- What is the difference between == and === in Verilog?
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
- verilog - What is `+:` and `-:`? - Stack Overflow
5 2 1 Vector bit-select and part-select addressing Bit-selects extract a particular bit from a vector net, vector reg, integer, or time variable, or parameter The bit can be addressed using an expression If the bit-select is out of the address bounds or the bit-select is x or z , then the value returned by the reference shall be x A bit-select or part-select of a scalar, or of a variable
- What is the difference between = and lt;= in Verilog?
What is the difference between = and <= in Verilog? Asked 9 years, 4 months ago Modified 2 years, 6 months ago Viewed 110k times
- verilog - What is the difference between single ( ) and double . . .
In IEEE 1800-2005 or later, what is the difference between amp; and amp; amp; binary operators? Are they equivalent? I noticed that these coverpoint definitions behave identically where a and b
- lt;= Assignment Operator in Verilog - Stack Overflow
25 "<=" in Verilog is called non-blocking assignment which brings a whole lot of difference than "=" which is called as blocking assignment because of scheduling events in any vendor based simulators
- operator in verilog - Stack Overflow
10 i have a verilog code in which there is a line as follows: parameter ADDR_WIDTH = 8 ; parameter RAM_DEPTH = 1 << ADDR_WIDTH; here what will be stored in RAM_DEPTH and what does the << operator do here
- vhdl - Verilog question mark (?) operator - Stack Overflow
I'm trying to translate a Verilog program into VHDL and have stumbled across a statement where a question mark (?) operator is used in the Verilog program The following is the Verilog code; 1 m
- verilog - Order of bits in reg declaration - Stack Overflow
We can access them by using array indices So, accessing the first digit of the second number is done by something like this a[0][1] assuming that this array is stored in a variable a Returning to Verilog, how would accessing elements change if I would swap values in reg or declare them in reverse order ( [0:7]), for example?
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