While loop - display the last value while statement is still true

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The example given on Mathworks doesn't make sense to me.
x=0
p=1-poisscdf(1,x);
while p<=0.1
x=x+0.01;
p=1-poisscdf(1,x);
end
display(x);
The x value given at the end of the loop is the first value that makes the statement false.What can I add to the script so that display(x) gives me the last value that makes the statement true?
*Edit:
I just tried:
x=0;
p=1-poisscdf(1,x);
while p2more<=0.1
x=x+0.01;
p=1-poisscdf(1,x);
if p>=0.1
break
end
display(x);
end
Now I get all the values of x where this statement is true. I just want the last value. :X

Accepted Answer

James Tursa
James Tursa on 21 Mar 2017
Edited: James Tursa on 21 Mar 2017
You could just use another variable. E.g., y
x=0
p=1-poisscdf(1,x);
while p<=0.1
y = x; % <-- save current value of x
x=x+0.01;
p=1-poisscdf(1,x);
end
display(y); % <-- display last value of x for which test was true
  3 Comments
James Tursa
James Tursa on 24 Mar 2017
Right after the test passes, you save the x value that made the test pass. But when you generate an x value that doesn't pass the test, you don't execute the y = x statement. So it is only the x values that pass the test that are saved in y. And the last one is what makes it to the display(y) statement.

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More Answers (1)

Jan
Jan on 21 Mar 2017
Edited: Jan on 21 Mar 2017
The trivial soultion is:
x = 0;
p = 1 - poisscdf(1, x);
while p <= 0.1
x = x + 0.01;
p = 1 - poisscdf(1, x);
end
x = x - 0.01;
display(x);
If the increasing of x is not a constant:
x = 0;
lastx = x;
p = 1 - poisscdf(1, x);
while p <= 0.1
lastx = x;
x = x + 0.01;
p = 1 - poisscdf(1, x);
end
display(lastx);
Now p is defined twice and redundacies in the code are prone to bugs: If one instance is modified later, that other might be forgotton. To avoid this:
x = 0;
ready = false
while ~ready
p = 1 - poisscdf(1, x);
ready = (p < 0.1);
lastx = x;
x = x + 0.01;
end
display(lastx);

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