Your code simulates two multiplexers. These are actually asynchronous components. The fact that Verilog requires data1_temp
and data2_temp
to be declared as reg
's is a quirk of Verilog syntax and your choice of coding style, and doesn't mean these signals would be the outputs of storage elements in a physical implementation.
If you want to capture these values in actual registers, you need to add those explicitly:
reg [7:0] data1, data2;
always @(posedge someclock) begin
data1 <= data1_tmp;
data2 <= data2_tmp;
end
But I would like to know what this mini register file would be made of in hardware. Particularly, the 4x8 bit array consisting of k0,k1,k2,k3.
You haven't shown how these variables are assigned, so it's not possible to say how they are implemented. As your code showed, just declaring them as reg
doesn't guarantee they are implemented with actual storage elements. If you assign them inside a block that begins always @(posedge clk)
then very likely they are flip-flops, but there are ways you could code them that would make them synthesize differently.
I thought when it came to registers and arrays like this, you need a clock to read out data, like RAM?
You need a clock to update a (physical) register. You can read it out at any time. For example:
wire [8:0] sum;
assign sum = k0 + k1;
is perfectly valid code. sum
will change whenever any of its inputs changes. If k0
and k1
are the outputs of flip-flops, their values will only change when there is a clock edge.
For another example, you could equally well describe your multiplexers with code like this:
reg [7:0] k0, k1, k2, k3;
wire [7:0] data1_tmp;
reg [1:0] reg1;
// k<n> and reg1 are assigned elsewhere.
assign data1_tmp = (reg1 == 0) ? k0 :
(reg1 == 1) ? k1 :
(reg1 == 2) ? k2 : k3;
how do I read from this tag_array and do the comparison all within the same clock cycle?
Let me repeat a key point for emphasis: You need to use a clock to assign a new value to a register (an actual hardware register or group of flip-flops). It's output is available at any time.
RAMs are different and how you access the contents of a RAM will depend on details of the type of RAM you use.
I got confused because frankly I don't know enough about memory hardware and how that's possible.
Another key strategy: When you are learning digital logic, I recommend you learn about the physical hardware first, and then work out or study how to simulate it in HDL second. So first, learn what a physical flip-flop is, then learn the standard Verilog methods of describing a flip-flop. Especially if you are trying to write HDL for synthesis, trying to write good code before you learn the capabilities of the underlying hardware will lead you down a lot of dead-end paths.
This is not one question but breaking down the main points:
<=
is a non-blocking assignment used when implying a flip-flop output.
=
is a blocking assignment used when implementing combinatorial output.
example usage:
input [10:0] in_data;
reg [11:0] flip_flop;
reg [11:0] next_data
//Flip-flop
always @(posedge clock) begin
flip_flop <= next_data;
end
//Combinatorial
always @* begin
next_data = in_data + 11'd1;
end
You defined 3 different data types:
input [15:0] me;
reg [15:0] p_array [7:0];
reg abc_pqr [2:0]; //Same as reg [0:0] abc_pqr [2:0]
me
is a standard 16 bit word. p_array
is an 8 deep memory of 16 bit words.
NB: it is typical to define as reg [15:0] p_array [0:7];
abc_pqr [2:0];
is a 3 deep 1bit memory.
You have :
abc_pqr[0] <= me[0]; //This is a 1 bit assignment:
abc_pqr[1] <= me[1];
abc_pqr[2] <= me[2]; //<-- corrected this to 2
Looks valid.
Then :
p_array[abc_pqr[0]] <= me[0];
p_array needs a [7:0]
wide index, you have only supplied 1 bit. and a p_array element is 16 bits wide your left hand side is again only 1 bit.
Best Answer
Xilinx presented this use case in the Synthesis User Guide for Vivado UG901 on page 126. This example also works for the Xilinx ISE tool chain, but I don't know the User Guide number.
The essential part of code is this:
There are two read functions:
- $readmemb -> binary
- $readmemh -> hexadecimal