我知道您可以使用此表将十进制转换为BCD:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
是否存在此转换的等式或您只需使用该表?我试图为这个转换写一些代码,但我不知道如何为它做数学。建议?
答案 0 :(得分:8)
你知道Binary numeral system,不是吗?
特别要看看this chapter。
编辑:还要注意KFro的评论,即数字的二进制ASCII表示的低半字节(= 4位)是BCD。这使得转换BCD< - > ASCII非常简单,因为您只需添加/删除前4位:
Number ASCII Code 0 0011 0000 1 0011 0001 ... 8 0011 1000 9 0011 1001
答案 1 :(得分:7)
#include <stdint.h>
/* Standard iterative function to convert 16-bit integer to BCD */
uint32_t dec2bcd(uint16_t dec)
{
uint32_t result = 0;
int shift = 0;
while (dec)
{
result += (dec % 10) << shift;
dec = dec / 10;
shift += 4;
}
return result;
}
/* Recursive one liner because that's fun */
uint32_t dec2bcd_r(uint16_t dec)
{
return (dec) ? ((dec2bcd_r( dec / 10 ) << 4) + (dec % 10)) : 0;
}
答案 2 :(得分:6)
这是来自微控制器世界....请注意,值在分区中舍入。例如91到BCD将是91/10 * 16 = 144 + 91%10 = 145.转换为二进制是10010001。
uint8_t bcdToDec(uint8_t val)
{
return ( (val/16*10) + (val%16) );
}
uint8_t decToBcd(uint8_t val)
{
return ( (val/10*16) + (val%10) );
}
答案 3 :(得分:5)
通常当有人说他们想要从十进制转换为BCD时,他们谈论的是十几位数。
BCD通常每个字节打包成两位十进制数字(因为0..9适合4位,如图所示),但我认为使用一个字节数组更为自然,每个十进制数字一个。 / p>
n位无符号二进制数将适合ceil(n * log_2(10))= ceil(n / log10(2))十进制数字。它也适用于ceil(n / 3)= floor((n + 2)/ 3))十进制数字,因为2 ^ 3 = 8小于10.
考虑到这一点,这里是我如何获得unsigned int的十进制数字:
#include <algorithm>
#include <vector>
template <class Uint>
std::vector<unsigned char> bcd(Uint x) {
std::vector<unsigned char> ret;
if (x==0) ret.push_back(0);
// skip the above line if you don't mind an empty vector for "0"
while(x>0) {
Uint d=x/10;
ret.push_back(x-(d*10)); // may be faster than x%10
x=d;
}
std::reverse(ret.begin(),ret.end());
// skip the above line if you don't mind that ret[0] is the least significant digit
return ret;
}
当然,如果您知道int类型的宽度,则可能更喜欢固定长度数组。如果你能记住第0位是最不重要的并且仅在输入/输出上反向的事实,也没有理由反转。在不使用固定位数的情况下,将最低有效位保持为第一位简化了数字运算操作。
如果要将“0”表示为单个“0”十进制数字而不是空数字字符串(两者都有效),那么您将专门检查x == 0.
答案 4 :(得分:1)
如果你想要每个字节有两个十进制数字,而“unsigned”是“unsigned long”的一半(如果你愿意,可以使用uint32和uint64 typedef):
unsigned long bcd(unsigned x) {
unsigned long ret=0;
while(x>0) {
unsigned d=x/10;
ret=(ret<<4)|(x-d*10);
x=d;
}
return ret;
}
这使您在最低有效半字节中具有最低有效(单位)十进制数字。你也可以执行一个固定数字(uint32为10)的循环,而不是只剩下0位时提前停止,这将允许优化器展开它,但如果你的数字通常很慢,那就慢了。
答案 5 :(得分:0)
此类内容适用于您的转化吗?
#include <string>
#include <bitset>
using namespace std;
string dec_to_bin(unsigned long n)
{
return bitset<numeric_limits<unsigned long>::digits>(n).to_string<char, char_traits<char>, allocator<char> >();
}
答案 6 :(得分:0)
此代码编码和解码。基准如下。
我使用了uint64_t来存储BCD。非常方便和固定的宽度,但对于大型桌子而言不是非常节省空间。将BCD数字2包装到char []中。
// -------------------------------------------------------------------------------------
uint64_t uint32_to_bcd(uint32_t usi) {
uint64_t shift = 16;
uint64_t result = (usi % 10);
while (usi = (usi/10)) {
result += (usi % 10) * shift;
shift *= 16; // weirdly, it's not possible to left shift more than 32 bits
}
return result;
}
// ---------------------------------------------------------------------------------------
uint32_t bcd_to_ui32(uint64_t bcd) {
uint64_t mask = 0x000f;
uint64_t pwr = 1;
uint64_t i = (bcd & mask);
while (bcd = (bcd >> 4)) {
pwr *= 10;
i += (bcd & mask) * pwr;
}
return (uint32_t)i;
}
// --------------------------------------------------------------------------------------
const unsigned long LOOP_KNT = 3400000000; // set to clock frequencey of your CPU
// --------------------------------------------------------------------------------------
int main(void) {
time_t start = clock();
uint32_t foo, usi = 1234; //456;
uint64_t result;
unsigned long i;
printf("\nRunning benchmarks for %u loops.", LOOP_KNT);
start = clock();
for (uint32_t i = 0; i < LOOP_KNT; i++) {
foo = bcd_to_ui32(uint32_to_bcd(i >> 10));
}
printf("\nET for bcd_to_ui32(uint_16_to_bcd(t)) was %f milliseconds. foo %u", (double)clock() - start, foo);
printf("\n\nRunning benchmarks for %u loops.", LOOP_KNT);
start = clock();
for (uint32_t i = 0; i < LOOP_KNT; i++) {
foo = bcd_to_ui32(i >> 10);
}
printf("\nET for bcd_to_ui32(uint_16_to_bcd(t)) was %f milliseconds. foo %u", (double)clock() - start, foo);
getchar();
return 0;
}
注意:强> 看起来,即使使用64位整数,也不可能向左移动超过32位,但幸运的是,它完全有可能乘以16的因子 - 这很快就会产生预期的效果。它也快得多。去图。
答案 7 :(得分:0)
我知道这已经得到了回答,但是我已经使用模板扩展了这个不同大小的无符号整数来构建特定的代码。
#include <stdio.h>
#include <unistd.h>
#include <stdint.h>
#define __STDC_FORMAT_MACROS
#include <inttypes.h>
constexpr int nBCDPartLength = 4;
constexpr int nMaxSleep = 10000; // Wait enough time (in ms) to check out the boundry cases before continuing.
