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How to assign a value to a Reg in Verilog?
I’m an FPGA noob trying to learn Verilog. How can I “assign” a value to a reg in an always block, either as an initial value, or as a constant. I’m trying to do something like this in the code below. I get an error because the 8 bit constant doesn’t count as input. I also don’t want to trigger the always off of a clock.
How to access 32 most significant bits in Verilog?
I have problem in accessing 32 most significant and 32 least significant bits in Verilog. I have written the following code but I get the error “Illegal part-select expression” The point here is that I don’t have access to a 64 bit register.
How to access 64 bit registers without Verilog?
So I will describe a solution that does not use any Verilog 64 bit registers. This will however not have any impact on the resulting hardware. It will only look different in the Verilog code. The idea is to access the MSB bits by shifting the result of A*B. The following naive version of this will not work:
Why does a * B not work in Verilog?
The reason why this does not work is that the width of A*B is determined by the left hand side of the assignment, which is 32 bits. Therefore the result of A*B will only contain the lower 32 bits of the results.
Is there an initial block for LedR in Verilog?
However, LEDR keeps being XXXX in my waveform timing diagram. I have tried using initial block, but it doesn’t work. (My IDE is Quartus Prime)
How to assign a synthesizable initial value to a Reg?
If you actually want a flop where you can change the value, the default would be in the reset clause. You should use what your FPGA documentation recommends. There is no portable way to initialize register values other than using a reset net. This has a hardware cost associated with it on most synthesis targets.
Is there a way to initialize register values?
There is no portable way to initialize register values other than using a reset net. This has a hardware cost associated with it on most synthesis targets. When a chip gets power all of it’s registers contain random values. It’s not possible to have an an initial value. It will always be random.