How do get around this limitation on array size?

Question

Zachary Duff 2018-6-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/404623-how-do-get-around-this-limitation-on-array-size

回答： Zachary Duff 2018-6-8

采纳的回答： Zachary Duff

在 MATLAB Online 中打开

R2017a, R2014, different computers, same issue.

Why does the computer (I'm pretty sure it's not MATLAB) limit me to arrays of length 32767?
How do I circumvent this feature?

For example, in my command window I can enter

v = rand(1000000,1);

and the variable is generated without a problem. However, when I run a script or function designed to generate variables with approximate length 10^6, I become automatically restricted to

>> size(z)
ans =
         32767           1
>> size(Y)
ans =
         32767           3

this doesn't make any sense to me, especially considering

>> whos
  Name           Size             Bytes  Class     Attributes
    T              1x1                  8  double              
    Y          32767x3             786408  double              
    sigma          1x1                  8  double              
    z          32767x1             262136  double              
>> memory
Maximum possible array:        5216 MB (5.470e+09 bytes) *
Memory available for all arrays:        5216 MB (5.470e+09 bytes) *
Memory used by MATLAB:        1180 MB (1.237e+09 bytes)
Physical Memory (RAM):        8124 MB (8.518e+09 bytes)
*  Limited by System Memory (physical + swap file) available.

Note that 32767 = 2^15-1

Complete Code:

clc
clear
close all
% Solve over time interval [0 T] with initial conditions [1,1,1]
% f is set of differential equations
% Y is array containing x, y, and z variables
% t is time variable
sigma = 10;
beta = 8/3;
rho = 28;  
tb = 20;
T = 2000;
% Lorenz system
R = @(Y) ... 
    [   -sigma*Y(1) + sigma*Y(2);       ... 
        rho*Y(1) - Y(2) - Y(1)*Y(3);    ... 
        -beta*Y(3) + Y(1)*Y(2)          ]; 
[t,Y] = rk4lorenz(R,[0 T],[1 1 1]);
z = Y(:,3);  
function [t,Y] = rk4lorenz(f, range, y0)
% RK4 Method for the Lorenz system
% range =  time interval
% y0    =  initial value vector
% h     =  time step
a = range(1)
b = range(2)
h = 1/128;
n = int16((b - a)/h) 
Y = zeros(n+1,3);
t = zeros(n+1,1);
t(1) = a;
Y(1,:) = y0; 
for i = 1:n  
    t(i+1) = t(i) + h;
    Y(i+1,:) = rk4step(Y(i,:),h,f);
end
end
function Y = rk4step(Y,h,f)
Y = Y';
s1 = f(Y);
s2 = f(Y + h * s1/2);
s3 = f(Y + h * s2/2);
s4 = f(Y + h * s3);
Y = Y + h*(f(Y) + 2*s2 + 2*s3 + s4)/6;
end