What does getAmountsOut/getAmountOut and getAmountsIn/getAmountIn exactly do? There isn't an explanation in the Pancakeswap docs, so I'm not sure how could I use it. What's the difference between the two and with the 's' or none?
The difference is that the one with the "s" return the series of outputs resulting from a path of swaps. Whereas the ones without the "s" return the output for a single swap.
So if I swap along the path ["ETHUSDT", "USDTUSDC"] then getAmountsOut will return the USDT output amount from the ETHUSDT swap and the USDC output amount from the "USDTUSDC" swap. The returned value will look like: [usdtAmount, usdcAmount].
If I swap "ETHUSDT" then getAmountOut will return usdtAmount.
The inverse is true for getAmountsIn/getAmountIn.
It's defined here: https://github.com/pancakeswap/pancake-swap-periphery/blob/master/contracts/libraries/PancakeLibrary.sol#L63
Related
I am writing a method which accepts a two dimensional array of doubles and an int row number as parameters and returns the highest value of the elements in the given row.
it looks like this:
function getHighestInRow(A, i)
return(maximum(A[:i,:]))
end
the issue i am having is when i slice the array with
A[:i,:]
I get an argument error because the :i makes i get treated differently.
the code works in the other direction with
A[:,i,:]
Is there a way to escape the colon? so that i gets treated as a variable after a colon?
You're doing something strange with the colon. In this case you're using the symbol :i not the value of i. Just getHighestInRow(A,i) = maximum(A[i,:]) should work.
Edit: As Dan Getz said in the comment on the question, getHighestInRow(A,i) = maximum(#view A[i,:]) is more efficient, though, as the slicing will allocate a temporary unnecessary array.
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 familiar with iterative methods on paper, but MATLAB coding is relatively new to me and I cannot seem to find a way to code this.
In code language...
This is essentially what I have:
A = { [1;1] [2;1] [3;1] ... [33;1]
[1;2] [2;2] [3;2] ... [33;2]
... ... ... ... ....
[1;29] [2;29] [3;29] ... [33;29] }
... a 29x33 cell array of 2x1 column vectors, which I got from:
[X,Y] = meshgrid([1:33],[1:29])
A = squeeze(num2cell(permute(cat(3,X,Y),[3,1,2]),1))
[ Thanks to members of stackOverflow who helped me do this ]
I have a function that calls each of these column vectors and returns a single value. I want to institute a 2-D 5-point stencil method that evaluates a column vector and its 4 neighbors and finds the maximum value attained through the function out of those 5 column vectors.
i.e. if I was starting from the middle, the points evaluated would be :
1.
A{15,17}(1)
A{15,17}(2)
2.
A{14,17}(1)
A{14,17}(2)
3.
A{15,16}(1)
A{15,16}(2)
4.
A{16,17}(1)
A{16,17}(2)
5.
A{15,18}(1)
A{15,18}(2)
Out of these 5 points, the method would choose the one with the largest returned value from the function, move to that point, and rerun the method. This would continue on until a global maximum is reached. It's basically an iterative optimization method (albeit a primitive one). Note: I don't have access to the optimization toolbox.
Thanks a lot guys.
EDIT: sorry I didn't read the iterative part of your Q properly. Maybe someone else wants to use this as a template for a real answer, I'm too busy to do so now.
One solution using for loops (there might be a more elegant one):
overallmax=0;
for v=2:size(A,1)-1
for w=2:size(A,2)-1
% temp is the horizontal part of the "plus" stencil
temp=A((v-1):(v+1),w);
tmpmax=max(cat(1,temp{:}));
temp2=A(v,(w-1):(w+1));
% temp2 is the vertical part of the "plus" stencil
tmpmax2=max(cat(1,temp2{:}));
mxmx=max(tmpmax,tmpmax2);
if mxmx>overallmax
overallmax=mxmx;
end
end
end
But if you're just looking for max value, this is equivalent to:
maxoverall=max(cat(1,A{:}));
I want to know how to store the characters or integers popped from stack to be stored so that I can compare it with the original string or int value.
For example :
n = 1221;
n = n / 1000; // so that i get the last digit in this case 1 and dividing the remainder
// further each time by 100, 10 and 1
If I store each number that I get in a variable-- for example, say that I store 1 which I got from the above division in a variable named s, and push it onto the stack.
After this, I pop the values back out; when I do so, how can I check weather it is equal to the original number? I can check it with a if condition, for eg if i have 3 did then
(i == p && check for other two numbers)
but i don`t want that i want to check for any size number.
Please don't send the source code on how to do it, just give me a few snippets or a few hints, thanks. Also please let me know how you came up for the solution.
Thanks!
Dont send the source code on how to do it
ok ;)
just give me few snippets or few hints
Recursion
and while you give the solution let me know how you came up for the solution, weather you had seen a program like this or know the algorithm or came up with a solution now when you saw the question. Thanks
i have been doing this too long to remember where i first saw it =\
You can use recursion to solve the problem, like Justin mentioned.
Once you start your recursion you will divide the number with an appropriate divisor(1000,100,10,1) store the quotient in the stack. You do this till the end.
Once you reach the 'unit' place, you store the unit digit, and now start popping out of the stack.
Have an if-else ladder to return an integer from the recursive function.
The ladder will have conditions of returning the int after left shifting and returning the variable shifted.
You can do your check in the main function.
1 ->1221(left shift 122 OR with 1)
2 ->122(left shift 12 OR with 2)
2 ->12(left shift 1 OR with 2)
1---->1
Hope this helps.
Thanks
Aditya
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;