Comparators and boundary specifications

The highest differing bit determines magnitude

Compare two unsigned^{iUnsigned A rule for interpreting a bit pattern as a nonnegative integer. The n-bit range is 0 through 2^n−1 and may differ from the signed interpretation of the same bits. Learn more} vectors from the MSB^{iMSB The bit with the highest place value. Its index depends on the vector's declaration direction, so do not assume it always has the largest index. Learn more} downward and find the first differing bit. The value with a 1 there is larger. With $e_i=\overline{a_i\oplus b_i}$, the four-bit greater-than condition is

Here, $\operatorname{gt}(a,b)$ is the Boolean comparison result. A lower-position comparison matters only when all higher bits agree. Equality is the AND of all $e_i$.

Signed comparison interprets the sign first

4'b1111 and 4'b0001 are unsigned 15 and 1, but signed -1 and 1. With differing signs, the negative value is smaller. The difference between logic [3:0] and logic signed [3:0] is the interpretation used by operations, not storage capacity.

logic [3:0] a, b;
logic unsigned_lt, signed_lt;
assign unsigned_lt = a < b;
assign signed_lt = $signed(a) < $signed(b);

This code uses operands of equal width. If widths differ, first choose the extension width before applying a simple cast.

The same two bit patterns can reverse order under different signedness contracts. Do not apply the unsigned highest-bit comparison rule directly to signed values.

State boundary conditions in words

For a buffer that holds 16 items, count==16 means full and count<16 means another item can be stored. A four-bit count cannot represent 16, so the full condition itself fails. This is why both comparator behavior and the representable range of its operands must be reviewed.

For threshold $T$, first check $T-1,T,T+1$, handling values outside the representable range separately. Construct tests that deliberately distinguish < from <=. Comparing only the value after unsigned wraparound cannot recover a carry that has already been lost.