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how to not use all input arguments in the function because some of the arguments are fixed?

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how to not use all input arguments in the function because some of the arguments are fixed?

Accepted Answer

DGM
DGM on 16 Feb 2022
Edited: DGM on 16 Feb 2022
If you're writing a function and want certain arguments to be optional (with internal defaults), read about varargin
From the scope of the function, varargin can be handled as a cell array. How you parse/validate its contents is up to your needs.
I generally assign all the parameters to their default values prior to parsing the inputs, overwriting the defaults as the user-defined values are collected from varargin.
  3 Comments
DGM
DGM on 16 Feb 2022
Unless area() is nested inside another function wherein B is defined, the above definition isn't available to it. It would either need to be explicitly passed to area(), or area() would need to internally define it such that it's a constant or a default for an optionally user-specified parameter.
As I doubt it really makes sense to have a rectangle area function that presumes the size of the rectangle, Steven's suggestion is probably more appropriate.
That said, I'll just offer this for sake of clarification anyway.
area(5,10)
ans = 50
area(5)
ans = 100
function [A] = area(varargin)
% A = area(height,{width})
% calculate the area of a rectangle
% if not specified, the width is assumed to be 20
B = 20; % default
narginchk(1,2);
switch nargin
case 1
h = varargin{1};
case 2
h = varargin{1};
B = varargin{2};
end
A = B*h;
end
Alfandias Saurena
Alfandias Saurena on 16 Feb 2022
So, B is defined inside the function.
this is another function i created. i define (Qn,S,B,M, and ks) for (rough_secant) function because in (secant) function its only need (y) variabel
clc;
clear all;
y=secant(1,@rough_secant);
function [dQ]=rough_secant(y)
Qn=84;
S=0.0005;
B=20;
m=0;
ks=0.006;
A=luas_penampang(B,m,y);
P=keliling_basah(B,m,y);
R=jari2_hidrolis(A,P);
C=chezy_rough(y,ks);
V=kecepatan_aliran(C,R,S);
Qc=V*A;
dQ=Qn-Qc;
end
function [C]=chezy_rough(y,ks)
C=18*log10(12*y/ks);
end
function [us]=shear_velocity(R,S)
us=sqrt(9.81*R*S);
end
function [C]=chezy_smooth(y,vsc,us)
C=18*log10(12*y/(3.3*vsc/us));
end
function [V]=kecepatan_aliran(C,R,S)
V=C*sqrt(R*S);
end
function [A]=luas_penampang(B,m,y)
A=(B+m*y)*y;
end
function [P]=keliling_basah(B,m,y)
P=B+2*sqrt(m^2+1).*y;
end
function [R]=jari2_hidrolis(A,P)
R=A/P;
end
function [y]=secant(x,F_x)
dx=0.01;
y=x;
f=F_x(y);
while abs(f)>0.0000001
f=F_x(y);
f1=F_x(y+dx);
f2=F_x(y-dx);
f3=(f1-f2)/(2*dx);
y2=y-f/f3;
y=y2;
end
end

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More Answers (1)

Steven Lord
Steven Lord on 16 Feb 2022
Edited: Steven Lord on 16 Feb 2022
You can use an anonymous function "adapter".
f = @(in1, in2) max(in1, in2); % I could have used @max
% but I wanted to be explicit
f_2p5 = @(x) f(x, 2.5); % Call f with the first input specified by
% the user and the second fixed by me as 2.5
f_2p5(1:5)
ans = 1×5
2.5000 2.5000 3.0000 4.0000 5.0000
f(1:5, 2.5)
ans = 1×5
2.5000 2.5000 3.0000 4.0000 5.0000

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