C语言 将整数从（纯）二进制转换为 BCD
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原文地址: http://stackoverflow.com/questions/13247647/
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Convert integer from (pure) binary to BCD
提问by Sagi
I'm to stupid right now to solve this problem...
为了解决这个问题，我现在很愚蠢......
I get a BCD number (every digit is an own 4Bit representation)
我得到一个 BCD 号码（每个数字都是自己的 4Bit 表示）
For example, what I want:
例如，我想要的：
- Input: 202 (hex) == 514 (dec)
Output: BCD 0x415
Input: 0x202
- Bit-representation: 0010 0000 0010 = 514
- 输入：202（十六进制）== 514（十进制）
输出：BCD 0x415
输入：0x202
- 位表示：0010 0000 0010 = 514
What have I tried:
我试过什么：
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;
}
But I know that's very wrong this would be 202 in Bit representation and then split into 5 nibbles and then represented as decimal number again
但我知道这是非常错误的，这将是 202 位表示，然后分成 5 个半字节，然后再次表示为十进制数
I can solve the problem on paper but I just cant get it in a simple C-Code
我可以在纸上解决问题，但我无法用简单的 C 代码来解决
回答by Daniel Gehriger
You got it the wrong way round. Your code is converting from BCD to binary, just as your question's (original) title says. But the input and output values you provided are correct only if you convert from binary to BCD. In that case, try:
你弄错了方法。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;
}
Proof: http://ideone.com/R0reQh
回答by enthusiasticgeek
Try the following.
请尝试以下操作。
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;
}
Also one may try the following variant (although maybe little inefficient)
也可以尝试以下变体（尽管可能效率低下）
/*
Copyright (c) 2016 enthusiasticgeek<[email protected]> 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 != 'char buffer[16];
sprintf(buffer, "%d", var);
sscanf(buffer, "%x", &var);
') // 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;
}
回答by Andrew
The real problem here is confusion of bases and units
这里真正的问题是基数和单位的混淆
The 202 should be HEX which equates to 514 decimal... and therefore the BCD calcs are correct
202 应该是十六进制，相当于十进制 514 ......因此 BCD 计算是正确的
Binary code decimal will convert the decimal (514) into three nibble sized fields: - 5 = 0101 - 1 = 0001 - 4 = 0100
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
The bigger problem was that you have the title the wrong way around, and you are converting Uint to BCD, whereas the title asked for BCD to Unint
更大的问题是你的标题错了，你正在将 Uint 转换为 BCD，而标题要求 BCD 为 Unint
回答by panda-34
A naive but simple solution:
一个天真但简单的解决方案：
#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);
}
回答by Ernesto Fl?res Barreira
This is the solution that I developed and works great for embedded systems, like Microchip PIC microcontrollers:
这是我开发的解决方案，非常适合嵌入式系统，例如 Microchip PIC 微控制器：
##代码##