8位BCD检查

Question

I've a 8-digit BCD number and need to check it out to see if it is a valid BCD number. How can I programmatically (C/C++) make this?

Ex: 0x12345678 is valid, but 0x00f00abc isn't.

Thanks in advance!

Answer 1

You need to check each 4-bit quantity to make sure it's less than 10. For efficiency you want to work on as many bits as you can at a single time.

Here I break the digits apart to leave a zero between each one, then add 6 to each and check for overflow.

uint32_t highs = (value & 0xf0f0f0f0) >> 4;
uint32_t lows = value & 0x0f0f0f0f;
bool invalid = (((highs + 0x06060606) | (lows + 0x06060606)) & 0xf0f0f0f0) != 0;

Edit: actually we can do slightly better. It doesn't take 4 bits to detect overflow, only 1. If we divide all the digits by 2, it frees a bit and we can check all the digits at once.

uint32_t halfdigits = (value >> 1) & 0x77777777;
bool invalid = ((halfdigits + 0x33333333) & 0x88888888) != 0;

Answer 2

The obvious way to do this is:

/* returns 1 if x is valid BCD */
int
isvalidbcd (uint32_t x)
{
    for (; x; x = x>>4)
    {
        if ((x & 0xf) >= 0xa)
            return 0;
    }
    return 1;
}

This link tells you all about BCD, and recommends something like this asa more optimised solution (reworking to check all the digits, and hence using a 64 bit data type, and untested):

/* returns 1 if x is valid BCD */
int
isvalidbcd (uint32_t x)
{
   return !!(((uint64_t)x + 0x66666666ULL) ^ (uint64_t)x) & 0x111111110ULL;
}

Answer 3

For a digit to be invalid, it needs to be 10-15. That in turn means 8 + 4 or 8+2 - the low bit doesn't matter at all.

So:

long mask8 = value & 0x88888888;
long mask4 = value & 0x44444444;
long mask2 = value & 0x22222222;
return ((mask8 >> 2) & ((mask4 >>1) | mask2) == 0;

Slightly less obvious:

long mask8 = (value>>2);
long mask42 = (value | (value>>1);
return (mask8 & mask42 & 0x22222222) == 0;

By shifting before masking, we don't need 3 different masks.

Answer 4

Inspired by @Mark Ransom

 bool invalid = (0x88888888 & (((value & 0xEEEEEEEE) >> 1) + (0x66666666 >> 1))) != 0;
 // or
 bool valid = !((((value & 0xEEEEEEEEu) >> 1) + 0x33333333) & 0x88888888);

Mask off each BCD digit's 1's place, shift right, then add 6 and check for BCD digit overflow.

---

How this works:
By adding +6 to each digit, we look for an overflow * of the 4-digit sum.

 abcd
+ 110
-----
*efgd

But the bit value of d does not contribute to the sum, so first mask off that bit and shift right. Now the overflow bit is in the 8's place. This all is done in parallel and we mask these carry bits with 0x88888888 and test if any are set.

 0abc
+  11
-----
 *efg