检查序列中的位模式

Question

So basically i need to check if a certain sequence of bits occurs in other sequence of bits(32bits). The function shoud take 3 arguments:

1. n right most bits of a value.
2. a value
3. the sequence where the n bits should be checked for occurance

The function has to return the number of bit where the desired sequence started. Example chek if last 3 bits of 0x5 occur in 0xe1f4.

void bitcheck(unsigned int source, int operand,int n)
{
   int i,lastbits,mask;
   mask=(1<<n)-1;
   lastbits=operand&mask;

   for(i=0; i<32; i++)
   {
      if((source&(lastbits<<i))==(lastbits<<i))
          printf("It start at bit number %i\n",i+n);
   }
}

Answer 1

I take you to mean that you want to take source as a bit array, and to search it for a bit sequence specified by the n lowest-order bits of operand . It seems you would want to perform a standard mask & compare; the only (minor) complication being that you need to scan. You seem already to have that idea.

I'd write it like this:

void bitcheck(uint32_t source, uint32_t operand, unsigned int n) {
    uint32_t mask = ~((~0) << n);
    uint32_t needle = operand & mask;
    int i;

    for(i = 0; i <= (32 - n); i += 1) {
        if (((source >> i) & mask) == needle) {
            /* found it */
            break;
        }
    }
}

There are some differences in the details between mine and yours, but the main functional difference is the loop bound: you must be careful to ignore cases where some of the bits you compare against the target were introduced by a shift operation, as opposed to originating in source , lest you get false positives. The way I've written the comparison makes it clearer (to me) what the bound should be.

I also use the explicit-width integer data types from stdint.h for all values where the code depends on a specific width. This is an excellent habit to acquire if you want to write code that ports cleanly.

Answer 2

Perhaps:

if((source&(maskbits<<i))==(lastbits<<i))

Because: finding 10 in 11 will be true for your old code. In fact, your original condition will always return true when 'source' is made of all ones.

Answer 3

Your loop goes too far, I'm afraid. It could, for example 'find' the bit pattern '0001' in a value ~0 , which consists of ones only.

This will do better (I hope):

void checkbit(unsigned value, unsigned pattern, unsigned n)
{
    unsigned size = 8 * sizeof value;
    if( 0 < n && n <= size)
    {
        unsigned mask = ~0U >> (size - n);
        pattern &= mask;

        for(int i = 0; i <= size - n; i ++, value >>= 1)
            if((value & mask) == pattern)
                printf("pattern found at bit position %u\n", i+n);
    }
}