I'm writing a program for my TI-nspire calculator in TI-BASIC, an optimised version of BASIC. From what I can tell, TI-BASIC is a compiled language. I have had more experience working with scripting languages, where you can define i as i+1, where the interpreter adds the previous value of i to 1 and makes that the new value of i. But since BASIC, from my understanding, is compiled, the calculator will set the value of i to the equation of i+1 and loop. Is there a way to set the value of i to the outcome instead of the equation?
You are wrong, it is perfectly fine to reference a variable in assigning a value to the same variable, it does not result in a loop. However, in TI-Basic you do not use the = operator to assign a value to a variable.
For z80 and 68k calculators use the →, character like this:
Local x
2→x
x+1→x
Return x
This returns 3. (Tested on a TI-89.)
On an TI-nspire use :=, like this:
Local x
x:=2
x:=x+1
Return x
This also returns 3.
Your understanding is wrong. Compilation does not change the semantics of an assignment. It is still an assignment.
And then, what number would the compiler use as the solution for i = i + 1?
Related
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.
Why doesn't this code work?
b if b = true
Error: undefined local variable or method `b'
But this does:
if b = true
b
end
Shouldn't they be the same?
This is a very good question. It has to do with the scoping of variables in Ruby.
Here is a post by Matz on the Ruby bug tracker about this:
local variable scope determined up to down, left to right. So a local variable first assigned in the condition of if modifier is not effective in the left side if body. It's a spec.
In the first version as soon as k is hit, the parser pukes because it hasn't been seen yet.
In the second version, k is part of an assignment expression, and is parsed differently.
I don't know the reason but the problem that the interpreter tries to lookup the variable k before evaluating the condition.
If you write it like this, there won't be any error and works as you expected:
k = nil
h = {k: 1}
v = k if k = h.delete(:k)
you have put only one '='
Try with '=='
Then you will get error
In second example, you are assigning 'true' to b.
Because the Ruby interpreter creates a local variable when it sees an assignment
In the second case, it hasn't yet seen the assignment, so the variable doesn't exist when the expression is parsed.
To be more precise, a method is first parsed into an internal representation, and then, perhaps, the code will eventually be called and actually executed.
Local variables are created in that parsing pass. It's a matter of declaration, it just means that the interpreter becomes aware of them. They won't be created in the sense of being given space or a value until the surrounding method is called by someone.
It has been a while since I've used Mathematica, and I looked all throughout the help menu. I think one problem I'm having is that I do not know what exactly to look up. I have a block of code, with things like appending lists and doing basic math, that I want to define as a single variable.
My goal is to loop through a sequence and when needed I wanted to call a block of code that I will be using several times throughout the loop. I am guessing I should just put it all in a loop anyway, but I would like to be able to define it all as one function.
It seems like this should be an easy and straightforward procedure. Am I missing something simple?
This is the basic format for a function definition in Mathematica.
myFunc[par1_,par2_]:=Module[{localVar1,localVar2},
statement1; statement2; returnStatement ]
Your question is not entirely clear, but I interpret that you want something like this:
facRand[] :=
({b, x} = Last#FactorInteger[RandomInteger[1*^12]]; Print[b])
Now every time facRand[] is called a new random integer is factored, global variables b and x are assigned, and the value of b is printed. This could also be done with Function:
Clear[facRand]
facRand =
({b, x} = Last#FactorInteger[RandomInteger[1*^12]]; Print[b]) &
This is also called with facRand[]. This form is standard, and allows addressing or passing the symbol facRand without triggering evaluation.
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.
i know there is a build-in function findall/3 in prolog,
and im trying to find the total numbers of hours(Thrs) and store them in a list, then sum the list up. but it doesnt work for me. here is my code:
totalLecHrs(LN,THrs) :-
lecturer(LN,LId),
findall(Thrs, lectureSegmentHrs(CC,LId,B,E,THrs),L),
sumList(L,Thrs).
could you tell me what's wrong with it? thanks a lot.
You need to use a "dummy" variable for Hours in the findall/3 subgoal. What you wrote uses THrs both as the return value for sumList/2 and as the variable to be listed in L by findall/3. Use X as the first argument of findall and in the corresponding subgoal lectureSegmentHrs/5 as the last argument.
It looks like the problem is that you're using the same variable (Thrs) twice for different things. However it's hard to tell as you've also used different capitalisation in different places. Change the findall line so that the initial variable has the same capitalisation in the lectureSegmentHrs call. Then use a different variable completely to get the final output value (ie the one that appears in sumList and in the return slot of the entire predicate).
You need to use a different variable because Prolog does not support variable reassignment. In a logical language, the notion of reassigning a variable is inherently impossible. Something like the following may seem sensible...
...
X = 10,
X = 11,
...
But you have to remember that , in Prolog is the conjunction operator. You're effectively telling Prolog to find a solution to your problem where X is both 10 and 11 at the same time. So it's obviously going to tell you that that can't be done.
Instead you have to just make up new variable names as you go along. Sometimes this does get a bit annoying but it's just goes with the territory of a logical languages.