How to fix error "Array indices must be positive integers or logical values."

I am fairly new to MATLAB, and I am trying to write a code in MATLAB for the Bisect method, like so:

% Bisect Method
% Computes approximate solution of f(x) = 0
% Input: function handle f: a, b, s.t. f(a) * f(b) < 0
%        and tolerance handle tol
% Output: Approximate solution xc
function xc = mybisect(f, a, b, tol)
format longg;

a_vals = zeros(a, length(tol));
b_vals = zeros(b, length(tol));
c_vals = zeros(1, length(tol));

if sign(f(a)) * sign(f(b)) >= 0
    error ('f(a)f(b) < 0 NOT SATISFIED!')
end

fa = f(a);
fb = f(b);

while (b-a) / 2 > tol
    c = (a+b) / 2; 
    fc = f(c); 
    c_vals(1i) = c;
    
    if (fc == 0) % c is a solution
        break
    end
    
    if sign(fc) * sign(fa) < 0 % a and c make new interval
        b = c;
        fb = fc;
        b_vals(2i) = b;
    else                       % b and c make new interval 
        a = c; 
        fa = fc;
        a_vals(2i) = a;
    end
    
    xc = (a + b) / 2;
    iter_nr = size(c_vals, 2);
end

fprintf('\n');     
disp("A-VALUES:")
perLine = 5;
fmt = [repmat('%8.10f ', 1, perLine), '\n'];
fprintf(fmt, a_vals);
if mod( length(a_vals), perLine) ~= 0
    fprintf('\n'); 
end
fprintf('\n'); 
disp("B-VALUES:")
perLine = 5;
fmt = [repmat('%8.10f ', 1, perLine), '\n'];
fprintf(fmt, b_vals);
if mod( length(b_vals), perLine) ~= 0
    fprintf('\n'); 
end
fprintf('\n'); 
disp("C-VALUES:")
perLine = 5;
fmt = [repmat('%8.10f ', 1, perLine), '\n'];
fprintf(fmt, c_vals);
if mod( length(c_vals), perLine) ~= 0
    fprintf('\n'); 
end
fprintf('\n'); 
disp("Number of iterations : " + iter_nr);
end

But I get this error-message:

Array indices must be positive integers or logical values.

Error in mybisect (line 24)
c_vals(1i) = c;

I don't understand why. How do I fix this?

[Torsten]

a_vals = zeros(a, length(tol));
b_vals = zeros(b, length(tol));
c_vals = zeros(1, length(tol));

a and b are real numbers - it's not possible to give an array the size of a real number.

tol is a scalar - thus length(tol) = 1. Also this setting in the dimensioning part does not make sense.

c_vals(1i) = c;
b_vals(2i) = b;
a_vals(2i) = a;

1i and 2i are not defined before. Further, variable names are not allowed to start with a number.

But worse: MATLAB interprets i in the expressions as the complex unit.

Thus complete confusion.

[Rik]

I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.

Answer 1

[Chunru]

function xc = mybisect(f, a, b, tol)
format longg;
a_vals = zeros(a, length(tol));
b_vals = zeros(b, length(tol));
c_vals = zeros(1, length(tol));
if sign(f(a)) * sign(f(b)) >= 0
    error ('f(a)f(b) < 0 NOT SATISFIED!')
end
fa = f(a);
fb = f(b);
i = 1;          % initialize i   <===============================
while (b-a) / 2 > tol
    c = (a+b) / 2; 
    fc = f(c); 
    %c_vals(1i) = c;             % 1i is the complex number 
    c_vals(i) = c;               % <===========================
    if (fc == 0) % c is a solution
        break
    end
    
    if sign(fc) * sign(fa) < 0 % a and c make new interval
        b = c;
        fb = fc;
        b_vals(2i) = b;
    else                       % b and c make new interval 
        a = c; 
        fa = fc;
        a_vals(2i) = a;
    end
    
    xc = (a + b) / 2;
    iter_nr = size(c_vals, 2);
    i = i+1;                   % <==========================================
end