Taking in a function as an argument
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Jakob Grunditz
le 8 Nov 2019
Commenté : Jakob Grunditz
le 8 Nov 2019
I'm trying to make a function which takes in a function and then uses it. Right now my code looks like this:
function out = func(inputfunc, constant1, constant2)
inputfuncval1 = inputfunc(constant1);
...code...
end
When I then try to call it with a function like this:
out = func(@(x) testfunc(x),x1 ,x2);
It interpretets my input as an array instead of a function.
Anyone got any ideas on how to solve this issue?
2 commentaires
Guillaume
le 8 Nov 2019
Can you give us the full text of the error you see. The code you show should work although
out = func(@testfunc, x1, x2);
would be simpler.
Réponse acceptée
Guillaume
le 8 Nov 2019
Seems clear enough:
Error in bisekt (line 14)
bisekt(m, x2)
Within bisekt, you call the function again, this time with input m which is a scalar value, not a function, and only one argument. So, the input f is then your m scalar, and of course, m(x2) is going to fail if x2 is anything but 1.
Perhaps, the call should be:
bisekt(f, m, x2);
same for the other recursions...
Plus de réponses (1)
M
le 8 Nov 2019
How is testfunc defined ?
Here is a simple working example:
function out = func(inputfunc, constant1, constant2)
out = inputfunc(constant1, constant2);
end
out = func(@(x,y) x+y,1,2)
out =
3
2 commentaires
Guillaume
le 8 Nov 2019
As I commented in your question, if the above doesn't work, give us the full text of the error message.
With the following function:
function out = func(inputfunc, constant1, constant2)
out(1) = inputfunc(constant1);
out(2) = inputfunc(constant2);
end
I get:
>> out = func(@(x) exp(x)+x^2-1, 1, 2)
out =
2.71828182845905 10.3890560989307
No problem there.
Voir également
Catégories
En savoir plus sur Creating and Concatenating Matrices dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!