I have a function which sets a variable based on another variable.
if(!button_x.on)
button_x.on = 1;
if(!button_y.on)
button_y.on = 1;
if(!button_z.on)
button_z.on = 1;
.
.
.
If it is x, y, z, …. is determined only at runtime. Having multiple such conditions for various(100s) different cases does not look good. Is there a better way to go about this in C?
EDIT:
I should have better framed my example above.
if (!structureA.visited)
visit_structureA(); // does some operation
if (!structureB.visited)
visit_structureB();
if (!structureC.visited)
visit_structureC();
.
.
.
The number of structures and the name of the structure is not known at compile time. But the structure names follow a specific pattern shown above. It is known only at runtime.
I tried using macros something like:
#define VISIT(str) \
if (!structure##str.visited) \
visit_structure##str();
//In the function:
// str = 'known at runtime'
VISIT(str);
But this would not work for obvious reason that preprocessor directives are replaced during compile time and not runtime. I am not sure if there is a better way for this?
In your example, you set a variable value according to the same variable, not another one, if this is the case, and you want to change it from 0 to 1 and vice versa, you can do it without condition:
button_x.on = !button_x.on;
If you have many of those with the same idea of behavior, consider using array and itertating it.
In C, the following condition:
if (!x)
x = 1;
is equivalent to:
x = 1;
if the variable is boolean (on/off), which I assume is the case if we are talking about buttons.
Related
Following to the question published in How expressive can we be with arrays in Z3(Py)? An example, I expressed the following formula in Z3Py:
Exists i::Integer s.t. (0<=i<|arr|) & (avg(arr)+t<arr[i])
This means: whether there is a position i::0<i<|arr| in the array whose value a[i] is greater than the average of the array avg(arr) plus a given threshold t.
The solution in Z3Py:
t = Int('t')
avg_arr = Int('avg_arr')
len_arr = Int('len_arr')
arr = Array('arr', IntSort(), IntSort())
phi_1 = And(0 <= i, i< len_arr)
phi_2 = (t+avg_arr<arr[i])
phi = Exists(i, And(phi_1, phi_2))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
Note that, (1) the formula is satisfiable and (2) each time I execute it, I get a different model. For instance, I just got: [avg_a = 0, t = 7718, len_arr = 1, arr = K(Int, 7719)].
I have three questions now:
What does arr = K(Int, 7719)] mean? Does this mean the array contains one Int element with value 7719? In that case, what does the K mean?
Of course, this implementation is wrong in the sense that the average and length values are independent from the array itself. How can I implement simple avg and len functions?
Where is the i index in the model given by the solver?
Also, in which sense would this implementation be different using sequences instead of arrays?
(1) arr = K(Int, 7719) means that it's a constant array. That is, at every location it has the value 7719. Note that this is truly "at every location," i.e., at every integer value. There's no "size" of the array in SMTLib parlance. For that, use sequences.
(2) Indeed, your average/length etc are not related at all to the array. There are ways of modeling this using quantifiers, but I'd recommend staying away from that. They are brittle, hard to code and maintain, and furthermore any interesting theorem you want to prove will get an unknown as answer.
(3) The i you declared and the i you used as the existential is completely independent of each other. (Latter is just a trick so z3 can recognize it as a value.) But I guess you removed that now.
The proper way to model such problems is using sequences. (Although, you shouldn't expect much proof performance there either.) Start here: https://microsoft.github.io/z3guide/docs/theories/Sequences/ and see how much you can push it through. Functions like avg will need a recursive definition most likely, for that you can use RecAddDefinition, for an example see: https://stackoverflow.com/a/68457868/936310
Stack-overflow works the best when you try to code these yourself and ask very specific questions about how to proceed, as opposed to overarching questions. (But you already knew that!) Best of luck..
I'm working on a frama-c plugin that should resolve the values of all kinds of varibles. I managed to dereference pointers and structs and typedefs and print the correspoinding values.
Now I'm struggling with getting the values of an array.
Here is my approach so far, description below:
| TArray (typ, exp, bitsSizeofTypCache, attributes) -> (
let len = Cil.lenOfArray exp in
let rec loc_rec act max =
if act < max then(
let packed = match exp with
| Some x -> x
in
let inc = Cil.increm packed act in
let new_offset = (Index(inc, offset)) in
rec_struct_solver typ (vi_name^"["^(string_of_int act)^"]") (lhost, new_offset);
loc_rec (act+1) max
);
in
loc_rec 0 len
)
I managed to get the length of the array by using Cil.lenOfArray with the expression-option when matching the type.
Now my approach is to go over the length of the array, increment the describing expression and modify the offset, and then handle the variable like a local variable (in the next recursion-step).
