The following code has to check for an 'e' value such that the gcd(h,e)=1. Where 1
module great(p,q,e,d);
input p,q;
output e,d;
reg e,d;
h=((p-1)*(q-1));
always
begin
for(e=2;e<h;e=e+1)
begin
g1=gcd(h,e);
if(g1==1)
return e;
If, by "return a value", you mean spit out a value you can use in another module, you would use the output of this module as your "return" value. But even ignoring the return e, I don't think your code will work if you tried to run it, because it is too much like a programming language. There's several major things wrong:
You already declared output e,d so you can't declare two reg with the same name. You probably want output reg e,d instead.
You didn't declare a type for h or g1.
You have a for loop for e but e can never be anything other than 0 or 1 because you didn't set a size for it, so by default it is only 1-bit long. Even if it was big enough that you could increment it past 1, it's a wire type by default, and you can't make those kind of increments to a wire directly.
I assume gcd is some module you made somewhere else, but this isn't how you interconnect modules together. You can't call it like it's a function. You have to use wire and reg to connect the inputs and outputs of two modules together, almost like you're plugging components in.
Those are what stick out the most to me, anyway. I think you are coding your Verilog as if it were Python and that's what's causing these misunderstandings. Verilog is very, very different.
Related
I'm new to coding and I recently started to take my baby steps in LUA. I have a small problem so it would be very helpful if you can help me. In my code, I need to code that
If x ~= 1 and x~=2 and x~=3 and x~=4 then (do something) end
is there a faster way not to hardcode that part, not to type the whole thing from x~=1 to x~=4?
Thank you!
If you need something like if x ~= 1 and x~=2 and x~=3 and x~=4 then (do something) end x is usually an integer.
Then
if x < 1 or x > 4 then
-- do your stuff here
end
Is what you are looking for. If you want to explicitly check wether x is unquald 1,2,3,4 you can simply do something like Egor suggested.
But as you see unless you can describe your conditions in a shorter mathematical way you still have separate unique conditions and you won't come around writing them down.
If you have to check those conditions repeatedly you can use a truth table like in Egor's example or you write a function that returns if that condition is met for its argument.
set var1 A
set var2 {A}
Is it possible to check if variable is list in TCL? For var1 and var2 llength gives 1. I am thinking that these 2 variables are considered same. They are both lists with 1 element. Am I right?
Those two things are considered to be entirely identical, and will produce identical bytecode (except for any byte offsets used for indicating where the content of constants are location, which is not information normally exposed to scripts at all so you can ignore it, plus the obvious differences due to variable names). Semantically, braces are a quoting mechanism and not an indicator of a list (or a script, or …)
You need to write your code to not assume that it can look things up by inspecting the type of a value. The type of 123 could be many different things, such as an integer, a list (of length 1), a unicode string or a command name. Tcl's semantics are based on you not asking what the type of a value is, but rather just using commands and having them coerce the values to the right type as required. Tcl's different to many other languages in this regard.
Because of this different approach, it's not easy to answer questions about this in general: the answers get too long with all the different possible cases to be considered in general yet most of it will be irrelevant to what you're really seeking to do. Ask about something specific though, and we'll be able to tell you much more easily.
You can try string is list $var1 but that will accept both of these forms - it will only return false on something that can't syntactically be interpreted as a list, eg. because there is an unmatched bracket like "aa { bb".
In the source-code for nanodns, there is an atypical use of the ternary operator in an attempt to reduce the size of the code:
/* If the incoming packet has an AR record (such as in an EDNS request),
* mark the reply as "NOT IMPLEMENTED"; using a?b:c form to save one byte*/
q[11]?q[3]|=4:1;
It’s not obvious what this line does. At first glance, it looks like it is assigning a value to one of two array elements, but it is not. Rather, it seems to be either or’ing an array element, or else, doing nothing (running the “command” 1).
It looks like it is supposed to be a replacement for this line of code (which is indeed one byte longer):
if(q[11])q[3]|=4;
The literal equivalent would be this:
if (q[11])
q[3]|=4;
else
1;
The ternary operator is typically used as part of an expression, so seeing it used as a standalone command seems odd. Coupled with the seemingly out of place 1, this line almost qualifies as obfuscated code.
I did a quick test and was able to compile and run a C(++) program with data constants as “command”, such as void main() {0; 'a'; "foobar"; false;}. It seems to be a sort of nop command, but I cannot find any information about such usage—Google isn’t very amenable to this type of search query).
Can anyone explain exactly what it is and how it works?
In C and C++ any expression can be made into a statement by putting ; at the end.
Another example is that the expression x = 5 can be made into a statement: x = 5; . Hopefully you agree that this is a good idea.
It would needlessly complicate the language to try and "ban" some subset of expressions from having ; come after them. This code isn't very useful but it is legal.
Please note that the code you linked to is awful and written by a really bad programmer. Particularly, the statement
"It is common practice in tiny C programs to define reused expressions
to make the code smaller"
is complete b***s***. That statement is where things started to go terribly wrong.
