我知道你可以使用这个表格将十进制转换为BCD:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
是否有一个公式可以用于此转换,或者只能使用表格? 我正在尝试编写一些代码来进行此转换,但我不确定如何进行计算。 有什么建议吗?
#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;
}
0xf0
到0xf9
之间。 - fuz这来自于微控制器世界... 请注意在除法中数值是四舍五入的。例如,将91转换为BCD码,应该计算91/10 * 16 = 144 + 91%10 = 145。转换为二进制即为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) );
}
#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;
}
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;
}
简单化它。
#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);
}
像这样的东西对于你的转换是否有效?
#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> >();
}
这段代码进行编码和解码。以下是基准测试结果。
我在这里使用了 uint64_t 来存储 BCD。非常方便且固定宽度,但对于大型表格来说不太空间有效。为此,请将 BCD 数字打包成 char[] 的 2 位。
// -------------------------------------------------------------------------------------
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: 似乎即使使用64位整数,左移超过32位也是不可能的,但幸运的是,完全可以通过某个16的倍数进行乘法运算 - 这正好具有所需的效果。而且速度更快。想想看。
我知道这个问题之前已经有答案了,但是我使用模板来构建特定代码,以处理不同大小的无符号整数。
#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 :
// https://dev59.com/iEnSa4cB1Zd3GeqPMk3p
// &&
// 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;
}
#define U16TOBCD(u) ((((u/1000)%10)<<12)|(((u/100)%10)<<8)|\
(((u/10)%10)<<4)|(u%10))
5
已经是0101
,不需要任何转换,7
是0111
等等。 - LukeH