Storing input variables x in a persistent variable called buffer

2 vues (au cours des 30 derniers jours)
Hi there,
I am trying yo create a function which input x as a scalar and stores in a persistent variable called buffer. and it calculates the average of input every time I input a new x. The function uses a “buffer” to hold previous inputs, and the buffer can hold a maximum of 25 inputs. I want to know how I can add input to the buffer which is a persistent variable and see the inputs stored in the buffer? Thanks

Réponse acceptée

Jan
Jan le 2 Mai 2015
The description sounds clean and clear enough to allow for a straight forward implementation. What should happen if the function is called the 26th time? Should the buffer wrap around in a first-in-first-out style?
What have you tried so far? Usually it is more efficient to posr, what you have written and ask specific questions. But it is saturday and I try it:
function YourAccumulator(x)
persistent buffer bufferIndex
if isempty(buffer) % Called the first time
bufferIndex = 0;
end
bufferIndex = bufferIndex + 1
buffer(bufferIndex) = x;
if bufferIndex == 25
bufferIndex = 1;
end
M = mean(buffer);
disp(M);
The mod() operator would help to remove the if part.
  1 commentaire
yashar khatib shahidi
yashar khatib shahidi le 4 Mai 2015
Modifié(e) : yashar khatib shahidi le 4 Mai 2015
Hi, Thanks for your reply. I have created a function that approximates pi and Instead of going to infinity, the function stops at the smallest k for which the approximation differs from pi (i.e., the value returned MATLAB’s built-in function) by no more than the positive scalar delta, which is the only input argument. The first output of the function is the approximate value of π, while the second is k. There is something wrong with this when I enter any delta it goes into infinite loop. Here is my code:
function [approx, k] = approximate_pi(delta)
approx = sqrt(12);
k = 0;
while abs(approx - pi) > 1e-2*delta
k = k + 1;
approx = approx + sqrt(12)*(((-3)^(-k))/(2*k+1))
end

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Language Fundamentals dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by