I have an array of a given dimension. I want to swap two given elements of the array. I tried to swap the two with the help of a temporary variable, but the result was bogus and confusing. Can someone suggest some other way of doing so?
Also , it would be really helpful if someone told me why my current program is not working.
Swapping using a temporary variable is the correct way.
There's no way to figure out what error you did to make it fail, since you didn't post any code.
In general for an array of type T we can swap the elements at positions i and j like so:
T x[N];
const T tmp = x[i];
x[i] = x[j];
x[j] = tmp;
This works for any type T, including structs.
Related
Working in MATLAB R2017a. I'm trying to optimise a piece of code I'm working on. It uses arrays to store field values on a grid.
In order to create a specific function in a field array I originally used the straight forward method of two for loops iterating over all the array elements. But i know for loops are slow so since then I came back and tried my best to remove them. However I could only manage to remove one of the loops; leaving me with this:
for n = 1:1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp * (1/(1+exp(-(t-stepuppos)))) * exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg))*exp(-((t-((ztemp-...
THzstartz)/vg))^2)/(bt^2));
end
The important bit here is the last 5 lines, but I figured the stuff before might be important for context. I've removed the for loop looping over the x coordinates. I want to vectorize the z/k for loop but I can't figure out how to distinguish between the dimensions with the array oporators.
Edit: THzamp, stepuppos, bx, bt, THzstartz, THzstartx are all just scalars, they control the function (Ez0) I'm trying to create. dX and t are also just scalars. Ez0 is a square array of size Ng.
What I want to achieve is to remove the for loop that loops over k, so that that the values of ztemp are defined in a vector (like xtemps already is), rather than individually in the loop. However, I don't know how I'd write the definition of Ez0 in that case.
First time posting here, if I'm doing it wrong let me know. If you need more info just ask.
It isn't clear if n is used in the other headers and as stated in the comments your sizes aren't properly defined so you'll have to ensure the sizes are correct.
However, you can give this vectorize code a try.
n = 1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp .* (1./(1+exp(-(t-stepuppos)))) .* exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg)).*exp(-((t-((ztemp-...
THzstartz)/vg)).^2)/(bt.^2));
end
So now t has the size K you'll need to ensure stepupposand (ztemp-THzstartz)/vg) have the same size K. Also you can take a look at vectors vs array operators here.
I have a doubt on how the machines stores a two dimensional array in memory. I'll present you my code in order to be clearer.
I'm defining a two dimensional array in this way, in my main loop:
int main()
{
int i;
internalNode**tab =(internalNode **)malloc(sizeof(internalNode *)* DIM);
for (i=0; i<DIM ; i++)
tab[i] = (internalNode *)malloc(sizeof(internalNode) * DIM);
//CODE
CalculusOnGrid(tab,DIM);
}
Where DIM is a user defined variable and internalNode is a structure.
In the function CalculusOnGrid i'm going to do this calculus on the grid ( my two dimensional array):
for(i=1;i<DIM-1;i++)
for(j=1;j<DIM-j;i++)
tab[i][j].temperature_new = 0.25*tab[i+1][j].temperature+tab[i-1][j].temperature + tab[i][j+1].temperature + tab[i][j-1].temperature);
So i'm going to search for the 4 neighbors of my current point (i,j) of the grid.
Here there is my question: I'm going to do a Cache Miss on the upper and below element ( that's to say tab[i+1][] and tab[i-1][]) or on the right and left elements? (that's to say tab[][j+1] and tab[][j-1])
What's your suggestion for speeding up my code and reduce the number of Cache misses?
I hope that the question is proposed in a clear way. If this is not the case, ask me whatever you want!
Thank you!
Alessandro
Cache misses is one of many reasons why you should avoid using pointer-based lookup tables to emulate dynamic arrays.
Instead, use a 2D array:
internalNode (*tab)[DIM] = malloc( sizeof(internalNode[DIM][DIM]) );
free(tab);
Now the memory will be adjacent and performance should be much better.
So what I have is an array that's size is decided by me and then the elements in the array are randomly generated. It's supposed to take an integer array,its size, and an integer number
and find how many times the number is present in the array and return that count at the end.I keep trying stuff and nothing seems to be getting me anywhere close to an answer. I was just trying to see if someone could point me in the right direction on where to start
count_numbers(int array[], int size, int z)
Hhave you tried running a loop through the array and trying a match expression to the array value in another loop. This seems like a logic question rather than actual code related. Maybe a search around the internet looking at how to count in arrays could help you.
This should point you in the right direction...
for (int i = 0; i < arraySize; i++) {
if (array[i] == z /*z being your search value**/) {
you may have to alter this a little
//dosomething
// e.g. increment a count here
}
else
do-nothing essentially.
There is a method for checking array size - so don't worry about defining it's size. have a look at the java method for this and use it.
Hope this helps
This is for a Deal or No Deal game.
So in my main function I'm calling my casesort method as such:
casesort(cases);
My method looks like this, I already realize it's not the most efficient sort but I'm going with what I know:
void casesort(float cases[10])
{
int i;
int j;
float tmp;
float zero = 0.00;
for (i = 0; i < 10; i++)
{
for (j = 0; j < 10; j++)
{
if (cases[i] < cases[j])
{
tmp = cases[i];
cases[i] = cases[j];
cases[j] = tmp;
}
}
}
//Print out box money amounts
printf("\n\nHidden Amounts: ");
for (i = 0; i < 10; i++)
{
if (cases[i] != zero)
printf("[$%.2f] ", cases[i]);
}
}
So when I get back to my main it turns out the array is sorted. I thought void would prevent the method returning a sorted array. I need to print out actual case numbers, I do this by just skipping over any case that is populated with a 0.00. But after the first round of case picks I get "5, 6, 7, 8, 9, 10" printing out back in my MAIN. I need it to print the cases according to what has been picked. I feel like it's a simple fix, its just that my knowledge of the specifics of C is still growing. Any ideas?
