error in arrays line 7
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function V = fuelvol2(h)
global r H L
validInput = true; %to test each value in the vector h is valid
for i = 1:length(h)
Nh = h(i);
if (Nh < 0 && (Nh > h + 2*r))
validInput = false;
break;
end
end
if validInput == true % if it is valid vector the calculate volume using vectorized operations
d = h-H;
V = 2*r*L*H + (r^2 * acos((r-d)/r) - (r-d).*sqrt(2*r.*d -d.^2))*L;
else
disp("one of more values in vector his/are valid ")
end
I keep getting tis error each time i try to run the code
Not enough input arguments.
Error in fuelvol2 (line 7)
for i = 1:length(h)
6 件のコメント
Adam Danz
2019 年 9 月 29 日
Function calls are case sensitive. There is no matlab function named Length() but there is a function with the lower case length(). However, it's recommended to use numel() instead of length().
回答 (2 件)
Adam Danz
2019 年 9 月 29 日
It's getting hard to follow the discussion because we're jumping around between different similarly named functions. fuelvol1, fuelvol2, main_fuelvol12.
This error appears to be in fuelvol1 but we don't have access to that function. However, I think the error can be fixed by replacing && with &.
0 件のコメント
Walter Roberson
2019 年 9 月 30 日
編集済み: Walter Roberson
2019 年 9 月 30 日
for i = 1:length(h)
So you are expecting h to be a non-scalar.
if (Nh < 0 && (Nh > h + 2*r))
You add the non-scalar h to something and compare it to the scalar Nh, producing a non-scalar result on the right hand side of the && . However, the && operator can only be used when both sides are scalars.
Changing to & is not the correct solution. The correct solution is to stop using single-letter variable names that differ only in upper versus lower case, because it is too easy to get the wrong variable name. You do not want h there, you want H .
In your other question, H and 2*r were both positive, and Nh > some_positive_value cannot be simultaneously true with Nh < zero.
1 件のコメント
Adam Danz
2019 年 9 月 30 日
Just curious, from where did you get "for i = 1:length(h)"? I see it in fuelvol2() but the error is in fuelvol1() which we don't have access to, unless I'm missing something.
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