nand2tetris | Answers to projects 6 , 7 , and 8 of nand2tetris

I believe you have typos in your Not components. The in should be in the form in=in[x], not in=[x] as you have it currently.

Conversely, your Nand components are properly formatted.

Because all the calculations are being computed concurrently. In an hdl each instance of a block/entity/module create an instance of all the behaviors inside. You didn't specify which one you're using, but almost all behave the same.

Do you have any free opcodes in aaaa?

If so, you might add an "is-greater-than-0" and an "is-greater-than-or-equal-to-0" opcodes.  These would check, say, the D register's value and replace it with a boolean 1 or 0.

So, your compare sequence is to subtract two items, then use one of these opcodes, and your conditional jump on zero.

You can swap the operands to subtraction to get the other relations (less than, less than zero).

You can also subtract 1 from a boolean to reverse it (make true 0 and make false non-zero).

Alternatively, you can shift away all but the sign bit, i.e. logical shift right by 15 bits will move the sign bit to the low bit position — that will give you 0 for >= 0 and 1 for < 0.  (An arithmetic shift by 15 bits right would give 0 for >= 0 and -1 for < 0.)

As @Peter points out, compare instructions can operate without risk of overflow.  MIPS provides for a single compare, slt, which stand for set less than.  MIPS also provides for branch on condition true as well as branch on condition false.  That, combined with the ability to swap the operands, allows for all 4 relational operations (<, <=, >, >=).

For something as simple as a CPU from nand2tetris you'll be just ok with block RAMs, there's plenty of it on DE10Nano, likely enough for all your needs. Plus some more distributed memory.

In case if you still want an access to DDR, DE10Nano is an SoC, with a hard DDR controller managed by the processor subsystem. It's very easy to interface with it over an Avalon bus (don't bother with AXI unless you really need maximum possible performance).

For the ROM, just use LUTs. A simple static case in Verilog will be translated into an efficient LUT-based ROM.

For accessing HDMI on DE10Nano, you can take a look at this example: https://github.com/combinatorylogic/soc/blob/a1d282d793548030cba940496bed90ff3a29c0ba/backends/c2/hw/de10nano/vga1080p.v (you can also take a look at the DDR access in the same project). Before you can use HDMI you'll need to set up the ADV7513 chip over i2c, see a copy of a library from Terasic under the same project.

For a monochrome 800x600 video you'll be ok with a block RAM. For higher resolutions, as in the example above, you'll have to use DDR.

I find that it helps to break things down, and in complex situations, write down the truth table for the various signals (including the intermediate signals). Sometimes this will lead to an "aha" moment where you see a connection you can use to simplify things.

You may also find it helpful to draw a graph that maps the flow of the signals through the circuit.

Finally, the adage that "first make it work, then make it pretty" applies. Once you have something working, you can look at the design and find simplifications and optimizations. A good example of this is building the XOR circuit. Once you have the straightforward version, if you look at it hard, you can find the clever optimization.

I suppose it is just one of those things where you have to practice until you get the knack of it. You may find it useful to revisit earlier projects with an eye to making them cleaner and easier for you to understand. There are often several ways to build the required circuit, but some ways may be more comprehensible to you. Also, I would recommend you get in the habit of commenting your designs; it is helpful to have a reminder what you were thinking when you did something.

The y & twoToThe[0] expressions evaluates to a new integer (as a result of a bitwise and operation between the bits of 5 and the bits of twoToThe[i]), not to a boolean as I think you were expecting. Check it:

gcc warns about unknows escape sequence \- Demo.

You have to escape \,

Your logic for selecting what input to pass through appears to be incorrect. You should test it by creating a truth table for finalSel, notFinalSel, aAndB, cAndd and out for each of the 4 control conditions.

In general, when doing these kinds of problems, the KISS principle holds; Keep It Simple and Stupid. You don't need any fancy logical manipulation of your sel[] bits, you can just use them directly. So once you get your version fixed (and understand where you went wrong), try doing a version that just consists of 3 Mux16's and nothing else. Once you have both versions working, you'll then understand the error that caused you to go down the wrong path in your first attempt, and that will be a valuable lesson going forward.

Have fun!

There are 2 main parts to consider here:

1. call foo 2 : this instruction tells vm to call funcion foo that takes 2 arguments ( which should be pushed on top of the stack before this call ).
   Calling any function means that you should take these steps as follow: push return address on top of stack (SP++), then push LCL, ARG, THIS,THAT ( SP+4). At this point SP should equal 310.
2. function foo 4 : this is NOT the first instruction in function foo, but still it has an effect on SP as this means that functionfoo has 4 local variables. And these variables must be located somewhere. Where? On top of the stack. This means that before first 'real' instruction of foo gets executed, we must push 4 values onto the stack. What values? Well - according do vm specification it should be 0's, resulting in local variables being initiated to 0. This also means that we increase SP for every local variable ( SP + 4)

This leads to conclusion that SP, after calling foo but before executing 1st instruction of this function will have value of 314.