In my a for loop where I eg need to go from 1 to 11, and want to make my step size 4 what happens when the computer have to make the last step and it goes over 11?

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Christina Sørensen
评论： John D'Errico 2018-2-6
for 1:2:11 rest of code

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John D'Errico
John D'Errico 2018-2-6
编辑：John D'Errico 2018-2-6
Why not try it? You can see the behavior from one line of code.
1:4:11
ans =
1 5 9
So if the increment takes you past the endpoint, you do not get 11, or 13. You get up to the step that did not exceed the endpoint.
In fact, if the start point is itself above the end point, with a positive step, you get an empty. So the for loop would not iterate at all.
2:1:0
ans =
1×0 empty double row vector
As you can see, running this code never goes through the loop even once.
for i = 2:1:0
'sdfghwrhrt'
end
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Image Analyst
Image Analyst 2018-2-6
Then you need to create an array with linspace and index into it. For example:
numSteps = 30; % Whatever you want!
v = linspace(1, 11, numSteps);
for k = 1 : length(v)
thisV = v(k);
% Now put code to use thisV in your computations.
end
thisV will be 1 at the first iteration, and 11 at the last iteration. Basically linspace takes the first value, the last value, and the number of steps you need: v = linspace(startValue, endValue, numberOfSteps)
John D'Errico
John D'Errico 2018-2-6
If you always want some fixed number of steps, then linspace is of course the correct answer. But that won't insure a given increment. The increment from linspace will usually be some very arbitrary looking number.
If you know the desired increment, then colon is correct. But colon cannot insure that it hits the end point. Thus a call like
0:2:5
Will ALWAYS generate even numbers. That the end point is an odd number will never cause it to generate an odd number.

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