What is the difference between == and === in Verilog?

Tags:

hdl

Some data types in Verilog, such as reg, are 4-state. This means that each bit can be one of 4 values: 0,1,x,z.

With the "case equality" operator, ===, x's are compared, and the result is 1.

With ==, the result of the comparison is not 0, as you stated; rather, the result is x, according to the IEEE Std (1800-2009), section 11.4.5 "Equality operators":

For the logical equality and logical inequality operators (== and !=), if, due to unknown or high-impedance bits in the operands, the relation is ambiguous, then the result shall be a 1-bit unknown value (x).

In Verilog:

== tests logical equality (tests for 1 and 0, all other will result in x)
=== tests 4-state logical equality (tests for 1, 0, z and x)

== For comparing bits (0 or 1) === For comparing all 4 states (0, 1, x, z)

== can be synthesized into a hardware (x-nor gate), but === can't be synthesized as x is not a valid logic level in digital, it is infact having voltages in between 0 and 1. And z is not itself any logic, it shows disconnection of the circuit.

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What is the difference between == and === in Verilog?

Tags:

verilog

hdl

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