I realize similar questions have been asked here and here -- but they weren't very clear and I just wanted to see if anything has changed since these were asked over a year ago.
Scenario: I have 3 values displayed to the user: A and B are returned from the API, and X is supposed to be the result of subtracting A - B. The values of A and B are floats on the backend, but are parsed to an Integer on the frontend so the user never sees the decimal values, thus Protractor doesn't know their real value (whether they were rounded up or down). So, my value of X will frequently be +/- 1 from the expected value.
Use Case: expect(someVar).toEqual(x || y);
Question: Has Jasmine implemented anything like an OR statement to execute in one expect? Otherwise I need to create a custom matcher.
Would the toBeWithinRange jasmine matcher, coming from the jasmine-matchers help:
expect(someVar).toBeWithinRange(expectedValue - 1, expectedValue + 1);
There is also a related thread with several more options:
Expect item in array
Related
Assume I made a function like below in swi-prolog.
function1(param) :- VALUE is 0, findValue(VALUE), write(VALUE).
However 0 is always printed out. findValue function is logically correct.'
Is it impossible to use calculated VALUE in function1?
Omit VALUE is 0.
By stating it, the function findValue will set the rest of the variables according to VALUE, because the claims you provide first are the axiom-level true ones, and the rest correspond to them.
For example, take
func(X):- X is 0.
If you query func(X). it will result in X=0 because it assumes you want to make an assignment.
However, if you query with a number like func(0). it will check if X==0 or not, resulting in a boolean answer.
I am new to using Halide and I am playing around with implementing algorithms first. I am trying to write a function which, depending on the value of the 8 pixels around it, either skips to the next pixel or does some processing and then moves on to the next pixel. When trying to write this I get the following compiler error:
84:5: error: value of type 'Halide::Expr' is not contextually convertible to 'bool'
if(input(x,y) > 0)
I have done all the tutorials and have seen that the select function is an option, but is there a way to either compare the values of a function or store them somewhere?
I also may be thinking about this problem wrong or might not be implementing it with the right "Halide mindset", so any suggestions would be great. Thank you in advance for everything!
The underlying issue here is that, although they are syntactically interleaved, and Halide code is constructed by running C++ code, Halide code is not C++ code and vice versa. Halide code is entirely defined by the Halide::* data structures you build up inside Funcs. if is a C control flow construct; you can use it to conditionally build different Halide programs, but you can't use it inside the logic of the Halide program (inside an Expr/Func). select is to Halide (an Expr which conditionally evaluates to one of two values) as if/else is to C (a statement which conditionally executes one of two sub-statements).
Rest assured, you're hardly alone in having this confusion early on. I want to write a tutorial specifically addressing how to think about staged programming inside Halide.
Until then, the short, "how do I do what I want" answer is as you suspected and as Khouri pointed out: use a select.
Since you've provided no code other than the one line, I'm assuming input is a Func and both x and y are Vars. If so, the result of input(x,y) is an Expr that you cannot evaluate with an if, as the error message indicates.
For the scenario that you describe, you might have something like this:
Var x, y;
Func input; input(x,y) = ...;
Func output; output(x,y) = select
// examine surrounding values
( input(x-1,y-1) > 0
&& input(x+0,y-1) > 0
&& ...
&& input(x+1,y+1) > 0
// true case
, ( input(x-1,y-1)
+ input(x+0,y-1)
+ ...
+ input(x+1,y+1)
) / 8
// false case
, input(x,y)
);
Working in Halide definitely requires a different mindset. You have to think in a more mathematical form. That is, a statement of a(x,y) = b(x,y) will be enforced for all cases of x and y.
Algorithm and scheduling should be separate, although the algorithm may need to be tweaked to allow for better scheduling.
I am using
rand(200)
in my Rails application.
When I run it in console it always returns random number, but if I use it in application line:
index = rand(200)
index is always the same number.
Why is that and how to overcome this?
Simple pseudo-random number generators actually generate a fixed sequence of numbers. The particular sequence you get is determined by the initial "seed" value. My suspicion is that you are always getting the first number in the same sequence. Therefore I suggest we try to change the sequence by calling srand every time before calling rand, thus changing the seed value every time. The docs explain that, when called without a parameter, srand generates a new seed based on current circumstances (e.g. the time on the clock). Thus you should get a difference sequence and hence a different random number:
srand
rand(200)
Now, you may ask - why are you always getting the same sequence? I have no idea! As someone else suggested in one of the comments, the behavior you are seeing is the behavior one would expect if you had other code, anywhere, that calls srand with the same, fixed value every time. So it might be good to look for that.
