segfault on the strcat():
int int_to_bin(int k)
{
char *bin;
bin = (char *)malloc(sizeof(char));
while(k>0) {
strcat(bin, k%2);
k = k/2;
bin = (char *)realloc(bin, sizeof(char) * (sizeof(bin)+1));
}
bin[sizeof(bin)-1] = '\0';
return atoi(bin);
}
How do I convert an integer to binary in C?
If you want to transform a number into another number (not number to string of characters), and you can do with a small range (0 to 1023 for implementations with 32-bit integers), you don't need to add char* to the solution
unsigned int_to_int(unsigned k) {
if (k == 0) return 0;
if (k == 1) return 1; /* optional */
return (k % 2) + 10 * int_to_int(k / 2);
}
HalosGhost suggested to compact the code into a single line
unsigned int int_to_int(unsigned int k) {
return (k == 0 || k == 1 ? k : ((k % 2) + 10 * int_to_int(k / 2)));
}
You need to initialise bin, e.g.
bin = malloc(1);
bin[0] = '\0';
or use calloc:
bin = calloc(1, 1);
You also have a bug here:
bin = (char *)realloc(bin, sizeof(char) * (sizeof(bin)+1));
this needs to be:
bin = (char *)realloc(bin, sizeof(char) * (strlen(bin)+1));
(i.e. use strlen, not sizeof).
And you should increase the size before calling strcat.
And you're not freeing bin, so you have a memory leak.
And you need to convert 0, 1 to '0', '1'.
And you can't strcat a char to a string.
So apart from that, it's close, but the code should probably be more like this (warning, untested !):
int int_to_bin(int k)
{
char *bin;
int tmp;
bin = calloc(1, 1);
while (k > 0)
{
bin = realloc(bin, strlen(bin) + 2);
bin[strlen(bin) - 1] = (k % 2) + '0';
bin[strlen(bin)] = '\0';
k = k / 2;
}
tmp = atoi(bin);
free(bin);
return tmp;
}
Just use itoa to convert to a string, then use atoi to convert back to decimal.
unsigned int_to_int(unsigned int k) {
char buffer[65]; /* any number higher than sizeof(unsigned int)*bits_per_byte(8) */
return atoi( itoa(k, buffer, 2) );
}
The working solution for Integer number to binary conversion is below.
int main()
{
int num=241; //Assuming 16 bit integer
for(int i=15; i>=0; i--) cout<<((num >> i) & 1);
cout<<endl;
for(int i=0; i<16; i++) cout<<((num >> i) & 1);
cout<<endl;
return 0;
}
You can capture the cout<< part based on your own requirement.
Well, I had the same trouble ... so I found this thread
I think the answer from user:"pmg" does not work always.
unsigned int int_to_int(unsigned int k) {
return (k == 0 || k == 1 ? k : ((k % 2) + 10 * int_to_int(k / 2)));
}
Reason: the binary representation is stored as an integer. That is quite limited.
Imagine converting a decimal to binary:
dec 255 -> hex 0xFF -> bin 0b1111_1111
dec 1023 -> hex 0x3FF -> bin 0b11_1111_1111
and you have to store this binary representation as it were a decimal number.
I think the solution from Andy Finkenstadt is the closest to what you need
unsigned int_to_int(unsigned int k) {
char buffer[65]; // any number higher than sizeof(unsigned int)*bits_per_byte(8)
return itoa( atoi(k, buffer, 2) );
}
but still this does not work for large numbers.
No suprise, since you probably don't really need to convert the string back to decimal. It makes less sense. If you need a binary number usually you need for a text somewhere, so leave it in string format.
simply use itoa()
char buffer[65];
itoa(k, buffer, 2);
You can use function this function to return char* with string representation of the integer:
char* itob(int i) {
static char bits[8] = {'0','0','0','0','0','0','0','0'};
int bits_index = 7;
while ( i > 0 ) {
bits[bits_index--] = (i & 1) + '0';
i = ( i >> 1);
}
return bits;
}
It's not a perfect implementation, but if you test with a simple printf("%s", itob(170)), you'll get 01010101 as I recall 170 was. Add atoi(itob(170)) and you'll get the integer but it's definitely not 170 in integer value.