I think this idea basically makes sense, but I don't know if the increment is done correctly (value ok, or multiplied by some size or something), or other things don't work.
The program compiles (with the warning that the matching doesn't include all cases, which is irrelevant, since I can only work with expressions, not with NONE), but doesn't output the (correct) results.
Is this the nearly the right approach, or am I doing it completely wrong? Can someone give me some hints on how to get the array's values?
If something is unclear (since it is hard to describe what I want), please let me know, I will modify my question.
EDIT
The expected result of a Code like this
int arr[3];
arr[0]=0;
arr[1]=1;
arr[2]=2;
arr[0]=3;
Should be something like that:
arr[0]=0;3
arr[1]=1
arr[2]=2
I simply want to get all the values at every index of the array over the program.
While I get only empty results, like arr[1]={ } (also for the other Indizes), so I simply don't get results for this Kind of access I use.
Your original code queries the value for the index Index(inc, offset), where inc is Cil.increm packed act, act is the current index, and packed is the size of the array. So you are basically making queries for size+0, size+1 ... size-+(size-1). All these offsets are invalid, because they are out-of-bounds. This is why you obtain the value Cvalue.V.bottom which pretty-prints as .
The simplest fix to your original code would have been to replace packed by a call to Cil.zero Cil_datatype.Location.unknown, but your own fix is fine.
I figured out how to do it:
The trick was, that with Cil.integer, a new constant exp can be built!
With Cil.integer Cil_datatype.Location.unknown act, I created a new exp.
With this exp, I was able to build an Index-Offset. Then I added this new offset to the actual offset of the array. This new offset was used to build a new lval.
By doing this, I got the access to the arrays indizes.
Now my output looks ok:
arrayTest:arr[0]---> 0; 3
arrayTest:arr[1]---> 1
arrayTest:arr[2]---> 2
Is there a way to pass an argument multiple times to different arrays?
What I want to do is:
r = '1:10:end'; % This doesn't work for me
plot(x1(r), y1(r));
plot(x2(r), y2(r));
...
and pass r to different arrays (with different lengths) in many plot functions. I tried with [r] but no success.
As I understand it, you want to plot every 10th element of possibly different sized arrays. There are a few ways you could do this. One way would be to write a short function to filter your arrays for you, for instance:
plot_10 = #(x,y) plot(x(1:10:end),y(1:10:end));
plot_10(x1,y1);
plot_10(x2,y2);
...
EDIT: Just an additional thought. If you wanted to enable the extended functionality of plot (e.g. passing line/colour arguments, etc). You could do something like this:
plot_10 = #(x,y,varargin) plot(x(1:10:end),y(1:10:end),varargin{:});
plot_10(x1,t1,'k+');
To use the "end" operator, it needs to be inside an array access call;
n = 10;
r = 1 : 1 : n;
r(1:end) % is legal
r(1:floor(end/2)) % is legal
So you could do something like this:
s = rand(1,2*n);
s(r)
% to compare...
s( 1:1:n )
I'm seeing an issue when I try and reference an object property after having used a dot notation to apply a method.
it only occurs when I try to index the initial object
classdef myclassexample
properties
data
end
methods
function obj = procData(obj)
if numel(obj)>1
for i = 1:numel(obj)
obj(i) = obj(i).procData;
end
return
end
%do some processing
obj.data = abs(obj.data);
end
end
end
then assigning the following
A = myclassexample;
A(1).data= - -1;
A(2).data = -2;
when calling the whole array and collecting the property data it works fine
[A.procData.data]
if i try and index A then i only get a scalar out
[A([1 2]).procData.data]
even though it seems to do fine without the property call
B = A([1 2]).procData;
[B.data]
any ideas?
I would definitely call this a bug in the parser; A bug because it did not throw an error to begin with, and instead allowed you to write: obj.method.prop in the first place!
The fact that MATLAB crashed in some variations of this syntax is a serious bug, and should definitely be reported to MathWorks.
Now the general rule in MATLAB is that you should not "index into a result" directly. Instead, you should first save the result into a variable, and then index into that variable.
This fact is clear if you use the form func(obj) rather than obj.func() to invoke member methods for objects (dot-notation vs. function notation):
>> A = MyClass;
>> A.procData.data % or A.procData().data
ans =
[]
>> procData(A).data
Undefined variable "procData" or class "procData".
