Repeated using of the for-loop with different steps ...

I use a for loop to solve a problem with a time step. Inside a for-loop there is equation 'X' that depends on the time-step size. This time step is defined outside the for-loop. Upon each iteration a difference between an actual and previous value of 'X'-equation should not exceed a certain value (for example 5). If this value is exceeded the for loop should break iteration and the next value of the time-step should be taken form defined time-step range. After that a for-loop should run again but with a newly chosen time-step value inside.

[Stephen23]

An explanation is nice, but seeing your code is even better. Please do either of these:

• edit your question and actually upload your complete code (if it is large)
• edit your question and include your code in the text of the question (if it is small). Use the text formatting button above the text box: {} Code.

[Oleg]

could you explain why 'break' command does not terminate a for-loop in this code? z defines a difference between b-results and if this difference exceeds 12 then for-loop should be stopped. In my case it does not happen.

   a = 1:100;
    t_del = 0.5;
    b = zeros(1,length(a));
    c = 0;
    n = 12;
    for i = 1:length(a)
        c = c + 2*a(i)*t_del;
        b(i) = (c);
        z = diff(b)
        if z > n
            break
        end
    end

Answer 1

[Star Strider]

I would use a while loop:

a = 1:100;
t_del = 0.5;
b = zeros(1,length(a));
c(1) = 0;
n = 12;
z = 0;
k = 1;
while z < n
    k = k + 1;
    c(k) = c(k-1) + 2*a(k-1)*t_del;
    b(k) = (c(k));
    z = diff(b);
end

I am not certain what you want, so you may have to experiment with it.

EDIT —

The likely reason that break does not do what you want it to has to do with the condition in your if statement, and the way MATLAB evaluates conditions on vectors. (Your ‘z’ variable is a vector.)

With one change, this version of your original code works:

a = 1:100;
t_del = 0.5;
b = zeros(1,length(a));
c = 0;
n = 12;
for k = 1:length(a)
    c = c + 2*a(k)*t_del;
    b(k) = (c);
    z = diff(b)
    if any(z > n)
        break
    end
end

Note the use of the any function as the condition in your if statement.

[Image Analyst]

"break" does break out of the loop. "continue" skips the the bottom and keeps going with the next iteration (if it was not already the last iteration).

[Star Strider]

Correct. I tested it in my code to be sure, then when I confirmed my impression of break (your experience as well), I looked closer at Oleg’s code to see if I could discover the problem, and found the error. I edited my Answer to reflect this change, and fixed Oleg’s code.

[Oleg]

Star Strider, could you please give any suggestions on how I can re-run the loop with another t_del value. For example t_del = (0.5;0.4;0.3). How can chose the next value from the list after the for-loop is stopped and than run it with a new t_del value ?

[Star Strider]

If I understand correctly what you want to do, this should work for a vector of ‘t_del’:

a = 1:100;
t_del = [0.5; 0.4; 0.3];
b = zeros(1,length(a));
c = 0;
n = 12;
for k1 = 1:length(t_del)
    for k2 = 1:length(a)
        c = c + 2*a(k2)*t_del(k1);
        b(k1,k2) = (c);
        z = diff(b(k1,:));
        if any(z > n)
            break
        end
    end
end

It adds a loop for the individual ‘t_del’ values, but its behaviour is otherwise unchanged, although the ‘b’ and ‘z’ calculations now have to take that into account. Note that ‘b’ becomes a matrix, not a vector. Each row is for the corresponding value of ‘t_del’.

[Oleg]

Should I use continue command if t_del value satisfies the condition (z<n) and I don't want to take any other value from a t_del vector but continue iteration with only suitable t_del.

   if any(z > n)
       break
    else
       continue
    end

[Guillaume]

continue skips the rest of the body of a for loop and jumps directly to the end statement. Therefore, continue as the last statement in the loop has no effect and is just confusing. Don't do it.

[Star Strider]

@Guillame — Spot on. Thanks.