The size of the source code has no relation to the size of the compiler executable, nor any relation to that executable's memory consumption, nor any relation to program performance. The only thing it affects is the size of the source code files on the programmers computer, expressed in bytes.
Unless you are programming on some 8086 computer from mid-80s with very limited hard drive space, you never need to "reduce the size of the code". Instead, write readable code.
That being said, since q is an array of characters , the code you linked is equivalent to
if(q[11])
{
(int)(q[3] |= 4);
}
else
{
1;
}
Where 1 is a statement with no side effect, it will get optimized away. It was only placed there because the ?: operator demands a 3rd operator.
The only difference between if statements and the ?: operator is subtle: the ?: implicitly balances the type between the 2nd and 3rd operand.
To increase readability and produce self-documenting code, the code should get rewritten to something like
if (q[AR_INDEX] != 0)
{
q[REPLY_INDEX] |= NOT_IMPLEMENTED;
}
As a side note, there is a bug here: q[2]|=128;. q is of type char, which has implementation-defined signedness, so this line is potentially disastrous. The core problem is that you should never use the char type for bit-wise operations or any form of arithmetic, which is a classic beginner mistake. It must be replaced with uint8_t or unsigned char.
I've got experience in a lot of other programming languages, but I'm having a lot of difficulty with Stata syntax. I've got a statement that evaluates with no problem if I put in values, but I can't figure out why it's not evaluating variables like I expect it to.
gen j=5
forvalues i = 1(1)5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
replace j=`j'-1
}
If I replace i and j with 1 and 5 respectively, like I'm expecting to happen from the code above, then it works fine, but I get an if not found error otherwise, which hasn't produced meaningful results when Googled. Does anyone see what I don't see? I hate to brute-force something that could so simply be done with a loop.
Easy to understand once you approach it the right way!
Problem 1. You never defined local macro j. That in itself is not an error, but it often leads to errors. Macros that don't exist are equivalent to empty strings, so Stata sees in this example the code
if TrustBusiness_local2==`j'
as
if TrustBusiness_local2==
which is illegal; hence the error message.
Problem 2. There is no connection of principle between a variable you called j and a local macro called j but referenced using single quotes. A variable in Stata is a variable (namely, column) in your dataset; that doesn't mean a variable otherwise in the sense of any programming language. Variables meaning single values can be held in Stata within scalars or within macros. Putting a constant into a variable, Stata sense, is legal, but usually bad style. If you have millions of observations, for example, you now have a column j with millions of values of 5 within it.
Problem 3. You could, legally, go
local j "j"
so that now the local macro j contains the text "j", which depending on how you use it could be interpreted as a variable name. It's hard to see why you would want to do that here, but it would be legal.
Problem 4. Your whole example doesn't even need a loop as it appears to mean
replace TrustBusiness_local= 6 - TrustBusiness_local2 if inlist(TrustBusiness_local2, 1,2,3,4,5)
and, depending on your data, the if qualifier could be redundant. Flipping 5(1)1 to 1(1)5 is just a matter of subtracting from 6.
Problem 5. Your example written as a loop in Stata style could be
local j = 5
forvalues i = 1/5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
local j=`j'-1
}
and it could be made more concise, but given Problem 4 that no loop is needed, I will leave it there.
Problem 6. What you talking about are, incidentally, not if statements so far as Stata is concerned, as the if qualifier used in your examples is not the same as the if command.
The problem of translating one language's jargon into another can be challenging. See my comments at http://www.stata.com/statalist/archive/2008-08/msg01258.html After experience in other languages, the macro manipulations of Stata seemed at first strange to me too; they are perhaps best understood as equivalent to shell programming.
I wouldn't try to learn Stata by Googling. Read [U] from beginning to end. (A similar point was made in the reply to your previous question at use value label in if command in Stata but you don't want to believe it!)
In the Stack Overflow podcast #36 (https://blog.stackoverflow.com/2009/01/podcast-36/), this opinion was expressed:
Once you understand how easy it is to set up a state machine, you’ll never try to use a regular expression inappropriately ever again.
I've done a bunch of searching. I've found some academic papers and other complicated examples, but I'd like to find a simple example that would help me understand this process. I use a lot of regular expressions, and I'd like to make sure I never use one "inappropriately" ever again.
A rather convenient way to help look at this to use python's little-known re.DEBUG flag on any pattern:
>>> re.compile(r'<([A-Z][A-Z0-9]*)\b[^>]*>(.*?)</\1>', re.DEBUG)
literal 60
subpattern 1
in
range (65, 90)
max_repeat 0 65535
in
range (65, 90)
range (48, 57)
at at_boundary
max_repeat 0 65535
not_literal 62
literal 62
subpattern 2
min_repeat 0 65535
any None
literal 60
literal 47
groupref 1
literal 62
The numbers after 'literal' and 'range' refer to the integer values of the ascii characters they're supposed to match.
