我现在很愚蠢，无法解决这个问题...
我得到一个 BCD 数字（每个数字都是自己的 4 位表示）
例如，我想要什么：
输出：BCD 0x415
输入：0x202
我尝试了什么：
unsigned int uiValue = 0x202;
unsigned int uiResult = 0;
unsigned int uiMultiplier = 1;
unsigned int uiDigit = 0;
// get the dec bcd value
while ( uiValue > 0 )
{
uiDigit= uiValue & 0x0F;
uiValue >>= 4;
uiResult += uiMultiplier * uiDigit;
uiMultiplier *= 10;
}
但我知道这是非常错误的，这将是位表示的 202，然后分成 5 个半字节，然后再次表示为十进制数
我可以在纸上解决问题，但我无法用简单的 C 代码解决
你弄错了。正如您的问题（原始）标题所说，您的代码正在从BCD 转换为 binary 。但是您提供的输入和输出值只有在您从binary 转换为 BCD时才是正确的。在这种情况下，请尝试：
#include <stdio.h>
int main(void) {
int binaryInput = 0x202;
int bcdResult = 0;
int shift = 0;
printf("Binary: 0x%x (dec: %d)\n", binaryInput , binaryInput );
while (binaryInput > 0) {
bcdResult |= (binaryInput % 10) << (shift++ << 2);
binaryInput /= 10;
}
printf("BCD: 0x%x (dec: %d)\n", bcdResult , bcdResult );
return 0;
}
证明：http: //ideone.com/R0reQh
试试下面的。
unsigned long toPackedBcd (unsigned int val)
{
unsigned long bcdresult = 0; char i;
for (i = 0; val; i++)
{
((char*)&bcdresult)[i / 2] |= i & 1 ? (val % 10) << 4 : (val % 10) & 0xf;
val /= 10;
}
return bcdresult;
}
也可以尝试以下变体（尽管可能效率低下）
/*
Copyright (c) 2016 enthusiasticgeek<enthusiasticgeek@gmail.com> Binary to Packed BCD
This code may be used (including commercial products) without warranties of any kind (use at your own risk)
as long as this copyright notice is retained.
Author, under no circumstances, shall not be responsible for any code crashes or bugs.
Exception to copyright code: 'reverse string function' which is taken from http://stackoverflow.com/questions/19853014/reversing-a-string-in-place-in-c-pointers#19853059
Double Dabble Algorithm for unsigned int explanation
255(binary) - base 10 -> 597(packed BCD) - base 16
H| T| U| (Keep shifting left)
11111111
1 1111111
11 111111
111 11111
1010 11111 <-----added 3 in unit's place (7+3 = 10)
1 0101 1111
1 1000 1111 <-----added 3 in unit's place (5+3 = 8)
11 0001 111
110 0011 11
1001 0011 11 <-----added 3 in ten's place (6+3 = 9)
1 0010 0111 1
1 0010 1010 1 <-----added 3 in unit's place (7+3 = 10)
10 0101 0101 -> binary 597 but bcd 255
^ ^ ^
| | |
2 5 5
*/
#include <stdio.h>
#include <string.h>
//Function Prototypes
unsigned int binaryToPackedBCD (unsigned int binary);
char * printPackedBCD(unsigned int bcd, char * bcd_string);
// For the following function see http://stackoverflow.com/questions/19853014/reversing-a-string-in-place-in-c-pointers#19853059
void reverse(char *str);
//Function Definitions
unsigned int binaryToPackedBCD (unsigned int binary) {
const unsigned int TOTAL_BITS = 32;
/*Place holder for bcd*/
unsigned int bcd = 0;
/*counters*/
unsigned int i,j = 0;
for (i=0; i<TOTAL_BITS; i++) {
/*
Identify the bit to append to LSB of 8 byte or 32 bit word -
First bitwise AND mask with 1.
Then shift to appropriate (nth shift) place.
Then shift the result back to the lsb position.
