Converting an integer to binary in C

Question

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)
{
          


        

Answer 1

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.

Answer 2

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)));
}

Answer 3

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;
     }

Answer 4

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

Answer 5

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.

Answer 6

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]);

    }

}