Undefined operand in function_handle
信息
此问题已关闭。 请重新打开它进行编辑或回答。
显示 更早的评论
When I enter ' H(3) ' in the command line, the following error is shown:
Undefined operator '-' for input arguments of type 'function_handle'.
Error in
Trial_H>@(m)d1+s*(cos(gama)+sum(cos(gama+(pi/2-phi_i))))+((2*r)-omega)*sin(B_j(m))+(r-omega)*sum(sin((1:m-1)*pi-sum(B_j(1:m-1))))
Here is the code:
% Variables
L = 10;
R = 1;
r = R/(m+1);
S = sqrt(2);
s = S/(m+1);
d1 = 1;
d2 = 0.25;
omega = R/(3*(m+1));
gama = atan(R/d1);
% Functions
phi_i = @(ii, m) atan( (d1+(ii*d2)./(m+1) )./(r/3) );
B_j = @(ii, m) 2*phi_i(ii, m);
H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2-phi_i(1:m)))) ) + ( (2*r)-omega)*sin(B_j(m) ) + (r-omega)*sum(sin((1:m-1)*pi - sum(B_j(1:m-1))));
0 个评论
回答（1 个）
Tommy
2020-4-7
Where you call phi_i or B_j within the definition of H, you need to supply the arguments ii and m. For example,
..(pi/2-phi_i)..
is problematic because MATLAB thinks you are trying to subtract the function phi_i from pi/2. If you pass arguments,
..(pi/2-phi_i(1,m))..
for example, MATLAB will instead subtract the output of phi_i from pi/2.
7 个评论
Jay
2020-4-7
Tommy
2020-4-7
Run this from the command window:
>> B_j(3)
Not enough input arguments.
Error in answers>@(ii,m)2*phi_i(ii,m)
The function B_j takes two inputs, ii and m, and calls phi_i(ii, m). If you call B_j(3), you are only giving B_j the first input, ii. It does not know what m is, and when it tries to call phi_i, it cannot. You need to pass both arguments into B_j:
>> B_j(1,3)
ans =
2.9851
Now run this from the command window:
>> 2+phi_i
Undefined operator '+' for input arguments of type
'function_handle'.
phi_i is a function. You cannot add or subtract anything to/from it. You can, however, do this:
>> 2+phi_i(1,3)
ans =
3.4925
Fill all of your function calls with both arguments and you shouldn't get any errors.
Jay
2020-4-7
Tommy
2020-4-7
That is where you define B_j. You call B_j twice within H() (and you call phi_i once):
H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2- ...
phi_i ... HERE
))) ) + ( (2*r)-omega)*sin( ...
B_j(m) ... HERE
) + (r-omega)*sum(sin((1:m-1)*pi - sum( ...
B_j(1:m-1) ... HERE
)));
All three of those lines need to be changed, as described above.
Something like this would work:
H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2- ...
phi_i(1,m) ... HERE
))) ) + ( (2*r)-omega)*sin( ...
B_j(1,m) ... HERE
) + (r-omega)*sum(sin((1:m-1)*pi - sum( ...
B_j(1,1:m-1) ... HERE
)));
I have used 1 for ii in all three lines, but I do not know if that is the value for ii that you want to use.
Jay
2020-4-7
Tommy
2020-4-7
You cannot use m only, because your definition of B_j
B_j = @(ii, m) 2*phi_i(ii, m)
requires both ii and m.
Steven Lord
2020-4-7
Calling B_j, which requires two inputs, with just one would be like me asking you "The first number is 7. What is the sum of the two numbers?" There's not enough information to answer the question.
此问题已关闭。
关闭：
2021-8-20
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