You could use this function to get array of bits from integer.
int* num_to_bit(int a, int *len){
int arrayLen=0,i=1;
while (i<a){
arrayLen++;
i*=2;
}
*len=arrayLen;
int *bits;
bits=(int*)malloc(arrayLen*sizeof(int));
arrayLen--;
while(a>0){
bits[arrayLen--]=a&1;
a>>=1;
}
return bits;
}
void intToBin(int digit) {
int b;
int k = 0;
char *bits;
bits= (char *) malloc(sizeof(char));
printf("intToBin\n");
while (digit) {
b = digit % 2;
digit = digit / 2;
bits[k] = b;
k++;
printf("%d", b);
}
printf("\n");
for (int i = k - 1; i >= 0; i--) {
printf("%d", bits[i]);
}
}
You can convert decimal to bin, hexa to decimal, hexa to bin, vice-versa etc by following this example.
CONVERTING DECIMAL TO BIN
int convert_to_bin(int number){
int binary = 0, counter = 0;
while(number > 0){
int remainder = number % 2;
number /= 2;
binary += pow(10, counter) * remainder;
counter++;
}
}
Then you can print binary equivalent like this:
printf("08%d", convert_to_bin(13)); //shows leading zeros
Result in string
The following function converts an integer to binary in a string (n is the number of bits):
// Convert an integer to binary (in a string)
void int2bin(unsigned integer, char* binary, int n=8)
{
for (int i=0;i<n;i++)
binary[i] = (integer & (int)1<<(n-i-1)) ? '1' : '0';
binary[n]='\0';
}
Test online on repl.it.
Source : AnsWiki.
Result in string with memory allocation
The following function converts an integer to binary in a string and allocate memory for the string (n is the number of bits):
// Convert an integer to binary (in a string)
char* int2bin(unsigned integer, int n=8)
{
char* binary = (char*)malloc(n+1);
for (int i=0;i<n;i++)
binary[i] = (integer & (int)1<<(n-i-1)) ? '1' : '0';
binary[n]='\0';
return binary;
}
This option allows you to write something like printf ("%s", int2bin(78)); but be careful, memory allocated for the string must be free later.
Test online on repl.it.
Source : AnsWiki.
Result in unsigned int
The following function converts an integer to binary in another integer (8 bits maximum):
// Convert an integer to binary (in an unsigned)
unsigned int int_to_int(unsigned int k) {
return (k == 0 || k == 1 ? k : ((k % 2) + 10 * int_to_int(k / 2)));
}
Test online on repl.it
Display result
The following function displays the binary conversion
// Convert an integer to binary and display the result
void int2bin(unsigned integer, int n=8)
{
for (int i=0;i<n;i++)
putchar ( (integer & (int)1<<(n-i-1)) ? '1' : '0' );
}
Test online on repl.it.
Source : AnsWiki.
You can add the functions to the standard library and use it whenever you need.
Here is the code in C++
#include <stdio.h>
int power(int x, int y) //calculates x^y.
{
int product = 1;
for (int i = 0; i < y; i++)
{
product = product * x;
}
return (product);
}
int gpow_bin(int a) //highest power of 2 less/equal to than number itself.
