Suppose I have non-scalar structure
res = struct(); res(1).name = 'hello'; res(2).name = 'world';
Now I want to copy entire content of name field to another field, say tag.
Neither of the following worked:
>> res.tag = res.name;
Scalar structure required for this assignment.
>> [res.tag] = [res.name];
Insufficient number of outputs from right hand side of equal sign to satisfy assignment.
>> {res.tag} = {res.name};
{res.tag} = {res.name};
↑
Error: The expression to the left of the equals sign is not a valid target for an assignment.
Any other ideas?
Use
[res(:).tag] = res(:).name;
or more simply, as you have discovered yourself:
[res.tag] = res.name;
The syntax with square brackets on the left-hand side is similar to that used for capturing several outputs returned by a function: [out1, out2] = fun(...); see MATLAB special characters.
Actually, the syntax res.tag produces a comma-separated list; and [...] is standard for assigning values to each element in one such list; see Assigning output from a comma-separated list.
The right-hand side of the assignment should be another comma-separated list. If it is a single element, or you want to specify a list manually, you need deal:
values = {10, 20};
[res.test] = values{:}; % works. {:} produces a comma-separated list
[res.test] = 10,20; % doesn't work. Use `deal`
[res.test] = deal(10,20); % works
[res.test] = 10; % doesn't work, unless `res` is scalar. Use `deal`
[res.test] = deal(10); % also works. 10 is repeated as needed
The reason why your attempt [res.tag] = [res.name]; doesn't work is that [res.name] on the right-hand side concatenates the results of the comma-separated list res.name into one array, and so it's the same case as [res.test] = 10; above.
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
I'am learning OCaml and currently i'am trying to undertand how iteration works in OCaml and how to create a matrix. I want an array 5 x 5 filled with 0. I know there is an issue with shared references so i created a new array at each iteration however iam having issues in other places, specifically at line 6. Let me know of other issues like indentation practices.
open Array;;
let n = ref 5 and i = ref 0 in
let m = Array.make !n 0 in
while !i < !n do
m.(!i) <- Array.make !n 0;;
i := !i + 1;;
done
m;;
You are using ;; too much. Contrary to popular belief, ;; is not part of ordinary OCaml syntax (in my opinion anyway). It's just a special way to tell the toplevel (the REPL) that you want it to evaluate what you've typed so far.
Leave the ;; after open Array. But change all but the last ;; to ; instead.
(Since you reference the Array module by name in your code, which IMHO is good style, you can also just leave out the open Array;; altogether.)
You want the last ;; because you do want the toplevel to evaluate what you've typed so far.
Your syntax error is caused by the fact that your overall code is like this
let ... in
let ... in
while ... do
...
done
m
The while is one expression (in OCaml everything is an expression) and m is another expression. If you want to have two expressions in a row you need ; between them. So you need ; after done.
You also have a type error. When you create m you're creating an array of ints (your given initial value is 0). So you can't make it into a matrix (an array of arrays) later in the code.
Also (not trying to overload you with criticisms :-) this code reads like imperative code. It's not particularly idiomatic OCaml code. In most people's code, using ref is pretty rare. One immediate improvement I see would just be to say let n = 5. You're not changing the value of n anywhere that I see (though maybe this is part of a larger chunk of code). Another improvement would be to use for instead of while.
Finally, you can do this entire operation in one function call:
let n = 5 in
let m = Array.init n (fun i -> Array.make n 0) in
m
Using explicit loops is actually also quite rare in OCaml (at least in my code).
Or you could try this:
let n = 5 in
let m = Array.make_matrix n n 0 in
m
Updated Question; Original below.
I am trying to create an array which represents a grid of cells, which have tuples containing the walls they are surrounded by.
I have come up with this:
rooms(Array) ->
Size = array:size(Array),
if
Size == ?HSIZE * ?VSIZE ->
Array;
true ->
HFactor = Size rem ?VSIZE,
VFactor = Size div ?HSIZE,
Room = {1+HFactor+11*VFactor,
7+HFactor+11*VFactor,
12+HFactor+11*VFactor,
6+HFactor+11*VFactor},
rooms(array:set(Size, Room, Array))
end.
When I run this with rooms(array:new()). I get the following array back:
{array,25,100,undefined,
{{{1,7,12,6},
{2,8,13,7},
{3,9,14,8},
{4,10,15,9},
{5,11,16,10},
{12,18,23,17},
{13,19,24,18},
{14,20,25,19},
{15,21,26,20},
{16,22,27,21}},
{{23,29,34,28},
{24,30,35,29},
{25,31,36,30},
{26,32,37,31},
{27,33,38,32},
{34,40,45,39},
{35,41,46,40},
{36,42,47,41},
{37,43,48,42},
{38,44,49,43}},
{{45,51,56,50},
{46,52,57,51},
{47,53,58,52},
{48,54,59,53},
{49,55,60,54},
undefined,undefined,undefined,undefined,undefined},
10,10,10,10,10,10,10,10}}
Which is quite close to the desired result, but there are two things I can't quite put my finger on (The numbers are correct). Why does it look like it is split up into multiple subarrays? What are those undefineds and 10's doing there? These are mostly due to my lack of erlang knowledge, because array:get produces the expected results, but I couldn't find anything which explains where they come from.
Original Question
rooms(Array) ->
Size = array:size(Array),
if
Size == 5 ->
Array;
Size rem 5 == 0 ->
rooms(array:set(Size, array:new(), Array));
true ->
In_Array = array:get(array:size(Array), Array),
In_Size = array:size(In_Array),
Room = {1+In_Size+11*In_Size,
7+In_Size+11*In_Size,
12+In_Size+11*In_Size,
6+In_Size+11*In_Size},
New_In = array:set(In_Size, Room, In_Array),
rooms(array:set(Size, New_In, Array))
end.
I call it with rooms(array:new()). but the result is
** exception error: bad argument
in function array:size/1 (array.erl, line 317)
in call from framework_kamer:rooms/1 (framework_kamer.erl, line 195)
Which makes sense because In_Array is not an array, but undefined. However, I can't figure out why.
Side question, is there an easier/cleaner/better way to do this?
You are (in the second call of the recursion) trying
In_Array = array:get(array:size(Array), Array),
As array is zero-indexed this will always fail, as the access will be always be off by one. Change this line to
In_Array = array:get(array:size(Array) - 1, Array),
and you are fine.
A few comments on your code:
Conventional variable naming in Erlang would be CamelCase without underscores (i.e. InArray)
Expressing a 2-dimensional array as nested arrays is almost never a good idea. Linearalise it by writing simple wrappers that recalculate a one-dimensional index from x and y as index = y * max_x + x.
I am assigning a value to an array dataBase with the white pixel count of images using the function count. The function count() is working properly but not the assignment to the dataBase array. Please help me with this. The prototypes of functions are,
count(),
function whitecount = new(image1)
The error am receiving is,
??? In an assignment A(I) = B, the number of elements in B and
I must be the same.
Error in ==> init at 9
dataBase(i)= w;
complete function of init()
function dataBase = init();
dataBase=zeros(1,1000);
srcFiles = dir('DB\*.jpg');
for i=1: length(srcFiles)
filename = strcat('DB\',srcFiles(i).name);
w = count(filename);
dataBase(i)= w;
end
If count returns something other than a scalar value, the assignment will fail since dataBase(i) takes a scalar value only.
You have two options
Ensure that w is always a non-empty scalar
make dataBase into a cell array