I am using Arena simulation software. I need to define "if" in the expression section of the Process Module. I could not find any instruction to define "ifelse" function.
I never found IF statement in Arena simulation.
But don't worry. There is workaround with using expressions.
According to official Documentation:
Logical Expression Evaluation
ASSIGN: InventoryLevel = 50 *
(SystemStatus==Early) +
30 * (SystemStatus==Late);
Assign InventoryLevel a value of 50 if the variable SystemStatus is
equal to Early (logical expressions evaluate to 1 for TRUE, 0 for
FALSE). If SystemStatus equals Late, assign InventoryLevel to 30. If
neither is true, assign SystemStatus equal to 0. This single ASSIGN
block may have been used to replace a BRANCH block (to check the value
of SystemStatus) and two ASSIGN blocks (to assign the correct value).
Related
I am wondering why ballotARB(false) returns always 0 and not return the bitfield where the corresponding bits are set for all active invocations in the sub-group.
Because of that, I do not see how can I use ballotARB fonction in my program.
Is my understanding of this function false?
What should ballotARB(false) return, if not 0?
It does return the count of the true votes in the local subgroup. Since all your invocations vote false no matter what, the result must be 0. This is even explicitely stated in the ARB_shader_ballot extension spec:
The following trivial assumptions can be made:
ballotARB(true) returns bitfield where the corresponding bits are set for all active invocations in the sub-group.
ballotARB(false) returns zero.
So to make use of the votes, you should use it on an experession which is actually not dynamically uniform.
In Z3, how can we write a program to get the result from evaluation? By default model.eval(expression) will return another expression of the evaluation result. How can I assign the result to a type-specific data? Below is what I want to do in my program.
int a = model.eval(x + 1) // compiler error
Sometimes models are not complete. For instance, when nothing depends on the value of x, then Z3 may not assign any value at all to it, i.e., you're free to chose whatever value suits you. The eval function has a second argument, which, when set to true, will enable model completion, i.e., eval will substitute those don't-cares with some legal value (often 0).
Z3-ints are actual integers, not C/C++-ints smaller than 2^32-1, so the conversion is not performed automatically. If you know that in your application this will always be ok, and that eval will always return a numeral, then you can use Z3_get_numeral_int to perform that conversion.
I want to know wich of these options is correct.
An atomic register R initially holding value 33 is used by two process P and Q that perform the
following concurrent operations: P executes write(R,68) during time interval [2,6] and Q executes
read(R) during time interval [4,7] (the operations overlap in time). In this situation, since the
register provides the atomic semantics, it is guaranteed that:
(A) The read operation always returns value 68.
(B) The read operation always returns value 33.
(C) The read operation can return either value 33 or value 68.
(D) Nothing can be ensured, because the operations are concurrent.
I know atomic registers ensure that
if Ri → Rj then i j (if i is before j)
During a concurrent read and write, the read can return either the old value or the new value. To maintain the properties of an atomic register, during the write, there is a point in time:
before which 33 is always returned, and
after which 68 is always returned.
Therefore, option (C) is correct.
You can read an explanation here:
What's the difference between safe, regular and atomic registers?
The notable piece of information is
Readers that act at a point before that point will all read the old value and readers that act after that point will all read the new value
So by definition, since the read occurs before the write completes it should always see 33. If it were a regular register, it could flicker between the two.
I need to find in an image if a pixel has the maximum value compared to the 8 pixels around it.
I am not sure what is the most optimal way, so my idea is to use the if statement like this:
if(pixel > pixel1 && pixel > pixel2 && pixel > pixel3 && ... && pixel> pixel8)
My question is the following: if it found that for instance pixel is not bigger than pixel1, will it still check the rest of the statement or since it's only ANDs, it will already discard the instruction and go further?
And if the answer is the first one, that would make it very computationally heavy to check each pixel all the time, can somebody give me a hint as how to approach more efficiently this simple problem?
This is called Short Circuit Evaluation.
the second argument is only executed or evaluated if the first argument does not suffice to determine the value of the expression
Since the condition is &&, it will NOT check further if it gets a false in any of the conditions.
Similarly if the condition were ||, it would stop checking once it finds a true.
Btw, I am not absolutely certain of the precedence rules, and because of that I would surround each condition in parentheses just to be safe.
if((pixel > pixel1) && (pixel > pixel2) && ...
Edit: Operator precedence rules seem to indicate that the parentheses in this case are unnecessary.
No, it won't check the rest of the statements. C++ "short-circuits" conditional statements, ignoring the second operand to an && if the first is false (and ignoring the second operand to a || if the second is true).