*/
unsigned int binary_bit_to_lsb = (1<<(TOTAL_BITS-1-i)&binary)>>(TOTAL_BITS-1-i);
/*shift by 1 place and append bit to lsb*/
bcd = ( bcd<<1 ) | binary_bit_to_lsb;
/*printf("=> %u\n",bcd);*/
/*Don't add 3 for last bit shift i.e. in this case 32nd bit*/
if( i >= TOTAL_BITS-1) {
break;
}
/*else continue*/
/* Now, check every nibble from LSB to MSB and if greater than or equal 5 - add 3 if so */
for (j=0; j<TOTAL_BITS; j+=4) {
unsigned int temp = (bcd & (0xf<<j))>>j;
if(temp >= 0x5) {
/*printf("[%u,%u], %u, bcd = %u\n",i,j, temp, bcd);*/
/*Now, add 3 at the appropriate nibble*/
bcd = bcd + (3<<j);
// printf("Now bcd = %u\n", bcd);
}
}
}
/*printf("The number is %u\n",bcd);*/
return bcd;
}
char * printPackedBCD(unsigned int bcd, char * bcd_string) {
const unsigned int TOTAL_BITS = 32;
printf("[LSB] =>\n");
/* Now, check every nibble from LSB to MSB and convert to char* */
for (unsigned int j=0; j<TOTAL_BITS; j+=4) {
//for (unsigned int j=TOTAL_BITS-1; j>=4; j-=4) {
unsigned int temp = (bcd & (0xf<<j))>>j;
if(temp==0){
bcd_string[j/4] = '0';
} else if(temp==1){
bcd_string[j/4] = '1';
} else if(temp==2){
bcd_string[j/4] = '2';
} else if(temp==3){
bcd_string[j/4] = '3';
} else if(temp==4){
bcd_string[j/4] = '4';
} else if(temp==5){
bcd_string[j/4] = '5';
} else if(temp==6){
bcd_string[j/4] = '6';
} else if(temp==7){
bcd_string[j/4] = '7';
} else if(temp==8){
bcd_string[j/4] = '8';
} else if(temp==9){
bcd_string[j/4] = '9';
} else {
bcd_string[j/4] = 'X';
}
printf ("[%u - nibble] => %c\n", j/4, bcd_string[j/4]);
}
printf("<= [MSB]\n");
reverse(bcd_string);
return bcd_string;
}
// For the following function see http://stackoverflow.com/questions/19853014/reversing-a-string-in-place-in-c-pointers#19853059
void reverse(char *str)
{
if (str != 0 && *str != '\0') // Non-null pointer; non-empty string
{
char *end = str + strlen(str) - 1;
while (str < end)
{
char tmp = *str;
*str++ = *end;
*end-- = tmp;
}
}
}
int main(int argc, char * argv[])
{
unsigned int number = 255;
unsigned int bcd = binaryToPackedBCD(number);
char bcd_string[8];
printPackedBCD(bcd, bcd_string);
printf("Binary (Base 10) = %u => Packed BCD (Base 16) = %u\n OR \nPacked BCD String = %s\n", number, bcd, bcd_string);
return 0;
}
这里真正的问题是基础和单位的混淆
202 应该是十六进制，等于十进制 514 ......因此 BCD 计算是正确的
二进制代码十进制将十进制 (514) 转换为三个半字节大小的字段： - 5 = 0101 - 1 = 0001 - 4 = 0100
更大的问题是您的标题错误，并且您正在将 Uint 转换为 BCD，而标题要求将 BCD 转换为 Unint
我的 2 美分，我需要类似的 RTC 芯片，它使用 BCD 来编码时间和日期信息。想出了以下可以很好地满足要求的宏：
#define MACRO_BCD_TO_HEX(x) ((BYTE) ((((x >> 4) & 0x0F) * 10) + (x & 0x0F)))
#define MACRO_HEX_TO_BCD(x) ((BYTE) (((x / 10 ) << 4) | ((x % 10))))
一个天真的但简单的解决方案：
char buffer[16];
sprintf(buffer, "%d", var);
sscanf(buffer, "%x", &var);
这是我开发的解决方案，适用于嵌入式系统，例如 Microchip PIC 微控制器：
#include <stdio.h>
void main(){
unsigned int output = 0;
unsigned int input;
signed char a;
//enter any number from 0 to 9999 here:
input = 1265;
for(a = 13; a >= 0; a--){
if((output & 0xF) >= 5)
output += 3;
if(((output & 0xF0) >> 4) >= 5)
output += (3 << 4);
if(((output & 0xF00) >> 8) >= 5)
output += (3 << 8);
output = (output << 1) | ((input >> a) & 1);
}
printf("Input decimal or binary: %d\nOutput BCD: %X\nOutput decimal: %u\n", input, output, output);
}
这是我的字节转换版本：
//----------------------------------------------
// This function converts n bytes Binary (up to 8, but can be any size)
// value to n bytes BCD value or more.
