Convert integer to binary and store it in an integer array of specified size:c++

Question

I want to convert an integer to binary string and then store each bit of the integer string to an element of a integer array of a given size. I am sure that the input integer's binary expression won't exceed the size of the array specified. How to do this in c++?

Why would you want to do that? Ints are already natively an "array of bits", you can access each bit. — Mat
– Mat, Commented Dec 31, 2012 at 17:09
A "Binary string"? As in characters of 1s and 0s? What a strange task... — Mooing Duck
– Mooing Duck, Commented Dec 31, 2012 at 17:09
@Mat: reread the question, he wants to convert an integer into an array of int, where each integer in the array holds a bit from the original integer. — Mooing Duck
– Mooing Duck, Commented Dec 31, 2012 at 17:10
@MooingDuck: I understand. That's like a 32x or 64x storage increase. Doesn't change my question. — Mat
– Mat, Commented Dec 31, 2012 at 17:12

0 _ · Accepted Answer · 2017-10-06 07:48:16Z

12

Pseudo code:

int value = ????  // assuming a 32 bit int
int i;

for (i = 0; i < 32; ++i) {
    array[i] = (value >> i) & 1;
}

edited Oct 6, 2017 at 7:48

0 _

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answered Dec 31, 2012 at 17:14

mah

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5 Comments

Mats Petersson Over a year ago

Why not array[i] = (theValue >> i) & 1 - I'm sure the compiler does the same thing, but seeing that "there isn't going to be a branch in there" makes me happier.

James Over a year ago

<sarcasm>The question is tagged as C++ and so you must use templates, otherwise it's C. </sarcasm>

Marek Over a year ago

works good, however order of bits is reversed, so instead of array[i] I suggest using index array[31 - i]

jin zhenhui Over a year ago

may 32 be changed to sizeof(int) * 8

mah Over a year ago

@jinzhenhui my answer (8 years ago) was, as indicated, intended to be pseudo code. There are several variations one could apply when writing this for real, including your suggestion -- but I would not literally use sizeof(int). Instead, I would use sizeof(value) which: yes, is currently the sam thing. The benefit of specifying the variable, not its type, is that if the variable type should change in the future, you only need to change its type directly and other things that know about the variable instead do not become broken or require a change.

Community · Accepted Answer · 2020-06-20 09:12:55Z

7

You could use C++'s bitset library, as follows.

#include<iostream>
#include<bitset>

int main()
{
  int N;//input number in base 10
  cin>>N;
  int O[32];//The output array
  bitset<32> A=N;//A will hold the binary representation of N 
  for(int i=0,j=31;i<32;i++,j--)
  {
     //Assigning the bits one by one.
     O[i]=A[j];
  }
  return 0;
}

A couple of points to note here: First, 32 in the bitset declaration statement tells the compiler that you want 32 bits to represent your number, so even if your number takes fewer bits to represent, the bitset variable will have 32 bits, possibly with many leading zeroes. Second, bitset is a really flexible way of handling binary, you can give a string as its input or a number, and again you can use the bitset as an array or as a string.It's a really handy library. You can print out the bitset variable A as cout<<A; and see how it works.

edited Jun 20, 2020 at 9:12

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answered Dec 31, 2012 at 17:30

Aravind

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2 Comments

Mooing Duck Over a year ago

Well that's a good idea. +1 even though it competes with mine. But why do you only support 21 digits? Why not 32?

Aravind Over a year ago

Okay let's make it 32 then.

Mooing Duck · Accepted Answer · 2012-12-31 17:34:25Z

6

template<class output_iterator>
void convert_number_to_array_of_digits(const unsigned number, 
         output_iterator first, output_iterator last) 
{
    const unsigned number_bits = CHAR_BIT*sizeof(int);
    //extract bits one at a time
    for(unsigned i=0; i<number_bits && first!=last; ++i) {
        const unsigned shift_amount = number_bits-i-1;
        const unsigned this_bit = (number>>shift_amount)&1;
        *first = this_bit;
        ++first;
    }
    //pad the rest with zeros
    while(first != last) {
        *first = 0;
        ++first;
    }
}

int main() {
    int number = 413523152;
    int array[32];
    convert_number_to_array_of_digits(number, std::begin(array), std::end(array));
    for(int i=0; i<32; ++i)
        std::cout << array[i] << ' ';
}

Proof of compilation here

edited Dec 31, 2012 at 17:34

answered Dec 31, 2012 at 17:16

Mooing Duck

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2 Comments

James Over a year ago

Don't you mean (number >> i) & 1?

