# Double dabble

In computer science, the double dabble algorithm is used to convert binary numbers into decimal (in particular, binary-coded decimal, or BCD, notation). The algorithm operates as follows:

Suppose the original number to be converted is stored in a register that is n bits wide. Reserve a scratch space wide enough to hold both the original number and its BCD representation $n + (RoundUp(n/3))*4$ bits will be enough. It takes a maximum of 4 bits in binary to store each decimal digit.

Partition the scratch space into BCD digits (on the left) and the original register (on the right). For example, if the original number to be converted is eight bits wide, the scratch space would be partitioned as follows:

```100S TENS ONES ORIGINAL
0010 0100 0011 11110011
```

The diagram above shows the binary representation of 24310 in the original register, and the BCD representation of 243 on the left.

The scratch space is initialized to all zeros, and then the value to be converted is copied into the "original register" space on the right.

```0000 0000 0000 11110011
```

Then, the algorithm iterates n times. On each iteration, the entire scratch space is left-shifted one bit. However, before the left-shift is done, any BCD digit which is greater than 4 is incremented by 3. The increment ensures that a value of 5, incremented and left-shifted, becomes 16, thus correctly "carrying" into the next BCD digit.

The double-dabble algorithm, performed on the value 24310, looks like this:

```0000 0000 0000 11110011      Initialization
0000 0000 0001 11100110      Shift
0000 0000 0011 11001100      Shift
0000 0000 0111 10011000      Shift
0000 0000 1010 10011000      Add 3 to ONES, since it was 7
0000 0001 0101 00110000      Shift
0000 0001 1000 00110000      Add 3 to ONES, since it was 5
0000 0011 0000 01100000      Shift
0000 0110 0000 11000000      Shift
0000 1001 0000 11000000      Add 3 to TENS, since it was 6
0001 0010 0001 10000000      Shift
0010 0100 0011 00000000      Shift
```

Now eight shifts have been performed, so the algorithm terminates. The BCD digits to the left of the "original register" space display the BCD encoding of the original value 243.

Another example for the double dabble algorithm - value 6524410.

``` 104  103  102  101  100  ORIGINAL BINARY
0000 0000 0000 0000 0000 1111111011011100      Initialization
0000 0000 0000 0000 0001 1111110110111000      Shift left (1st)
0000 0000 0000 0000 0011 1111101101110000      Shift left (2nd)
0000 0000 0000 0000 0111 1111011011100000      Shift left (3rd)
0000 0000 0000 0000 1010 1111011011100000      Add 3 to 100, since it was 7
0000 0000 0000 0001 0101 1110110111000000      Shift left (4th)
0000 0000 0000 0001 1000 1110110111000000      Add 3 to 100, since it was 5
0000 0000 0000 0011 0001 1101101110000000      Shift left (5th)
0000 0000 0000 0110 0011 1011011100000000      Shift left (6th)
0000 0000 0000 1001 0011 1011011100000000      Add 3 to 101, since it was 6
0000 0000 0001 0010 0111 0110111000000000      Shift left (7th)
0000 0000 0001 0010 1010 0110111000000000      Add 3 to 100, since it was 7
0000 0000 0010 0101 0100 1101110000000000      Shift left (8th)
0000 0000 0010 1000 0100 1101110000000000      Add 3 to 101, since it was 5
0000 0000 0101 0000 1001 1011100000000000      Shift left (9th)
0000 0000 1000 0000 1001 1011100000000000      Add 3 to 102, since it was 5
0000 0000 1000 0000 1100 1011100000000000      Add 3 to 100, since it was 9
0000 0001 0000 0001 1001 0111000000000000      Shift left (10th)
0000 0001 0000 0001 1100 0111000000000000      Add 3 to 100, since it was 9
0000 0010 0000 0011 1000 1110000000000000      Shift left (11th)
0000 0010 0000 0011 1011 1110000000000000      Add 3 to 100, since it was 8
0000 0100 0000 0111 0111 1100000000000000      Shift left (12th)
0000 0100 0000 1010 0111 1100000000000000      Add 3 to 101, since it was 7
0000 0100 0000 1010 1010 1100000000000000      Add 3 to 100, since it was 7
0000 1000 0001 0101 0101 1000000000000000      Shift left (13th)
0000 1011 0001 0101 0101 1000000000000000      Add 3 to 103, since it was 8
0000 1011 0001 1000 0101 1000000000000000      Add 3 to 101, since it was 5
0000 1011 0001 1000 1000 1000000000000000      Add 3 to 100, since it was 5
0001 0110 0011 0001 0001 0000000000000000      Shift left (14th)
0001 1001 0011 0001 0001 0000000000000000      Add 3 to 103, since it was 6
0011 0010 0110 0010 0010 0000000000000000      Shift left (15th)
0011 0010 1001 0010 0010 0000000000000000      Add 3 to 102, since it was 6
0110 0101 0010 0100 0100 0000000000000000      Shift left (16th)
6    5    2    4    4
BCD
```