//----------------------------------------------
void bin2bcdn(void * val, unsigned int8 cnt)
{
unsigned int8 sz, y, buff[20]; // buff = malloc((cnt+1)*2);
if(cnt > 8) sz = 64; // 8x8
else sz = cnt * 8 ; // Size in bits of the data we shift
memset(&buff , 0, sizeof(buff)); // Clears buffer
memcpy(&buff, val, cnt); // Copy the data to buffer
while(sz && !(buff[cnt-1] & 0x80)) // Do not waste time with null bytes,
{ // so search for first significative bit
rotate_left(&buff, sizeof(buff)); // Rotate until we find some data
sz--; // Done this one
}
while(sz--) // Anyting left?
{
for( y = 0; y < cnt+2; y++) // Here we fix the nibbles
{
if(((buff[cnt+y] + 0x03) & 0x08) != 0) buff[cnt+y] += 0x03;
if(((buff[cnt+y] + 0x30) & 0x80) != 0) buff[cnt+y] += 0x30;
}
rotate_left(&buff, sizeof(buff)); // Rotate the stuff
}
memcpy(val, &buff[cnt], cnt); // Copy the buffer to the data
// free(buff); //in case used malloc
} // :D Done
long bin2BCD(long binary) { // double dabble: 8 decimal digits in 32 bits BCD
if (!binary) return 0;
long bit = 0x4000000; // 99999999 max binary
while (!(binary & bit)) bit >>= 1; // skip to MSB
long bcd = 0;
long carry = 0;
while (1) {
bcd <<= 1;
bcd += carry; // carry 6s to next BCD digits (10 + 6 = 0x10 = LSB of next BCD digit)
if (bit & binary) bcd |= 1;
if (!(bit >>= 1)) return bcd;
carry = ((bcd + 0x33333333) & 0x88888888) >> 1; // carrys: 8s -> 4s
carry += carry >> 1; // carrys 6s
}
}
简单的解决方案
#include <stdio.h>
int main(void) {
int binaryInput = 514 ; //0x202
int bcdResult = 0;
int digit = 0;
int i=1;
printf("Binary: 0x%x (dec: %d)\n", binaryInput , binaryInput );
while (binaryInput > 0) {
digit = binaryInput %10; //pick digit
bcdResult = bcdResult+digit*i;
i=16*i;
binaryInput = binaryInput/ 10;
}
printf("BCD: 0x%x (dec: %d)\n", bcdResult , bcdResult );
return 0;
}
二进制：0x202（十进制：514）
BCD：0x514（十进制：1300）
您还可以尝试以下方法：
在每次迭代中，剩余部分（表示为 nibble ）位于其相应的位置。
uint32_t bcd_converter(int num)
{
uint32_t temp=0;
int i=0;
while(num>0){
temp|=((num%10)<<i);
i+=4;
num/=10;
}
return temp;
}