Return type void has nothing to do with prevention of array from being sorted. It just says that function does not return anything.
You see that the passed array itself is affected because an array decays to a pointer when passed to a function. Make a copy of the array and then pass it. That way you have the original list.
In C, arrays are passed by reference. i.e. they're passed as pointer to the first element. So when you pass cases into your function, you're actually giving it the original array to modify. Try creating a copy and sorting the copy rather than the actual array. Creating a copy wouldn't be bad as you have only 10 floats.
Instead of rolling your own sort, consider using qsort() or std::sort() if you are actually using c++
There are 2 obvious solutions. 1) Make a copy of the array and sort the copy (easy, waste some memory, likely not a problem these days). 2) Create a parallel array of integers and perform an index sort, i.e., instead of sorting thing original, you sort the index and then dereference the array using the index when you want the sorted version, otherwise by the raw unsorted array.
Well, make a local copy of you input and sort it. Something like this:
void casesort(float cases[10])
{
float localCases[10];
memcopy(localCases, cases, sizeof(cases));
...
Then use localCases to do your sorting.
If you don't want the array contents to be affected, then you'll have to create a copy of the array and pass that to your sorting routine (or create the copy within the routine itself).
Arrays Are Differentâ„¢ in C; see my answer here for a more detailed explanation.
Is there any way to "vector" assign an array of struct.
Currently I can
edges(1000000) = struct('weight',1.0); //This really does not assign the value, I checked on 2009A.
for i=1:1000000; edges(i).weight=1.0; end;
But that is slow, I want to do something more like
edges(:).weight=[rand(1000000,1)]; //with or without the square brackets.
Any ideas/suggestions to vectorize this assignment, so that it will be faster.
Thanks in advance.
This is much faster than deal or a loop (at least on my system):
N=10000;
edge(N) = struct('weight',1.0); % initialize the array
values = rand(1,N); % set the values as a vector
W = mat2cell(values, 1,ones(1,N)); % convert values to a cell
[edge(:).weight] = W{:};
Using curly braces on the right gives a comma separated value list of all the values in W (i.e. N outputs) and using square braces on the right assigns those N outputs to the N values in edge(:).weight.
You can try using the Matlab function deal, but I found it requires to tweak the input a little (using this question: In Matlab, for a multiple input function, how to use a single input as multiple inputs?), maybe there is something simpler.
n=100000;
edges(n)=struct('weight',1.0);
m=mat2cell(rand(n,1),ones(n,1),1);
[edges(:).weight]=deal(m{:});
Also I found that this is not nearly as fast as the for loop on my computer (~0.35s for deal versus ~0.05s for the loop) presumably because of the call to mat2cell. The difference in speed is reduced if you use this more than once but it stays in favor of the for loop.
You could simply write:
edges = struct('weight', num2cell(rand(1000000,1)));
Is there something requiring you to particularly use a struct in this way?
Consider replacing your array of structs with simply a separate array for each member of the struct.
weights = rand(1, 1000);
If you have a struct member which is an array, you can make an extra dimension:
matrices = rand(3, 3, 1000);
If you just want to keep things neat, you could put these arrays into a struct:
edges.weights = weights;
edges.matrices = matrices;
But if you need to keep an array of structs, I think you can do
[edges.weight] = rand(1, 1000);
The reason that the structs in your example don't get initialized properly is that the syntax you're using only addresses the very last element in the struct array. For a nonexistent array, the rest of them get implicitly filled in with structs that have the default value [] in all their fields.
To make this behavior clear, try doing a short array with clear edges; edges(1:3) = struct('weight',1.0) and looking at each of edges(1), edges(2), and edges(3). The edges(3) element has 1.0 in its weight like you want; the others have [].
The syntax for efficiently initializing an array of structs is one of these.
% Using repmat and full assignment
edges = repmat(struct('weight', 1.0), [1 1000]);
% Using indexing
% NOTE: Only correct if variable is uninitialized!!!
edges(1:1000) = struct('weight', 1.0); % QUESTIONABLE
Note the 1:1000 instead of just 1000 when indexing in to the uninitialized edges array.
There's a problem with the edges(1:1000) form: if edges is already initialized, this syntax will just update the values of selected elements. If edges has more than 1000 elements, the others will be left unchanged, and your code will be buggy. Or if edges is a different type, you could get an error or weird behavior depending on its existing datatype. To be safe, you need to do clear edges before initializing using the indexing syntax. So it's better to just do full assignment with the repmat form.
BUT: Regardless of how you initialize it, an array-of-structs like this is always going to be inherently slow to work with for larger data sets. You can't do real "vectorized" operations on it because your primitive arrays are all broken up in to separate mxArrays inside each struct element. That includes the field assignment in your question – it is not possible to vectorize that. Instead, you should switch a struct-of-arrays like Brian L's answer suggests.
You can use a reverse struct and then do all operations without any errors
like this
x.E(1)=1;
x.E(2)=3;
x.E(2)=8;
x.E(3)=5;
and then the operation like the following
x.E
ans =
3 8 5
or like this
x.E(1:2)=2
x =
E: [2 2 5]
or maybe this
x.E(1:3)=[2,3,4]*5
x =
E: [10 15 20]
It is really faster than for_loop and you do not need other big functions to slow your program.