Sixteen shifts have been performed, so the algorithm terminates. The BCD digits is: 6*104 + 5*103 + 2*102 + 4*101 + 4*100 = 65244.

## C implementation

The double dabble algorithm might look like this when implemented in C. Notice that this implementation is designed to convert an "input register" of any width, by taking an array as its parameter and returning a dynamically allocated string. Also notice that this implementation does not store an explicit copy of the input register in its scratch space, as the description of the algorithm did; copying the input register into the scratch space was just a pedagogical device.

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

/*
This function takes an array of n unsigned integers,
each holding a value in the range [0, 65535],
representing a number in the range [0, 2**(16n)-1].
arr[0] is the most significant "digit".
This function returns a new array containing the given
number as a string of decimal digits.

For the sake of brevity, this example assumes that
calloc and realloc will never fail.
*/
void double_dabble(int n, const unsigned int *arr, char **result)
{
int nbits = 16*n;         /* length of arr in bits */
int nscratch = nbits/3;   /* length of scratch in bytes */
char *scratch = calloc(1 + nscratch, sizeof *scratch);
int i, j, k;
int smin = nscratch-2;    /* speed optimization */

for (i=0; i < n; ++i) {
for (j=0; j < 16; ++j) {
/* This bit will be shifted in on the right. */
int shifted_in = (arr[i] & (1 << (15-j)))? 1: 0;

/* Add 3 everywhere that scratch[k] >= 5. */
for (k=smin; k < nscratch; ++k)
scratch[k] += (scratch[k] >= 5)? 3: 0;

/* Shift scratch to the left by one position. */
if (scratch[smin] >= 8)
smin -= 1;
for (k=smin; k < nscratch-1; ++k) {
scratch[k] <<= 1;
scratch[k] &= 0xF;
scratch[k] |= (scratch[k+1] >= 8);
}

/* Shift in the new bit from arr. */
scratch[nscratch-1] <<= 1;
scratch[nscratch-1] &= 0xF;
scratch[nscratch-1] |= shifted_in;
}
}

/* Remove leading zeros from the scratch space. */
for (k=0; k < nscratch-1; ++k)
if (scratch[k] != 0) break;
nscratch -= k;
memmove(scratch, scratch+k, nscratch+1);

/* Convert the scratch space from BCD digits to ASCII. */
for (k=0; k < nscratch; ++k)
scratch[k] += '0';

/* Resize and return the resulting string. */
*result = realloc(scratch, nscratch+1);
return;
}

/*
This test driver should print the following decimal values:
246
16170604
1059756703745
*/
int main(void)
{
unsigned int arr[] = { 246, 48748, 1 };
char *text = NULL;
int i;
for (i=0; i < 3; ++i) {
double_dabble(i+1, arr, &text);
printf("%s\n", text);
free(text);
}
return 0;
}
```
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## Historical

In the 1960s, the term double dabble was also used for the mental algorithm used by programmers to convert a binary number to decimal. It is performed by reading the binary number from left to right, doubling if the next bit is zero, and doubling and adding one if the next bit is one. In the example above, 11110011, the thought process would be: "one, three, seven, fifteen, thirty, sixty, one hundred twenty-one, two hundred forty-three," the same result as that obtained above.

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