# how to use function handles

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G A on 29 Aug 2021
Commented: G A on 31 Aug 2021
Instead of the following, I want to use function handles and move if-statments out of the loop:
for n=1:10
if a==0
A = function2(n,a,b);
else
A = function1(n,a,b,c,d);
end
end
function A = function1(n,a,b,c,d)
A = n+a+b+c+d; % here it can be any expression
end
function A = function2(n,a,b)
A = n*a*b; % here it can be any expression
end
moving if-statement outside of the loop and using handles:
if a==0
func = @function2;
else
func = @function1;
end
and then I can solve my problem in two ways as
for n=1:10
A = func(n,a,b,c,d);
end
function A = function1(n,a,b,c,d)
A = n+a+b+c+d;
end
function A = function2(n,a,b,~,~)
A = n*a*b;
end
or as
for n=1:10
A = func(n,a,b,c,d)
end
function A = function1(n,a,varargin)
b = varargin{1};
c = varargin{2};
d = varargin{3};
A = n+a+b+c+d;
end
function A = function2(n,a,varargin)
b = varargin{1};
A = n*a*b;
end
Is there any more elegant way to do this?
G A on 29 Aug 2021
may be this way could be better:
function A = function1(n,S)
A = n + S.a + S.b + S.c + S.d;
end
function A = function2(n,S)
A = n*S.a*S.b;
end

Yongjian Feng on 29 Aug 2021
You don't need varargin
function A = function1(n,a,b, c, d)
A = n+a+b+c+d;
end
function A = function2(n,a,b, ~, ~)
A = n*a*b;
end
G A on 31 Aug 2021

the cyclist on 29 Aug 2021
I think the basic idea you need is
f1 = @(x) x;
f2 = @(x) x.^2;
f = @(a,x) (a~=0)*f1(x) + (a==0)*f2(x);
f(0,2)
ans = 4
f(1,2)
ans = 2
the cyclist on 29 Aug 2021
Searching keywords such as conditional anonymous function MATLAB turns up a lot of these same ideas. I did not find a great solution.