CODEWITHSUNDEEP
matlaboctave
user7423098
user7423098

Reputation: 321

How to call multiple dependent functions?

I want to call multiple dependent functions like this in the code.

function  [k11,k12] = k_fun()
% syntax is : function [outputs] = function-name(inputs)
  a=2.0;
  k11 = quad(B1,B1,a); %error
% For k11,  I want to call function b_fun and select only B1,B2 to pass to function quad.
  k12 = quad(B1,B2,a);
 endfunction

 function  [B] = b_fun(x)
  B1 = -0.5+x;
  B2 = 2.0*x;
  B3 = 0.5+x;
  B=[B1,B2,B3];
  endfunction


 function  [value] = quad(B_i,B_j,a)
   value=0
   points = [0.57,-0.57];
   wt=[1.0,1.0]
   for (ct=1:1:2)
    value = value + (B_i(points(ct))*B_j(points(ct))*a)*wt(ct);
   end
 endfunction

I want to run function k_fun. After a=2.0 It will go to k11 line. For k11, I want to get B1 and B2 from b_fun(), pass it to quad(). In quad(), the functions will be evaluated.

I am trying to do k12 = B1(0.57)*B2(0.57)*a*1 + B1(-0.57)*B2(-0.57)*a*1. These values, +-0.57, come in the function quad. I am not getting how to pass B1 and B2 in the line %error. I am getting stuck in invoking dependent functions. I want pretty much to keep same program format (variables defined in the respective functions) because the original program goes like this. Later, I want to translate this program to C++, so would like to use standard functions of programming languages instead of specific ones.

By the way, I started the program in Octave, so endfunction instead of end in MATLAB.

Upvotes: 2

Views: 85

Answers (2)

Wolfie
Wolfie

Reputation: 30047

Here's another method:

B = @(x) [-0.5 + x ; 
          2.0 .* x ;
          0.5 +  x ];

Now you can pass a row vector to B and get a column of outputs! So

points = [0.57,-0.57];

B(points) =
    0.0700   -1.0700
    1.1400   -1.1400
    1.0700   -0.0700

Then you say you want to actually compute:

k12 = B1(0.57)*B2(0.57)*a*1 + B1(-0.57)*B2(-0.57)*a*1

Which, under this new structure would be

k12 = B(1,1)*B(2,1)*a*1 + B(1,2)*B(2,2)*a*1

I'm not sure why you've left the *1s in, but I've kept them for completeness. In summary, you only have to have one B_output = B(points) variable, and B2 is now accessed via B_output(2,i), where i is the column index corresponding to the points index. If you only passed B(x) one point, you could say B2 = B_output(2)!

To summarise

Your entire code could be written:

a = 2.0;
B = @(x) [-0.5 + x ; 
          2.0 .* x ;
          0.5 +  x ];

k11 = quad(B, 1, 1, a);
k12 = quad(B, 1, 2, a);

function value = quad(B, i, j, a)

    value = 0;
    points = [0.57,-0.57];
    wt = [1.0,1.0];

    B_output = B(points);

    for ct = 1:2

        value = value + B_output(i, ct)*B_output(j, ct)*a*wt(ct);

    end

end

Upvotes: 0

Suever
Suever

Reputation: 65430

Your best bet is going to be to reorganize a little bit so that you have a separate version of b_fun for each B* variable. You can then create a function handle to these b_fun's and pass that to quad.

function [k11, k12] = k_fun()
    a = 2.0;
    k11 = quad(@b1_fun, @b1_fun, a);
    k12 = quad(@b1_fun, @b2_fun, a);
end

function result = b1_fun(x)
    result = x - 0.5;
end

function result = b2_fun(x)
    result = 2 * x;
end

Alternately, you can just create B1 and B2 anonymous functions since they are so straightforward

function [k11, k12] = k_fun()
    a = 2.0;

    B1 = @(x)x - 0.5;
    B2 = @(x) 2 * x;

    k11 = quad(B1, B1, a);
    k12 = quad(B1, B2, a);
end

Upvotes: 1

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