Instead, as you noted, you should use:
>> B = procData(A): % or: B = A.pocData;
>> [B.data]
FWIW, this is also what happens when working with plain structures and regular functions (as opposed to OOP objects and member functions), as you cannot index into the result of a function call anyway. Example:
% a function that works on structure scalar/arrays
function s = procStruct(s)
if numel(s) > 1
for i=1:numel(s)
s(i) = procStruct(s(i));
end
else
s.data = abs(s.data);
end
end
Then all the following calls will throw errors (as they should):
% 1x2 struct array
>> s = struct('data',{1 -2});
>> procStruct(s).data
Undefined variable "procStruct" or class "procStruct".
>> procStruct(s([1 2])).data
Undefined variable "procStruct" or class "procStruct".
>> feval('procStruct',s).data
Undefined variable "feval" or class "feval".
>> f=#procStruct; f(s([1 2])).data
Improper index matrix reference.
You might be asking yourself why they decided to not allow such syntax. Well it turns out there is a good reason why MATLAB does not allow indexing into a function call (without having to introduce a temporary variable that is), be it dot-indexing or subscript-indexing.
Take the following function for example:
function x = f(n)
if nargin == 0, n=3; end
x = magic(n);
end
If we allowed indexing into a function call, then there would be an ambiguity in how to interpret the following call f(4):
should it be interpreted as: f()(4) (that is call function with no arguments, then index into the resulting matrix using linear indexing to get the 4th element)
or should it interpreted as: f(4) (call the function with one argument being n=4, and return the matrix magic(4))
This confusion is caused by several things in the MATLAB syntax:
it allows calling function with no arguments simply by their name, without requiring the parentheses. If there is a function f.m, you can call it as either f or f(). This makes parsing M-code harder, because it is not clear whether tokens are variables or functions.
parentheses are used for both matrix indexing as well as function calls. So if a token x represents a variable, we use the syntax x(1,2) as indexing into the matrix. At the same time if x is the name of a function, then x(1,2) is used to call the function with two arguments.
Another point of confusion is comma-separated lists and functions that return multiple outputs. Example:
>> [mx,idx] = max(magic(3))
mx =
8 9 7
idx =
1 3 2
>> [mx,idx] = max(magic(3))(4) % now what?
Should we return the 4th element of each output variables from MAX, or 4th element from only the first output argument along with the full second output? What about when the function returns outputs of different sizes?
All of this still applies to the other types of indexing: f()(3)/f(3), f().x/f.x, f(){3}/f{3}.
Because of this, MathWorks decided avoid all the above confusion and simply not allow directly indexing into results. Unfortunately they limited the syntax in the process. Octave for example has no such restriction (you can write magic(4)(1,2)), but then again the new OOP system is still in the process of being developed, so I don't know how Octave deals with such cases.
For those interested, this reminds me of another similar bug with regards to packages and classes and directly indexing to get a property. The results were different whether you called it from the command prompt, from a script, or from a M-file function...
I have an array of 20 items long and I would like to make them an output so I can input it into another program.
pos = [0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,]
I would like to use this as inputs for another program
function [lowest1, lowest2, highest1, highest2, pos(1), pos(2),... pos(20)]
I tried this and it does not work is there another way to do this?
I'm a little confused why you'd want to do that. Why would you want 20 outputs when you could just return pos as a single output containing 20 elements?
However, that said, you can use the specially named variable varargout as the last output variable, and assign a cell to it, and the elements of the cell will be expanded into outputs of the function. Here's an example:
function [lowest1, lowest2, highest1, highest2, varargout] = myfun
% First set lowest1, lowest2, highest1, highest2, and pos here, then:
varargout = num2cell(pos);
If what you're trying to do is re-arrange your array to pass it to another Matlab function, here it is.
As one variable:
s=unique(pos);
q=[s(1) s(2) s(end-1) s(end) pos];
otherFunction(q);
As 24 variables:
s=unique(pos); otherFunction(s(1), s(2), s(end-1), s(end), pos(1), pos(2), pos(3), pos(4), pos(5), pos(6), pos(7), pos(8), pos(9), pos(10), pos(11), pos(12), pos(13), pos(14), pos(15), pos(16), pos(17), pos(18), pos(19), pos(20));
I strongly recommend the first alternative.
Here are two examples of how to work with this single variable. You can still access all of its parts.
Example 1: Take the mean of all of its parts.
function otherFunction(varargin)
myVar=cell2mat(varargin);
mean(myVar)
end
Example 2: Separate the variable into its component parts. In our case creates 24 variables named 'var1' to 'var24' in your workspace.
function otherFunction(varargin)
for i=1:nargin,
assignin('base',['var' num2str(i)],varargin{i});
end
end
Hope this helps.
Consider using a structure in order to return that many values from a function. Carefully chosen field names make the "return value" self declarative.
function s = sab(a,b)
s.a = a;
s.b = b;