how to use function handles

조회 수: 2 (최근 30일)
G A
G A 2021년 8월 29일
댓글: G A 2021년 8월 31일
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?
  댓글 수: 1
G A
G A 2021년 8월 29일
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

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채택된 답변

Yongjian Feng
Yongjian Feng 2021년 8월 29일
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
  댓글 수: 7
Stephen23
Stephen23 2021년 8월 30일
"what is the main difficulty to make (or reason not to make) it possible in the Editor to higlight structure name with its field throughout a function like it occurs with simple variables when you click on it or select it?"
You have to consider that variables in MATLAB are dynamically typed. In general it is difficult to determine the type of a variable until runtime (e.g. changes to function scoping, variables created dynamically, inputs are not typed, etc.)
G A
G A 2021년 8월 31일
Thanks! - and I will accept this answer together with comments.

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추가 답변 (1개)

the cyclist
the cyclist 2021년 8월 29일
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
  댓글 수: 4
G A
G A 2021년 8월 29일
편집: G A 2021년 8월 29일
Also, it is time consuming if two functions are executed instead of one. Better to put if-statement into a loop. I have half of answer to my question in my comment above: 0*inf = NaN, 0*A = 0, so to give the answer 0 or NaN, Matlab must execute f2 before multipliing its return by 0.
the cyclist
the cyclist 2021년 8월 29일
Searching keywords such as conditional anonymous function MATLAB turns up a lot of these same ideas. I did not find a great solution.

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