How do I add an offset to a value obtained from an IndexOf expression? That is, how do I do this?
> import z3
> s = 'hello'
> t = 'e'
> z3.simplify(z3.IndexOf(s, t, 0) + z3.IntVal(1))
z3.z3types.Z3Exception: Non-sequence passed as a sequence
I want to get the location after than of e.
On the other hand, switching the order works as expected
> z3.simplify(z3.IntVal(1) + z3.IndexOf(s, t, 0))
2
You found a bug in z3py!
The bug is on this line: https://github.com/Z3Prover/z3/blob/master/src/api/python/z3/z3.py#L10150
which reads:
return SeqRef(Z3_mk_seq_index(s.ctx_ref(), s.as_ast(), substr.as_ast(), offset.as_ast()), s.ctx)
Instead, it should say:
return ArithRef(Z3_mk_seq_index(s.ctx_ref(), s.as_ast(), substr.as_ast(), offset.as_ast()), s.ctx)
I've reported this at their bug tracker: https://github.com/Z3Prover/z3/issues/2159
Once you make that change in your local-copy of z3.py, your program should work as is. Or you can wait till they release a fix.
Related
Will following code be executed atomically?
const int oldId = id.exchange((id.load()+1) % maxId);
Where id is std::atomic<int>, and maxId is some integer value.
I searched google and stackoverflow for std::atomic modulo increment. And I found some topics but I can't find clear answer how to do that properly.
In my case even better would be to use:
const int newId = id.exchange((++id) % maxId);
But I am still not sure if it will be executed atomically.
No, this is not atomic, because the load() and the exchange() are separate operations, and nothing is preventing id from getting updated after the load, but before the exchange. In that case your exchange would write a value that has been calculated based on a stale input, so you end up with a missed update.
You can implement a modulo increment using a simple compare_exchange loop:
int val = id.load();
int newVal = (val + 1) % maxId;
while (!id.compare_exchange_weak(val, newVal) {
newVal = (val + 1) % maxId;
}
If the compare_exchange fails it performs a reload and populates val with the updated value. So we can re-calculate newVal and try again.
Edit:
The whole point of the compare-exchange-loop is to handle the case that between the load and the compare-exchange somebody might change id. The idea is to:
load the current value of id
calculate the new value
update id with our own value if and only if the value currently stored in id is the same one as we read in 1. If this is the case we are done, otherwise we restart at 1.
compare_exchange is allows us to perform the comparison and the conditional update in one atomic operation. The first argument to compare_exchange is the expected value (the one we use in our comparison). This value is passed by reference. So when the comparison fails, compare_exchange automatically reloads the current value and updates the provided variable (in our case val).
And since Peter Cordes pointed out correctly that this can be done in a do-while loop to avoid the code duplication, here it is:
int val = id.load();
int newVal;
do {
newVal = (val + 1) % maxId;
} while (!id.compare_exchange_weak(val, newVal);
So I have this piece of code and it is this:
do
local function index(n,m)
return n*(n+1)//2 + m
end
local binomtable = {}
function binom3(n,m)
if n<0 or m<0 or m>n then return 0 end
if n=0 or m=0 or m=n then return 1 end
local i = index(n,m)
local v = binomtable[i]
if v then return v end
v = binom3(n-1,m-1) + binom3(n-1,m)
binomtable[i] = v
return v
end
end
and I would like to know what
if v then return v end
means.
Thank you!
The short answer is that if v then return v end returns the value v if it is truthy, i.e., if it is neither false nor nil. Otherwise the function continues by calculating a value for v, storing that value in binomtable, and finally returning it. The more interesting question is, why is the function doing all of this?
In the posted code, binom3 is a recursive function. With the recursive calls v = binom3(n-1,m-1) + binom3(n-1,m) there will be a lot of duplication of effort, meaning a lot of wasted space and time. Consider:
binom3(4, 2)
--> binom3(3, 1) + binom3(3, 2)
--> binom3(2, 0) + binom3(2, 1) + binom3(2, 1) + binom3(2, 2)
--> 1 + binom3(1, 0) + binom3(1, 1) + binom3(1, 0) + binom3(1, 1) + 1
Note how in the second reduction there are two identical terms:
binom3(2, 1) + binom3(2, 1)
There is no reason to calculate the term binom3(2, 1) twice, and doing so means that the pair of terms:
binom3(1, 0) + binom3(1, 1)
also must be calculated twice, as seen in the third reduction. It would be smart to calculate binom3(2, 1) only once, and to save the result for later use in the larger calculation. When m and n are larger and the number of calculations explodes exponentially this becomes a very important issue for performance both in the amount of memory required and in the amount of time required.
The posted code is using memoization to improve performance. When a calculation is made, it is stored in the table binomtable. Before any calculation is made, binomtable is consulted. First, v is set to the value of binomtable[i]; if this value is any truthy value (any integer is a truthy in Lua), then that value is simply returned without the need for recursive calculation. Otherwise, if nil is returned (i.e., no value has yet been stored for the calculation), the function continues with a recursive calculation. After completing the calculation, the new value is stored in binomtable for use the next time it is needed. This strategy saves a lot of wasted computational effort, and can make a huge difference in the performance of such recursive algorithms.
