value must be a double scaler
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so I am trying to make a cubic equation calculater in matlab but i keep getting the error :
Error using matlab.ui.control.internal.model.AbstractNumericComponent/set.Value (line 111)
'Value' must be a double scalar.
I am really not sure what it means but here is the code:
vala = app.a.Value;
valb = app.b.Value;
valc = app.c.Value;
vald = app.d.Value;
zz1 =(-valb/3*vala)-(1/3*vala)*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald+sqrt(2*valb^3-9*vala*valb*valc+27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3-(1/3*vala)*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald-sqrt(2*valb^3-9*vala*valb*valc-27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3;
zz2 =(-valb/3*vala)+(1+1i*sqrt(3)/6*vala*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald+sqrt(2*valb^3-9*vala*valb*valc+27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3)+(1-1i*sqrt(3)/6*vala)*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald-sqrt(2*valb^3-9*vala*valb*valc+27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3;
zz3 =(-valb/3*vala)+(1-1i*sqrt(3)/6*vala*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald+sqrt(2*valb^3-9*vala*valb*valc+27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3)+(1+1i*sqrt(3)/6*vala)*(0.5*(2*valb^3-9*vala*valb*valc+27*vala^2*vald-sqrt(2*valb^3-9*vala*valb*valc+27*vala^2*vald)^2-(4*valb^2-3*vala*valc)^3)^0.5)^1/3;
app.x1.Value =zz1;
app.x2.Value =zz2;
app.x3.Value =zz3;
p.s if you know any simpler ways that would be big help
7 个评论
Rik
2021-3-7
Doesn't the error message explain what is the problem? You need to use a scalar double. That means you can only set a single value, not multiple.
What are the values of zz1, zz2, and zz3?
Mohamed Rashed
2021-3-7
Walter Roberson
2021-3-7
One of zz1, zz2, zz3 is empty, or non-scalar, or not a double .
Is that roots of a cubic ? I see you are using imaginary components. At the moment I do not know if you are permitted to set the object to complex values.
Mohamed Rashed
2021-3-8
Walter Roberson
2021-3-8
Use roots() to find roots of the cubic.
And experiment with assigning real(zz1) and so on to the fields to test out whether really it is complaining about complex values.
Mohamed Rashed
2021-3-8
Walter Roberson
2021-3-8
"puts in the polynomial x^2+x+1" -- puts in how? Are you expecting that app.b.Value will be a formula instead of a numeric value? If you are expecting a numeric value, then App Designer permits constructing a numeric edit field that does not permit arbitrary text.
采纳的回答
Aghamarsh Varanasi
2021-3-12
编辑：Aghamarsh Varanasi
2021-3-12
Hi Rashed,
As suggested by Walter, you can use the root function in MATLAB to find the roots of a cubic polynomial.
From the error that you have mentioned, it seems that you are using a Edit Field (numeric) for x1, x2 and x3. As you are solving a cubic equation, there might be a chance that the root could be a complex number. As a workaround I would suggest you to use Edit Field (Text) of MATLAB App Designer for fields x1, x2 ,x3 and use num2str to convert the roots to string.
So the code would be:
vala = app.a.Value;
valb = app.b.Value;
valc = app.c.Value;
vald = app.d.Value;
p = [vala valb valc vald];
r = root(p);
% here x1, x2 and x3 are Edit Field (Text)
app.x1.Value = num2str(r(1));
app.x2.Value = num2str(r(2));
app.x3.Value = num2str(r(3));
Hope this helps
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