The operators && and || are so-called 'short circuit operators' in C++ (and in most other languages as well). This means that evaluation will stop as soon as the result can be determined. For &&, this means that evaluation of other terms will stop if one term is false, because then the answer is false, independent of the other terms. Conversely, for || this means that evaluation of other terms will stop if one term is true.
See also this link.
Think of it not as a series but a grouping of expressions so && has just a left and right side, and is left-side associative.
If the left hand side evaluates to false it is guaranteed by the standard not to evaluate what is on the right hand side. The right hand side might even contain an access violation (and often does), e.g. checking if a pointer is non-null on the left side, then dereferencing it on the right.
Your operation is O(N) at worst. If you do this once, it is the optimal way, if you are going to do this a lot, you'd be better off finding the max value of your pixels then just checking against that one.
There is "short-circuit" in C++ that means when first condition satisfies if then the second condition will not checked.
For example if pixel > pixel1 results false the following conditions will be ignored.
I refer you to this "Short circuit evaluation"
While short circuit evaluation has been explained in other answers, it's worth pointing out that a comparison of two pixels may not be completely trivial. For example, you may wish to add red, green and blue pixel values after multiplying them by a weighting factor (as the human eye is more sensitive to some colours than others)... in that case, if you don't preserve the overall pixel value inside the object being compared (thereby using more memory both for that value and to somehow track when it's invalidated, + CPU time to check & regenerate it when necessary), then you'll have to perform this redundant calculation during every one of those comparisons. To avoid this, you might - for example - add a "get_brightness()" function that returns a user-defined type that can be compared efficiently with each of the other pixels.
So sample loop:
do i=1,1
print *,i
enddo
print *,i
gives me 2 as the final value of i. How can I set up Intel Fortran for Visual Studio on Windows to give me a final value of 1 for i?
This has been the way that Fortran loops work for decades and you can't simply change this with a compiler option. The Fortran standard clearly states:
8.1.4.4.1 Loop initiation
(2) The DO variable becomes defined with the value of the initial parameter m1.
(3) The iteration count is established and is the value of the expression
MAX (INT ((m2 – m1 + m3) / m3), 0)
Here m1, m2 and m3 are the three parameters in the loop-control: [,] var = m1,m2[,m3], Given your example of i=1,1 (m3 is implicitly 1 if omitted) the iteration count is MAX(INT((1-1+1)/1),0) which evaluates to 1, i.e. the loop should get executed once. i is initialised to 1 as per (2).
8.1.4.4.2 The execution cycle
The execution cycle of a DO construct consists of the following steps performed in sequence repeatedly until termination:
(1) The iteration count, if any, is tested. If the iteration count is zero, the loop terminates and the DO construct becomes inactive. If loop-control is [ , ] WHILE (scalar-logical-expr), the scalar-logicalexpr is evaluated; if the value of this expression is false, the loop terminates and the DO construct becomes inactive. If, as a result, all of the DO constructs sharing the do-term-shared-stmt are inactive, the execution of all of these constructs is complete. However, if some of the DO constructs sharing the do-term-shared-stmt are active, execution continues with step (3) of the execution cycle of the active DO construct whose DO statement was most recently executed.
Fortran tests if the remaining iteration count is greater than zero, not if the DO variable is less than (greater than) the end value.
(2) If the iteration count is nonzero, the range of the loop is executed.
(3) The iteration count, if any, is decremented by one. The DO variable, if any, is incremented by the value of the incrementation parameter m3.
The DO variable is always incremented as an iteration of the loop is being executed. Thus after the first execution i becomes incremented by 1 which evaluates to 2.
Except for the incrementation of the DO variable that occurs in step (3), the DO variable must neither be redefined nor become undefined while the DO construct is active.
8.1.4.4.4 Loop termination
When a DO construct becomes inactive, the DO-variable, if any, of the DO construct retains its last defined value.
The last defined value is 2. Thus after the DO loop has ended, i is equal to 2.
I've pulled the text out of ISO/IEC 1539:1991 (a.k.a. Fortran 90) but one can also find pretty much the same text in §11.10.3 of ISO/IEC 1539:1980 (a.k.a. ANSI X3J3/90.4 a.k.a. FORTRAN 77; sans the WHILE stuff which is not present in F77) as well as in §8.1.6.6 of ISO/IEC 1539-1:2010 (a.k.a. Fortran 2008).
You can't, because that's how DO works; it stops when the control variable exceeds the limit.
In general, in pretty much any language with a FOR/DO counting loop, you should only use the loop control variable inside the loop body, and treat it as undefined elsewhere, even if you can't actually limit its scope to the body.
In your case, I would use a different variable to keep track of the actual last value of i in any iteration:
lasti = 0
do i=1,1
print *,i
lasti = i
enddo
print *,lasti