Taking in a function as an argument

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Jakob Grunditz
Jakob Grunditz 2019-11-8
评论： Jakob Grunditz 2019-11-8
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 个评论
Guillaume
Guillaume 2019-11-8
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.
Jakob Grunditz
Jakob Grunditz 2019-11-8
To understand the error I must give you the whole code so... you're welcome
function svar = bisekt(f, x1, x2)
fx1 = f(x1);
fx2 = f(x2);
if fx1>0 && fx2<0
if abs(x1-x2)>10^-6
m = (x1+x2)/2;
fm = f(m);
if fm == 0
svar = m;
return
elseif fm>0
bisekt(m, x2)
elseif fm<0
bisekt(x1, m)
end
else
svar = x1;
return
end
elseif fx1<0 && fx2>0
if abs(x1-x2)>10^-6
m = (x1+x2)/2;
fm = f(m);
if fm == 0
svar = m;
return
elseif fm<0
bisekt(m, x2)
elseif fm>0
bisekt(x1, m)
end
else
svar = x1;
return
end
end
end
svar = bisekt(@(x) exp(x)+x^2-1, -2, -0.5)
error code
Array indices must be positive integers or logical values.
Error in bisekt (line 2)
fx1 = f(x1);
Error in bisekt (line 14)
bisekt(m, x2)

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采纳的回答

Guillaume
Guillaume 2019-11-8
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...

更多回答（1 个）

M
M 2019-11-8
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 个评论
Jakob Grunditz
Jakob Grunditz 2019-11-8
In my case I tried inputting the function:
inputfunc = @(x) exp(x)+x^2-1
that I then try to evaluate in two places constant1 and constant2 for later use in the function.
function out = func(inputfunc, constant1, constant2)
inputfuncval1 = inputfunc(constant1);
inputfuncval2 = inputfunc(constant2);
...code...
end
out = func(@(x) exp(x)+x^2-1, x1, x2)
Guillaume
Guillaume 2019-11-8
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.

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