For your specific question of what
if v then return v end
means, is that if v, a variable, is not nil or false it is to return the value of the v variable and stop executing that function.
--Similar
function myfunc(input)
local MyVar = "I am a string and am not nil!"
if MyVar then
return "hi"
else
return "hello"
end
print("I am not seen because I am unreachable code!")
end
if this function was called it would always return "hi" instead of "hello" because MyVar is true, because it has a value. Also the print function below that will never get called because it stops executing the function after a return is called.
Now for your codes case it is checking a table to see if it has an entry at a certain index and if it does it returns the value.
i am new with f# , will be great if some 1 can help , nearly half a day gone solving this problem Thank you
module Certificate =
type T = {
Id: int
IsECert: bool
IsPrintCert: bool
CertifiedBy: string
Categories: Category.T list
}
let createPending now toZonedDateTime toBeCertifiedByName (job: Models.Job.T) (certificateType: Models.CertificateType.T) (pendingCertificate: Models.PendingCertificate.T) visualization (categories: Category.T list) =
let forCompletion = Models.PendingCertificate.getCertificateForCompletion pendingCertificate
{
Id = forCompletion.Id |> CertificateId.toInt
IsECert = Models.PendingCertificate.isECertificate pendingCertificate
IsPrintCert = Models.PendingCertificate.isPrintCertificate pendingCertificate
CertifiedBy = toBeCertifiedByName
Categories = categories}
i am getting an error in "Incomplete structured construct at or before this point"
Your formatting is all off. I will assume here that this is just a result of posting to StackOverflow, and your actual code is well indented.
The error comes from the definition of createPending: this function does not have a result. All its body consists of defining a forCompletion value, but there is nothing after it. Here's a simpler example that has the same problem:
let f x =
let y = 5
This function will produce the same error, because it also doesn't have a result. In F#, every function has to return something. The body cannot contain only definitions of helper functions or values. For example, I could fix my broken function above like this:
let f x =
let y = 5
x + y
This function first defines a helper value y, then adds it to its argument x, and returns the result.
> f 2
> 7
>
> f 0
> 5
How exactly you need to fix your function depends on what exactly you want it to mean. I can't help you here, because you haven't provided that information.
I have some trouble compiling lodepng (http://lodev.org/lodepng/) for D into my project.
In Encode.d I have the following code, where the compiler does not expect the assert statements.
Removing this block solves the issue.
invariant
{
assert(compressionLevel >=0 && compressionLevel <= 9, "invalid zlib compression level");
assert(targetColorType == ColorType.Any ||
targetColorType == ColorType.RGB ||
targetColorType == ColorType.RGBA, "colortype is not supported");
}
In Decode.d I have even more trouble, with the error "no identifier for declarator inout(value)"
for the middle line in:
info.backgroundColor.length = chunk.data.length / 2;
foreach(index, inout value; info.backgroundColor)
value = chunk.data[index * 2];
Is there some trouble with old syntax here, and how do I fix it?
Is there some other way to create png images in D in a simple manner?
Not sure about the invariant problem, but the second problem should be solved by replacing the "inout" with "ref" (D2 syntax change).
I gave up on lodepng, and used the code on
http://www.dsource.org/projects/plot2kill/browser/trunk/png.d
which works after some minor changes.
I'm trying to get Velocity to output the following Javascript code:
if ((whichOne+1) <= numCallouts ) {
whichOne = whichOne + 1; } else {
whichOne = 1;
}
Whenever I try to get Velocity to print a > or a <, it represents it as a & gt; or & lt;, which doesn't help me since I'm trying to get it to produce Javascript. I've tried:
#set ( $gt = ">" )
But even that ends up as a & gt;
Thanks in advance.
It is not a default behavior, the only reason I can think of why this is happening is if you have event ReferenceInsertionEventHandler configured with EscapeHtmlReference either in your velocity.config or in the Velocity initialization code.
Here is more info about events
I've had the same issue with Velocity - however, the problem is that I was using Velocity as an third party embedded language, and didn't have access to change the Velocity settings.
Unfortunately the only solution I was able to find was to rewrite the code without using greater than/less than explicitly, which admittedly is awful, but it's all about getting it to work...
Here is an example workaround for conditionals where you are trying to see if one number is larger than another:
if (n1 > n2) //Doesn't work because velocity turns this into if (n1 > n2)
if (n1 != n2)
{
diff = n1 - n2;
abs = abs(n1 - n2);
if (diff / abs == 1) //Greater than
else //if == -1 then less than
}
else //Equal
Maybe you are able to use the alternate symbols as described here :
http://velocity.apache.org/engine/devel/vtl-reference-guide.html#aifelseifelse_-_Output_conditional_on_truth_of_statements
So try to use if (n1 gt n2).