Do you want to get a 4-bit 2-to-1 mux (with 4 inputs, 1 select and 4 
outputs), or do you want a single 4-to-1 mux (4 in, 2 select,1 out)?

BTW: do you program VHDL with pictures? No? So, why do you not simply 
attach the *.vhd files?

Sorry I have already solved that problem. I was using my pre-edited code 
haha. However, I receive errors when I synthesize my flip flop. I get 
errors stating:

WARNING:Xst:2170 - Unit MMFlipFlop : the following signal(s) form a 
combinatorial loop: Q.

WARNING:Xst:2170 - Unit MMFlipFlop : the following signal(s) form a 
combinatorial loop: M1.

Am I supposed to be receiving these combinatorial loop warnings? The 
flip flop performed correctly when I did test bench tests. I figure 
since I am doing structural VHDL, I should have 0 errors in order to 
avoid problems down the road. Attached are the VHDL files and a picture 
of my diagram.

mike mr wrote:
> Sorry I have already solved that problem.
And how?
(Just for information, if someone else needs help in fututre times...)


> However, I receive errors when I synthesize my flip flop. I get
> errors stating:
> WARNING:Xst:2170 - Unit MMFlipFlop : the following signal(s) form a
> combinatorial loop: M1.
These are no ERRORS but just simple WARNINGS. Thats a major difference!

> Am I supposed to be receiving these combinatorial loop warnings?
This messages come from your latches. A latch consists mainly of a 
combinatorial loop from the Output to the input. But usually such loops 
are made unintenionally, and so the synthesizer warns...

> The flip flop performed correctly when I did test bench tests. I figure
> since I am doing structural VHDL, I should have 0 errors in order to
> avoid problems down the road.
You get this error because you are doing structural VDHL. Everyone 
else on the world would write something like this to get a D-FF:

1

process (ck) begin

2

  if rising_edge(ck) then 

3

     q <= d;

4

  end if;

5

end process;

With this the synthesizer immediatelly knows which component to use...

With your structural description the synthesizer tries to build a D-FF 
out of a bunch of LUTs, not being able to recognize that there are 
thousands of components laying around doing exactly what you want deep 
inside.

As I already said: tell your teacher its a fairly stupid job waisting 
time with a VHDL description style from the last millenium. The tools do 
not expect such an old-fashioned coding style and they will produce 
inefficient results. Why learning such things?

> And how?
> (Just for information, if someone else needs help in future times...)

Well it turns out that I didn't add the a gate for the inverter in the 2
to 1 MUX. Also in my 4 bit 2 to 1 MUX I should have had:


gate1  :  mm2to1MUX  PORT MAP(D0=>'0',D1=>D1(3),S=>S,Z=>Z(3));
gate2  :  mm2to1MUX  PORT MAP(D0=>'0',D1=>D1(2),S=>S,Z=>Z(2));
gate3  :  mm2to1MUX  PORT MAP(D0=>'0',D1=>D1(1),S=>S,Z=>Z(1));
gate4  :  mm2to1MUX  PORT MAP(D0=>'0',D1=>D1(0),S=>S,Z=>Z(0));



> This messages come from your latches. A latch consists mainly of a
> combinatorial loop from the Output to the input. But usually such loops
> are made unintenionally, and so the synthesizer warns...

I figured as much thanks for the info! This warning doesn't give me any
irregular ouputs so this shouldn't cause any errors in other components
when I use this 4 bit 2 to 1 MUX?

> As I already said: tell your teacher its a fairly stupid job waisting
> time with a VHDL description style from the last millenium. The tools do
> not expect such an old-fashioned coding style and they will produce
> inefficient results. Why learning such things?

I couldn't agree with you more! However, he will not use behavioral VHDL
as he believes the class will get a better understanding by doing
components individually and understanding how everything connects to
each other.

I'm trying to connect the 6 bit MUX(Ijust incremented it from 4 to 6) D0 
signal to ground(0). However when I try and do so I get a bunch of 
warnings:

1

WARNING:Xst:647 - Input <D0> is never used.

2

Unit <mm6bitMUX> synthesized.

3

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate1>.

4

   It will be removed from the design.

5

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate1>.

6

   It will be removed from the design.

7

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate2>.

8

   It will be removed from the design.

9

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate2>.

10

   It will be removed from the design.

11

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate3>.

12

   It will be removed from the design.

13

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate3>.

14

   It will be removed from the design.

15

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate4>.

16

   It will be removed from the design.

17

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate4>.

18

   It will be removed from the design.

19

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate5>.

20

   It will be removed from the design.

21

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate5>.

22

   It will be removed from the design.

23

WARNING:Xst:1290 - Hierarchical block <gate1> is unconnected in block <gate6>.

24

   It will be removed from the design.

25

WARNING:Xst:1290 - Hierarchical block <gate2> is unconnected in block <gate6>.

26

   It will be removed from the design.

What would the correct way to connect an input to ground? Attached is 
the overall program counter diagram that I need to make. As you can see 
D0 would be connected to '0' and D1 would be the output signal coming 
from the adder.

mike mr wrote:
> gate1  :  mm2to1MUX  PORT MAP(D0=>'0',D1=>D1(3),S=>S,Z=>Z(3));
>
> I'm trying to connect the 6 bit MUX(Ijust incremented it from 4 to 6) D0
> signal to ground(0). However when I try and do so I get a bunch of
> warnings:
>  ......
> What would the correct way to connect an input to ground?
You are doing it the correct way, and the toolchain correctly recognizes 
that it is not necessary to implement a static-'0' path in the mux. So 
this path is optimised away.

To sum it up: thats the information I give my trainees, and thats the 
information your teacher should give you. To bad that he doesn't...