// Convert from an integer to a BCD value.
// some ideas for this code are from :
// http://stackoverflow.com/questions/1408361/unsigned-integer-to-bcd-conversion
// &&
// http://stackoverflow.com/questions/13587502/conversion-from-integer-to-bcd
// Compute the last part of the information and place it into the result location.
// Decrease the original value to place the next lowest digit into proper position for extraction.
template<typename R, typename T> R IntToBCD(T nValue)
{
int nSizeRtn = sizeof(R);
char acResult[nSizeRtn] {};
R nResult { 0 };
int nPos { 0 };
while (nValue)
{
if (nPos >= nSizeRtn)
{
return 0;
}
acResult[nPos] |= nValue % 10;
nValue /= 10;
acResult[nPos] |= (nValue % 10) << nBCDPartLength;
nValue /= 10;
++nPos;
}
nResult = *(reinterpret_cast<R *>(acResult));
return nResult;
}
int main(int argc, char **argv)
{
//uint16_t nValue { 10 };
//printf("The BCD for %d is %x\n", nValue, IntToBCD<uint32_t, uint16_t>(nValue));
// UINT8_MAX = (255) - 2 bytes can be held in uint16_t (2 bytes)
// UINT16_MAX = (65535) - 3 bytes can be held in uint32_t (4 bytes)
// UINT32_MAX = (4294967295U) - 5 bytes can be held in uint64_t (8 bytes)
// UINT64_MAX = (__UINT64_C(18446744073709551615)) - 10 bytes can be held in uint128_t (16 bytes)
// Test edge case for uint8
uint8_t n8Value { UINT8_MAX - 1 };
printf("The BCD for %u is %x\n", n8Value, IntToBCD<uint16_t, uint8_t>(n8Value));
// Test edge case for uint16
uint16_t n16Value { UINT16_MAX - 1 };
printf("The BCD for %u is %x\n", n16Value, IntToBCD<uint32_t, uint16_t>(n16Value));
// Test edge case for uint32
uint32_t n32Value { UINT32_MAX - 1 };
printf("The BCD for %u is %" PRIx64 "\n", n32Value, IntToBCD<uint64_t, uint32_t>(n32Value));
// Test edge case for uint64
uint64_t n64Value { UINT64_MAX - 1 };
__uint128_t nLargeValue = IntToBCD<__uint128_t, uint64_t>(n64Value);
uint64_t nTopHalf = uint64_t(nLargeValue >> 64);
uint64_t nBottomHalf = uint64_t(nLargeValue);
printf("The BCD for %" PRIu64 " is %" PRIx64 ":%" PRIx64 "\n", n64Value, nTopHalf, nBottomHalf);
usleep(nMaxSleep);
// Test all the values
for (uint8_t nIdx = 0; nIdx < UINT8_MAX; ++nIdx)
{
printf("The BCD for %u is %x\n", nIdx, IntToBCD<uint16_t, uint8_t>(nIdx));
}
for (uint16_t nIdx = 0; nIdx < UINT16_MAX; ++nIdx)
{
printf("The BCD for %u is %x\n", nIdx, IntToBCD<uint32_t, uint16_t>(nIdx));
}
for (uint32_t nIdx = 0; nIdx < UINT32_MAX; ++nIdx)
{
printf("The BCD for %u is %" PRIx64 "\n", nIdx, IntToBCD<uint64_t, uint32_t>(nIdx));
}
for (uint64_t nIdx = 0; nIdx < UINT64_MAX; ++nIdx)
{
__uint128_t nLargeValue = IntToBCD<__uint128_t, uint64_t>(nIdx);
uint64_t nTopHalf = uint64_t(nLargeValue >> 64);
uint64_t nBottomHalf = uint64_t(nLargeValue);
printf("The BCD for %" PRIu64 " is %" PRIx64 ":%" PRIx64 "\n", nIdx, nTopHalf, nBottomHalf);
}
return 0;
}
答案 8 :(得分:0)
这是uint16_t的宏,因此它在编译时得到评估(假设你是一个预定义的常量)。这与从上到9999的dec2bcd()一致。
#define U16TOBCD(u) ((((u/1000)%10)<<12)|(((u/100)%10)<<8)|\
(((u/10)%10)<<4)|(u%10))
答案 9 :(得分:0)
简化它。
#include <math.h>
#define uint unsigned int
uint Convert(uint value, const uint base1, const uint base2)
{
uint result = 0;
for (int i = 0; value > 0; i++)
{
result += value % base1 * pow(base2, i);
value /= base1;
}
return result;
}
uint FromBCD(uint value)
{
return Convert(value, 16, 10);
}
uint ToBCD(uint value)
{
return Convert(value, 10, 16);
}