{
int i, z, t;
for (i = 0;; i++)
{
t = power(2, i);
z = a / t;
if (z == 0)
{
break;
}
}
return (i - 1);
}
void bin_write(int x)
{
//printf("%d", 1);
int current_power = gpow_bin(x);
int left = x - power(2, current_power);
int lower_power = gpow_bin(left);
for (int i = 1; i < current_power - lower_power; i++)
{
printf("0");
}
if (left != 0)
{
printf("%d", 1);
bin_write(left);
}
}
void main()
{
//printf("%d", gpow_bin(67));
int user_input;
printf("Give the input:: ");
scanf("%d", &user_input);
printf("%d", 1);
bin_write(user_input);
}
#define BIT_WIDTH 32
char *IntToBin(unsigned n, char *buffer) {
char *ptr = buffer + BIT_WIDTH;
do {
*(--ptr) = (n & 1) + '0';
n >>= 1;
} while(n);
return ptr;
}
#define TEST 1
#if TEST
#include <stdio.h>
int main() {
int n;
char buff[BIT_WIDTH + 1];
buff[BIT_WIDTH] = '\0';
while(scanf("%d", &n) == 1)
puts(IntToBin(n, buff));
return 0;
}
#endif
short a;
short b;
short c;
short d;
short e;
short f;
short g;
short h;
int i;
char j[256];
printf("BINARY CONVERTER\n\n\n");
//uses <stdlib.h>
while(1)
{
a=0;
b=0;
c=0;
d=0;
e=0;
f=0;
g=0;
h=0;
i=0;
gets(j);
i=atoi(j);
if(i>255){
printf("int i must not pass the value 255.\n");
i=0;
}
if(i>=128){
a=1;
i=i-128;}
if(i>=64){
b=1;
i=i-64;}
if(i>=32){
c=1;
i=i-32;}
if(i>=16){
d=1;
i=i-16;}
if(i>=8){
e=1;
i=i-8;}
if(i>=4){
f=1;
i=i-4;}
if(i>=2){
g=1;
i=i-2;}
if(i>=1){
h=1;
i=i-1;}
printf("\n%d%d%d%d%d%d%d%d\n\n",a,b,c,d,e,f,g,h);
}
I recently read a sample job interview question:
Write a function to convert an integer
to a string. Assume you do not have
access to library functions i.e.,
itoa(), etc...
How would you go about this?
fast stab at it: (edited to handle negative numbers)
int n = INT_MIN;
char buffer[50];
int i = 0;
bool isNeg = n<0;
unsigned int n1 = isNeg ? -n : n;
while(n1!=0)
{
buffer[i++] = n1%10+'0';
n1=n1/10;
}
if(isNeg)
buffer[i++] = '-';
buffer[i] = '\0';
for(int t = 0; t < i/2; t++)
{
buffer[t] ^= buffer[i-t-1];
buffer[i-t-1] ^= buffer[t];
buffer[t] ^= buffer[i-t-1];
}
if(n == 0)
{
buffer[0] = '0';
buffer[1] = '\0';
}
printf(buffer);
A look on the web for itoa implementation will give you good examples. Here is one, avoiding to reverse the string at the end. It relies on a static buffer, so take care if you reuse it for different values.
char* itoa(int val, int base){
static char buf[32] = {0};
int i = 30;
for(; val && i ; --i, val /= base)
buf[i] = "0123456789abcdef"[val % base];
return &buf[i+1];
}
The algorithm is easy to see in English.
Given an integer, e.g. 123
divide by 10 => 123/10. Yielding, result = 12 and remainder = 3
add 30h to 3 and push on stack (adding 30h will convert 3 to ASCII representation)
repeat step 1 until result < 10
add 30h to result and store on stack
the stack contains the number in order of | 1 | 2 | 3 | ...
I would keep in mind that all of the digit characters are in increasing order within the ASCII character set and do not have other characters between them.
I would also use the / and the% operators repeatedly.
How I would go about getting the memory for the string would depend on information you have not given.
Assuming it is in decimal, then like this:
int num = ...;
char res[MaxDigitCount];
int len = 0;
for(; num > 0; ++len)
{
res[len] = num%10+'0';
num/=10;
}
res[len] = 0; //null-terminating
//now we need to reverse res
for(int i = 0; i < len/2; ++i)
{
char c = res[i]; res[i] = res[len-i-1]; res[len-i-1] = c;
}
An implementation of itoa() function seems like an easy task but actually you have to take care of many aspects that are related on your exact needs. I guess that in the interview you are expected to give some details about your way to the solution rather than copying a solution that can be found in Google (http://en.wikipedia.org/wiki/Itoa)
Here are some questions you may want to ask yourself or the interviewer:
Where should the string be located (malloced? passed by the user? static variable?)
