Do constants created in primary function transfer over to subsidiary functions created?

Question

Jaider 2022-2-20

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1654480-do-constants-created-in-primary-function-transfer-over-to-subsidiary-functions-created

评论： Jaider 2022-2-21

So bisect is my primary function where I do my algorithm and call upon the other functions to be used. Is it necessary for me to pass all the constants created under bisect to all the subsidiary functions when I call them? Or can I safely ignore this and just say f(x) instead of f(x,h,xp,..)?

function bisect
%constants for routing problem
h=0.01;
xp=5;
yp=4;
xc=2;
yc=2;
r=1;
rho=10;
%numerical scheme constants
a=-5;
b=5;
n=20; % this will be the number of steps in the bisection method .
p=(a+b)/2;
omega=f(a);
eta=f(b);
gamma=f(p);
for i=1:n
  z=(a+p)/2;
  y=(p+b)/2;
  alpha=f(z);
  beta=f(y);
  fmin=min(omega,eta,gamma,alpha,beta);
  if (fmin==omega)&&(fmin==alpha)
      b=p;
      p=z;
      eta=gamma;
      gamma=alpha;
  elseif fmin==gamma
      a=z;
      b=y;
      omega=alpha;
      gamma=beta;
  elseif (fmin==beta)&&(fmin==eta)
      a=p;
      p=y;
      omega=gamma;
      gamma=beta;
  end
  fprintf(1,'i= %4i a= %18.14f b=%18.14f p=%18.14f \n',i,a,b,p);
  x=abs(b-a);
end 
fprintf(1,'i= %4i a= %18.14f b=%18.14f p=%18.14f \n',i,a,b,p);
function y=f(x,xc,yc,xp,yp,h,rho,r)
y=x+((x-xp).^2+yp.^2).^(1/2)+2*rho*r((-1)*((x-xp).*yc-yp.*(x-xc)).^(2)/((x-xp).^2+yp.^2)).^(1/2);
function y=df(x,h)
y=(f(x+h)-f(x-h))./(2.*h);
function y=ddf(x,h)
y=(f(x+h)-2.*f(x)+f(x-h))/(h.^2);

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Answer 1

John D'Errico 2022-2-20

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https://ww2.mathworks.cn/matlabcentral/answers/1654480-do-constants-created-in-primary-function-transfer-over-to-subsidiary-functions-created#answer_900680

编辑：John D'Errico 2022-2-20

在 MATLAB Online 中打开

There are TWO distinct type of subsidiary functions in MATLAB, sub-functions, and nested functions.

Consider first the function subexample. As you can see, in the workspace of the sub-function, y does not exist. x exists for the sub-function, only because I passed it in.

subexample(1)
sub-functions cannot see the caller workspace
  Name      Size            Bytes  Class     Attributes

  x         1x1                 8  double              

Whos saw ONLY x, the argument passed in.

But now consider the function nestedexampe.

nestedexample(1)
Nested functions can see the caller workspace
  Name      Size            Bytes  Class     Attributes

  x         1x1                 8  double              
  y         1x1                 8  double              

y, in the nested function workspace, before it was changed in the nested function.2
y, in the nested function workspace, after it was changed in the nested function.5
y, in the caller workspace, after it was changed in the nested function.5

Do you see that subb can see y? In fact, the caller workspace can see that y has been modified. The difference is where suba and subb are placed, in terms of the end statement for the caller function. In fact, If subb needs to use the variable x, I did not even need to pass it in.

function nestedexample(x)
  % do stuff
  y = 2;
  
  subb
  
  disp("y, in the caller workspace, after it was changed in the nested function." + y)
  
function subb
  % in this function, y does exist in the workspace
  disp("Nested functions can see the caller workspace")
  whos 
  % we can use y here.
  disp("y, in the nested function workspace, before it was changed in the nested function." + y)
  y = y + 3;
  disp("y, in the nested function workspace, after it was changed in the nested function." + y)
end
end

function subexample(x)
  % do stuff
  y = 2;
  suba(x)
end
function suba(x)
  % in this function, y does not exist in the workspace
  disp("sub-functions cannot see the caller workspace")
  whos 
end