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bool X86IsCanonicalAddress(void *a) {
	void *b = 0xffff000000000000;
	if ((a & b) == b)
		return true;

	if ((a ^ b) & b == b)
		return true;

	return false;
}

I think all you need to do for this one is isolate the high 16 bits to see if they are all 0 or 1.

bool X86IsCanonicalAddress(int a) {
	// Return true if lowest 48 bits are equal and highest 16 bits are all 1 or all 0
	// Assumes compiler treats int as unsigned 64bit and does arithmetic shifting

	// Isolate the highest 16 bits. 
	// If all 1, result will be -1 due to arithmetic type shifting. 
	// If all 0, result is 0
	int high16 = a >> 48;

	if(high16 == -1 || high16 == 0) {
		return true;
	}

	return false;
}

Can you please explain the code above?

Simple:

bool isCanonical(unsigned long long add) {

	unsigned long long int mask = 0xFFFF000000000000;

	return (!(add & mask));

}

BigInteger bigInt = new BigInteger(inputString, 2);
        return  new BigInteger(binStr, 2);.shiftRight(48).equals(new BigInteger("1111111111111111", 2)) || bigInt.shiftRight(48).equals(new BigInteger("0000000000000000", 2));

I think the question might be missing a detail. According to the wiki, canonical address definition is like this:
"... the AMD specification requires that bits 48 through 63 of any virtual address must be copies of bit 47"