Try Random.rand(). For example
Random.rand(200)
Or if you're working with an array you could use sample.
[*1..200].sample
rand(200) is run once and that value is assigned to your index variable. So 'index' will always be that number.
If you want index to change, you will need to continually run rand on it.
Here's a simple way to do that:
def randomIndex(num)
index = rand(num)
return index
end
randomIndex(200)
=> // this value will change
As the title says I'm curious about the difference between "call-by-reference" and "call-by-value-return". I've read about it in some literature, and tried to find additional information on the internet, but I've only found comparison of "call-by-value" and "call-by-reference".
I do understand the difference at memory level, but not at the "conceptual" level, between the two.
The called subroutine will have it's own copy of the actual parameter value to work with, but will, when it ends executing, copy the new local value (bound to the formal parameter) back to the actual parameter of the caller.
When is call-by-value-return actually to prefer above "call-by-reference"? Any example scenario? All I can see is that it takes extra memory and execution time due to the copying of values in the memory-cells.
As a side question, is "call-by-value-return" implemented in 'modern' languages?
Call-by-value-return, from Wikipedia:
This variant has gained attention in multiprocessing contexts and Remote procedure call: if a parameter to a function call is a reference that might be accessible by another thread of execution, its contents may be copied to a new reference that is not; when the function call returns, the updated contents of this new reference are copied back to the original reference ("restored").
So, in more practical terms, it's entirely possible that a variable is in some undesired state in the middle of the execution of a function. With parallel processing this is a problem, since you can attempt to access the variable while it has this value. Copying it to a temporary value avoids this problem.
As an example:
policeCount = 0
everyTimeSomeoneApproachesOrLeaves()
calculatePoliceCount(policeCount)
calculatePoliceCount(count)
count = 0
for each police official
count++
goAboutMyDay()
if policeCount == 0
doSomethingIllegal()
else
doSomethingElse()
Assume everyTimeSomeoneApproachesOrLeaves and goAboutMyDay are executed in parallel.
So if you pass by reference, you could end up getting policeCount right after it was set to 0 in calculatePoliceCount, even if there are police officials around, then you'd end up doing something illegal and probably going to jail, or at least coughing up some money for a bribe. If you pass by value return, this won't happen.
Supported languages?
In my search, I found that Ada and Fortran support this. I don't know of others.
Suppose you have a call by reference function (in C++):
void foobar(int &x, int &y) {
while (y-->0) {
x++;
}
}
and you call it thusly:
int z = 5;
foobar(z, z);
It will never terminate, because x and y are the same reference, each time you decrement y, that is subsequently undone by the increment of x (since they are both really z under the hood).
By contrast using call-by-value-return (in rusty Fortran):
subroutine foobar(x,y):
integer, intent(inout) :: x,y
do while y > 0:
y = y - 1
x = x + 1
end do
end subroutine foobar
If you call this routine with the same variable:
integer, z = 5
call foobar(z,z)
it will still terminate, and at the end z will be changed have a value of either 10 or 0, depending on which result is applied first (I don't remember if a particular order is required and I can't find any quick answers to the question online).
Kindly go to the following link , the program in there can give u an practical idea regarding these two .
Difference between call-by-reference and call-by-value
I have been trying to do ruby string comparision which doesnt seem to work
max == "value"
if user.name == max
I also tried using the eql method but nothing seems to work
max.eql(user.name)
This is not working althought the values are same.
What could be the reason?
This is because of white spaces. Try doing
if user.name.strip == max
strip will remove all the white spaces
Ruby use the same semantics as C when it comes to assignment versus comparison.
x = y
will assign x the value of y Even if this is done inside an if expression.
The second attempt to use eql (which really should be eql?) will fail, as x.eql?(y) returns true if the x and y are the same object. It is not sufficient that they have the same value.
Note, that in a language like Ruby, many variables can be bound to the same object. If you update the object destructively, this will be reflected on all variables bound to the same object. On the other hand, it will not affect variables bound the another object, even if that object happened to have an equal value as the first object.
Update: The poster changed the question after this answer was posted.