Sam's answer 引用的删除页面仍然是archived by the Way Back Machine。幸运的是，即使是附加的 M 文件 colonop 也在那里。而且似乎这个函数仍然匹配 MATLAB 所做的（我在 R2017a 上）：
>> all(0:step:5 == colonop(0,step,5))
ans =
logical
1
>> all(-pi:pi/21:pi == colonop(-pi,pi/21,pi))
ans =
logical
1
我将在这里复制该函数对一般情况的作用（有一些用于生成整数向量和处理特殊情况的快捷方式）。我正在用更有意义的变量名替换函数的变量名。输入是start、step 和stop。
首先它计算start 和stop 之间的步数。如果最后一步超过stop 超过一个容差，则不采取：
n = round((stop-start)/step);
tol = 2.0*eps*max(abs(start),abs(stop));
sig = sign(step);
if sig*(start+n*step - stop) > tol
n = n - 1;
end
这解释了问题中提到的最后一个观察结果。
接下来，它计算最后一个元素的值，并确保它不超过stop 的值，即使它在之前的计算中允许超过它。
last = start + n*step;
if sig*(last-stop) > -tol
last = stop;
end
这就是为什么问题中向量A中的lasat值实际上将stop值作为最后一个值。
接下来，它分两部分计算输出数组，正如宣传的那样：数组的左半部分和右半部分是独立填充的：
out = zeros(1,n+1);
k = 0:floor(n/2);
out(1+k) = start + k*step;
out(n+1-k) = last - k*step;
请注意，它们不是通过递增填充，而是通过计算一个整数数组并将其乘以步长，就像linspace 一样。这解释了问题中关于数组E 的观察。不同之处在于数组的右半部分是通过从last 值中减去这些值来填充的。
作为最后一步，对于奇数大小的数组，中间值是单独计算的，以确保它正好位于两个端点的中间：
if mod(n,2) == 0
out(n/2+1) = (start+last)/2;
end
完整的函数colonop复制到底部。
请注意，分别填充数组的左侧和右侧并不意味着步长的误差应该是完全对称的。这些误差由舍入误差给出。但是，如果步长未完全达到stop 点，就像问题中的数组A 一样，它确实会有所不同。在这种情况下，略短的步长是在数组的中间，而不是在末尾：
>> step=1/3;
>> A = 0 : step : 5-2*eps(5);
>> A/step-(0:15)
ans =
1.0e-14 *
Columns 1 through 10
0 0 0 0 0 0 0 -0.0888 -0.4441 -0.5329
Columns 11 through 16
-0.3553 -0.3553 -0.5329 -0.5329 -0.3553 -0.5329
但即使在准确到达stop 点的情况下，中间也会累积一些额外的错误。以问题中的数组C 为例。 linspace 不会发生这种错误累积：
C = 0:1/3:5;
lims = eps(C);
subplot(2,1,1)
plot(diff(C)-1/3,'o-')
hold on
plot(lims,'k:')
plot(-lims,'k:')
plot([1,15],[0,0],'k:')
ylabel('error')
title('0:1/3:5')
L = linspace(0,5,16);
subplot(2,1,2)
plot(diff(L)-1/3,'x-')
hold on
plot(lims,'k:')
plot(-lims,'k:')
plot([1,15],[0,0],'k:')
title('linspace(0,5,16)')
ylabel('error')
colonop:
function out = colonop(start,step,stop)
% COLONOP Demonstrate how the built-in a:d:b is constructed.
%
% v = colonop(a,b) constructs v = a:1:b.
% v = colonop(a,d,b) constructs v = a:d:b.
%
% v = a:d:b is not constructed using repeated addition. If the
% textual representation of d in the source code cannot be
% exactly represented in binary floating point, then repeated
% addition will appear to have accumlated roundoff error. In
% some cases, d may be so small that the floating point number
% nearest a+d is actually a. Here are two imporant examples.
%
% v = 1-eps : eps/4 : 1+eps is the nine floating point numbers
% closest to v = 1 + (-4:1:4)*eps/4. Since the spacing of the
% floating point numbers between 1-eps and 1 is eps/2 and the
% spacing between 1 and 1+eps is eps,
% v = [1-eps 1-eps 1-eps/2 1 1 1 1 1+eps 1+eps].
%
% Even though 0.01 is not exactly represented in binary,
% v = -1 : 0.01 : 1 consists of 201 floating points numbers
% centered symmetrically about zero.
%
% Ideally, in exact arithmetic, for b > a and d > 0,
% v = a:d:b should be the vector of length n+1 generated by
% v = a + (0:n)*d where n = floor((b-a)/d).
% In floating point arithmetic, the delicate computatations
% are the value of n, the value of the right hand end point,
% c = a+n*d, and symmetry about the mid-point.
if nargin < 3
stop = step;
step = 1;
end
tol = 2.0*eps*max(abs(start),abs(stop));
sig = sign(step);
% Exceptional cases.
if ~isfinite(start) || ~isfinite(step) || ~isfinite(stop)
out = NaN;
return
elseif step == 0 || start < stop && step < 0 || stop < start && step > 0
% Result is empty.
out = zeros(1,0);
return
end
% n = number of intervals = length(v) - 1.
if start == floor(start) && step == 1
% Consecutive integers.
n = floor(stop) - start;
elseif start == floor(start) && step == floor(step)
% Integers with spacing > 1.
q = floor(start/step);
r = start - q*step;
n = floor((stop-r)/step) - q;
else
% General case.
n = round((stop-start)/step);
if sig*(start+n*step - stop) > tol
n = n - 1;
end
end
% last = right hand end point.
last = start + n*step;
if sig*(last-stop) > -tol
last = stop;
end
% out should be symmetric about the mid-point.
out = zeros(1,n+1);
k = 0:floor(n/2);
out(1+k) = start + k*step;
out(n+1-k) = last - k*step;
if mod(n,2) == 0
out(n/2+1) = (start+last)/2;
end