Sure, although you'll need more complicated examples to truly understand how REs work. Consider the following RE:
^[A-Za-z][A-Za-z0-9_]*$
which is a typical identifier (must start with alpha and can have any number of alphanumeric and undescore characters following, including none). The following pseudo-code shows how this can be done with a finite state machine:
state = FIRSTCHAR
for char in all_chars_in(string):
if state == FIRSTCHAR:
if char is not in the set "A-Z" or "a-z":
error "Invalid first character"
state = SUBSEQUENTCHARS
next char
if state == SUBSEQUENTCHARS:
if char is not in the set "A-Z" or "a-z" or "0-9" or "_":
error "Invalid subsequent character"
state = SUBSEQUENTCHARS
next char
Now, as I said, this is a very simple example. It doesn't show how to do greedy/nongreedy matches, backtracking, matching within a line (instead of the whole line) and other more esoteric features of state machines that are easily handled by the RE syntax.
That's why REs are so powerful. The actual finite state machine code required to do what a one-liner RE can do is usually very long and complex.
The best thing you could do is grab a copy of some lex/yacc (or equivalent) code for a specific simple language and see the code it generates. It's not pretty (it doesn't have to be since it's not supposed to be read by humans, they're supposed to be looking at the lex/yacc code) but it may give you a better idea as to how they work.
Make your own on the fly!
http://osteele.com/tools/reanimator/???
This is a really nicely put together tool which visualises regular expressions as FSMs. It doesn't have support for some of the syntax you'll find in real-world regular expression engines, but certainly enough to understand exactly what's going on.
Is the question "How do I choose the states and the transition conditions?", or "How do I implement my abstract state machine in Foo?"
How do I choose the states and the transition conditions?
I usually use FSMs for fairly simple problems and choose them intuitively. In my answer to another question about regular expressions, I just looked at the parsing problem as one of being either Inside or outside a tag pair, and wrote out the transitions from there (with a beginning and ending state to keep the implementation clean).
How do I implement my abstract state machine in Foo?
If your implementation language supports a structure like c's switch statement, then you switch on the current state and process the input to see which action and/or transition too perform next.
Without switch-like structures, or if they are deficient in some way, you if style branching. Ugh.
Written all in one place in c the example I linked would look something like this:
token_t token;
state_t state=BEGIN_STATE;
do {
switch ( state.getValue() ) {
case BEGIN_STATE;
state=OUT_STATE;
break;
case OUT_STATE:
switch ( token.getValue() ) {
case CODE_TOKEN:
state = IN_STATE;
output(token.string());
break;
case NEWLINE_TOKEN;
output("<break>");
output(token.string());
break;
...
}
break;
...
}
} while (state != END_STATE);
which is pretty messy, so I usually rip the state cases out to separate functions.
I'm sure someone has better examples, but you could check this post by Phil Haack, which has an example of a regular expression and a state machine doing the same thing (there's a previous post with a few more regex examples in there as well I think..)
Check the "HenriFormatter" on that page.
I don't know what academic papers you've already read but it really isn't that difficult to understand how to implement a finite state machine. There are some interesting mathematics but to idea is actually very trivial to understand. The easiest way to understand an FSM is through input and output (actually, this comprises most of the formal definition, that I won't describe here). A "state" is essentially just describing a set of input and outputs that have occurred and can occur from a certain point.
Finite state machines are easiest to understand via diagrams. For example:
alt text http://img6.imageshack.us/img6/7571/mathfinitestatemachinedco3.gif
All this is saying is that if you begin in some state q0 (the one with the Start symbol next to it) you can go to other states. Each state is a circle. Each arrow represents an input or output (depending on how you look at it). Another way to think of an finite state machine is in terms of "valid" or "acceptable" input. There are certain output strings that are NOT possible certain finite state machines; this would allow you to "match" expressions.
Now suppose you start at q0. Now, if you input a 0 you will go to state q1. However, if you input a 1 you will go to state q2. You can see this by the symbols above the input/output arrows.
Let's say you start at q0 and get this input
0, 1, 0, 1, 1, 1
This means you have gone through states (no input for q0, you just start there):
q0 -> q1 -> q0 -> q1 -> q0 -> q2 -> q3 -> q3
Trace the picture with your finger if it doesn't make sense. Notice that q3 goes back to itself for both inputs 0 and 1.
Another way to say all this is "If you are in state q0 and you see a 0 go to q1 but if you see a 1 go to q2." If you make these conditions for each state you are nearly done defining your state machine. All you have to do is have a state variable and then a way to pump input in and that is basically what is there.
Ok, so why is this important regarding Joel's statement? Well, building the "ONE TRUE REGULAR EXPRESSION TO RULE THEM ALL" can be very difficult and also difficult to maintain modify or even for others to come back and understand. Also, in some cases it is more efficient.
Of course, state machines have many other uses. Hope this helps in some small way. Note, I didn't bother going into the theory but there are some interesting proofs regarding FSMs.