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In computer science, the double dabble algorithm is used to convert binary numbers into binary-coded decimal (BCD) notation.〔.〕〔Binary-to-BCD Converter: “Double-Dabble Binary-to-BCD Conversion Algorithm,” originally at http://edda.csie.dyu.edu.tw/course/fpga/Binary2BCD.pdf, as cited by . (Archived from original ), January 31, 2012.〕 It is also known as the shift and add 3 algorithm, and can be implemented using a small number of gates in computer hardware, but at the expense of high latency.〔.〕 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; bits will be enough. It takes a maximum of 4 bits in binary to store each decimal digit. Then 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 The algorithm then 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 2 4 3 BCD 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 #include #include / * This function takes an array of n unsigned integers, each holding a value in the range (65535 ), representing a number in the range (2 * *(16n)-1 ). arr() 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) } / * Remove leading zeros from the scratch space. */ for (k=0; k < nscratch-1; ++k) if (scratch() != 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() += '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) return 0; } 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Double dabble」の詳細全文を読む スポンサード リンク
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