Should I support signed numbers?
Should i support floating point?
Should I support other bases rather then 10?
Do we need any input checking?
Is the output string limited in legth?
And so on.
Convert integer to string without access to libraries
Convert the least significant digit to a character first and then proceed to more significant digits.
Normally I'd shift the resulting string into place, yet recursion allows skipping that step with some tight code.
Using neg_a in myitoa_helper() avoids undefined behavior with INT_MIN.
// Return character one past end of character digits.
static char *myitoa_helper(char *dest, int neg_a) {
if (neg_a <= -10) {
dest = myitoa_helper(dest, neg_a / 10);
}
*dest = (char) ('0' - neg_a % 10);
return dest + 1;
}
char *myitoa(char *dest, int a) {
if (a >= 0) {
*myitoa_helper(dest, -a) = '\0';
} else {
*dest = '-';
*myitoa_helper(dest + 1, a) = '\0';
}
return dest;
}
void myitoa_test(int a) {
char s[100];
memset(s, 'x', sizeof s);
printf("%11d <%s>\n", a, myitoa(s, a));
}
Test code & output
#include "limits.h"
#include "stdio.h"
int main(void) {
const int a[] = {INT_MIN, INT_MIN + 1, -42, -1, 0, 1, 2, 9, 10, 99, 100,
INT_MAX - 1, INT_MAX};
for (unsigned i = 0; i < sizeof a / sizeof a[0]; i++) {
myitoa_test(a[i]);
}
return 0;
}
-2147483648 <-2147483648>
-2147483647 <-2147483647>
-42 <-42>
-1 <-1>
0 <0>
1 <1>
2 <2>
9 <9>
10 <10>
99 <99>
100 <100>
2147483646 <2147483646>
2147483647 <2147483647>
The faster the better?
unsigned countDigits(long long x)
{
int i = 1;
while ((x /= 10) && ++i);
return i;
}
unsigned getNumDigits(long long x)
{
x < 0 ? x = -x : 0;
return
x < 10 ? 1 :
x < 100 ? 2 :
x < 1000 ? 3 :
x < 10000 ? 4 :
x < 100000 ? 5 :
x < 1000000 ? 6 :
x < 10000000 ? 7 :
x < 100000000 ? 8 :
x < 1000000000 ? 9 :
x < 10000000000 ? 10 : countDigits(x);
}
#define tochar(x) '0' + x
void tostr(char* dest, long long x)
{
unsigned i = getNumDigits(x);
char negative = x < 0;
if (negative && (*dest = '-') & (x = -x) & i++);
*(dest + i) = 0;
while ((i > negative) && (*(dest + (--i)) = tochar(((x) % 10))) | (x /= 10));
}
If you want to debug, You can split the conditions (instructions) into
lines of code inside the while scopes {}.
I came across this question so I decided to drop by the code I usually use for this:
char *SignedIntToStr(char *Dest, signed int Number, register unsigned char Base) {
if (Base < 2 || Base > 36) {
return (char *)0;
}
register unsigned char Digits = 1;
register unsigned int CurrentPlaceValue = 1;
for (register unsigned int T = Number/Base; T; T /= Base) {
CurrentPlaceValue *= Base;
Digits++;
}
if (!Dest) {
Dest = malloc(Digits+(Number < 0)+1);
}
char *const RDest = Dest;
if (Number < 0) {
Number = -Number;
*Dest = '-';
Dest++;
}
for (register unsigned char i = 0; i < Digits; i++) {
register unsigned char Digit = (Number/CurrentPlaceValue);
Dest[i] = (Digit < 10? '0' : 87)+Digit;
Number %= CurrentPlaceValue;
CurrentPlaceValue /= Base;
}
Dest[Digits] = '\0';
return RDest;
}
#include <stdio.h>
int main(int argc, char *argv[]) {
char String[32];
puts(SignedIntToStr(String, -100, 16));
return 0;
}
This will automatically allocate memory if NULL is passed into Dest. Otherwise it will write to Dest.