Mooing Duck Over a year ago

@James: Thanks. First I posted code, then I posted code that compiles, and now it compiles and executes and seems to be working.

Alfred · Accepted Answer · 2012-12-31 17:37:36Z

2

You can do like this:

while (input != 0) {

        if (input & 1)
            result[index] = 1; 
        else
            result[index] =0;
   input >>= 1;// dividing by two
   index++;
}

edited Dec 31, 2012 at 17:37

answered Dec 31, 2012 at 17:16

Alfred

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1 Comment

Mooing Duck Over a year ago

I don't think that's quite right... (1) you never appear to change index, (2) even then it's still wrong.

RageD · Accepted Answer · 2012-12-31 17:15:41Z

1

As Mat mentioned above, an int is already a bit-vector (using bitwise operations, you can check each bit). So, you can simply try something like this:

// Note: This depends on the endianess of your machine
int x = 0xdeadbeef; // Your integer?
int arr[sizeof(int)*CHAR_BIT];
for(int i = 0 ; i < sizeof(int)*CHAR_BIT ; ++i) {
  arr[i] = (x & (0x01 << i)) ? 1 : 0; // Take the i-th bit
}

answered Dec 31, 2012 at 17:15

RageD

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1 Comment

RageD Over a year ago

Made the correction - thanks ;) (always forget about CHAR_BIT)

Grijesh Chauhan · Accepted Answer · 2014-04-09 17:27:56Z

Decimal to Binary: Size independent

Two ways: both stores binary represent into a dynamic allocated array bits (in msh to lsh).

First Method:

#include<limits.h> // include for CHAR_BIT
int* binary(int dec){
  int* bits = calloc(sizeof(int) * CHAR_BIT, sizeof(int));
  if(bits == NULL) return NULL;
  int i = 0;

  // conversion
  int left = sizeof(int) * CHAR_BIT - 1; 
  for(i = 0; left >= 0; left--, i++){
    bits[i] = !!(dec & ( 1u << left ));      
  }

  return bits;
}

Second Method:

#include<limits.h> // include for CHAR_BIT
int* binary(unsigned int num)
{
   unsigned int mask = 1u << ((sizeof(int) * CHAR_BIT) - 1);   
                      //mask = 1000 0000 0000 0000
   int* bits = calloc(sizeof(int) * CHAR_BIT, sizeof(int));
   if(bits == NULL) return NULL;
   int i = 0;

   //conversion 
   while(mask > 0){
     if((num & mask) == 0 )
         bits[i] = 0;
     else
         bits[i] = 1;
     mask = mask >> 1 ;  // Right Shift 
     i++;
   }

   return bits;
}

Brandon · Accepted Answer · 2012-12-31 18:37:18Z

0

I know it doesn't add as many Zero's as you wish for positive numbers. But for negative binary numbers, it works pretty well.. I just wanted to post a solution for once :)

int BinToDec(int Value, int Padding = 8)
{
    int Bin = 0;

    for (int I = 1, Pos = 1; I < (Padding + 1); ++I, Pos *= 10)
    {
        Bin += ((Value >> I - 1) & 1) * Pos;
    }
    return Bin;
}

answered Dec 31, 2012 at 18:37

Brandon

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Comments

perry_the_python · Accepted Answer · 2020-03-12 05:33:29Z

0

This is what I use, it also lets you give the number of bits that will be in the final vector, fills any unused bits with leading 0s.

std::vector<int> to_binary(int num_to_convert_to_binary, int num_bits_in_out_vec)
{
    std::vector<int> r;

    // make binary vec of minimum size backwards (LSB at .end() and MSB at .begin())
    while (num_to_convert_to_binary > 0)
    {
        //cout << " top of loop" << endl;
        if (num_to_convert_to_binary % 2 == 0)
            r.push_back(0);
        else
            r.push_back(1);
        num_to_convert_to_binary = num_to_convert_to_binary / 2;
    }

    while(r.size() < num_bits_in_out_vec)
        r.push_back(0);

    return r;
}

answered Mar 12, 2020 at 5:33

perry_the_python

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