value must be a double scaler

31 次查看(过去 30 天)
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
Mohamed Rashed
Mohamed Rashed 2021-3-8
do you maybe now where i could find how to display an error message on the app aswell like if someone puts in the polynomial x^2+x+1 or other impossible equations?
Walter Roberson
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.

请先登录,再进行评论。

采纳的回答

Aghamarsh Varanasi
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

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Startup and Shutdown 的更多信息

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by