Here's a simple approach, but I suspect if you turn this in as-is without understanding and paraphrasing it, your teacher will know you just copied off the net:
char *pru(unsigned x, char *eob)
{
do { *--eob = x%10; } while (x/=10);
return eob;
}
char *pri(int x, char *eob)
{
eob = fmtu(x<0?-x:x, eob);
if (x<0) *--eob='-';
return eob;
}
Various improvements are possible, especially if you want to efficiently support larger-than-word integer sizes up to intmax_t. I'll leave it to you to figure out the way these functions are intended to be called.
Slightly longer than the solution:
static char*
itoa(int n, char s[])
{
int i, sign;
if ((sign = n) < 0)
n = -n;
i = 0;
do
{
s[i++] = n % 10 + '0';
} while ((n /= 10) > 0);
if (sign < 0)
s[i++] = '-';
s[i] = '\0';
reverse(s);
return s;
}
Reverse:
int strlen(const char* str)
{
int i = 0;
while (str != '\0')
{
i++;
str++;
}
return i;
}
static void
reverse(char s[])
{
int i, j;
char c;
for (i = 0, j = strlen(s)-1; i<j; i++, j--) {
c = s[i];
s[i] = s[j];
s[j] = c;
}
}
And although the decision davolno long here are some useful features for beginners. I hope you will be helpful.
This is the shortest function I can think of that:
Correctly handles all signed 32-bit integers including 0, MIN_INT32, MAX_INT32.
Returns a value that can be printed immediatelly, e.g.: printf("%s\n", GetDigits(-123))
Please comment for improvements:
static const char LARGEST_NEGATIVE[] = "-2147483648";
static char* GetDigits(int32_t x) {
char* buffer = (char*) calloc(sizeof(LARGEST_NEGATIVE), 1);
int negative = x < 0;
if (negative) {
if (x + (1 << 31) == 0) { // if x is the largest negative number
memcpy(buffer, LARGEST_NEGATIVE, sizeof(LARGEST_NEGATIVE));
return buffer;
}
x *= -1;
}
// storing digits in reversed order
int length = 0;
do {
buffer[length++] = x % 10 + '0';
x /= 10;
} while (x > 0);
if (negative) {
buffer[length++] = '-'; // appending minus
}
// reversing digits
for (int i = 0; i < length / 2; i++) {
char temp = buffer[i];
buffer[i] = buffer[length-1 - i];
buffer[length-1 - i] = temp;
}
return buffer;
}
//Fixed the answer from [10]
#include <iostream>
void CovertIntToString(unsigned int n1)
{
unsigned int n = INT_MIN;
char buffer[50];
int i = 0;
n = n1;
bool isNeg = n<0;
n1 = isNeg ? -n1 : n1;
while(n1!=0)
{
buffer[i++] = n1%10+'0';
n1=n1/10;
}
if(isNeg)
buffer[i++] = '-';
buffer[i] = '\0';
// Now we must reverse the string
for(int t = 0; t < i/2; t++)
{
buffer[t] ^= buffer[i-t-1];
buffer[i-t-1] ^= buffer[t];
buffer[t] ^= buffer[i-t-1];
}
if(n == 0)
{
buffer[0] = '0';
buffer[1] = '\0';
}
printf("%s", buffer);
}
int main() {
unsigned int x = 4156;
CovertIntToString(x);
return 0;
}
This function converts each digits of number into a char and chars add together
in one stack forming a string. Finally, string is formed from integer.
string convertToString(int num){
string str="";
for(; num>0;){
str+=(num%10+'0');
num/=10;
}
return str;
}
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);
}
}