Variable 'D' is not fully defined on some execution paths. Why?
Mostrar comentarios más antiguos
Hello. I'm using the following code in embedded matlab in simulink. And I,m getting the message that Variable 'D' is not fully defined on some execution paths. And i don't know what is the problem in code.
function D = mpp(V,I,T)
Dinit = 0.7;
delD = 0.001;
Dmax = 0.9;
Dmin = 0.1;
persistent P0 V0 D0 n;
if isempty(P0)
P0 = 0;
V0 = 0;
n = 1;
D0 = Dinit;
end
P = V*I;
dP = P - P0;
dV = V - V0;
if(T>0.02*n)
n=n+1;
if(dP*dV > 0)
D = D0 - delD;
elseif(dP*dV < 0)
D = D0 + delD;
end
else
D = D0;
end
if D>=Dmax | D<=Dmin
D = D0;
end
V0 = V;
P0 = P;
D0 = D;
Respuestas (1)
Walter Roberson
el 2 de Jun. de 2017
if(dP*dV > 0)
D = D0 - delD;
elseif(dP*dV < 0)
D = D0 + delD;
end
does not assign to D if dp*dV == 0
4 comentarios
Soham Delvadia
el 2 de Jun. de 2017
Walter Roberson
el 3 de Jun. de 2017
Editada: Walter Roberson
el 3 de Jun. de 2017
What did you change the code to? The code you posted has a problem in the situation that dP == 0 or dV == 0.
小东 刘
el 13 de Oct. de 2022
hello! now I have the same problem. How did you change it ?
Look at the simple example below. What does fun343107 do when x is exactly equal to 0? That behavior is undefined, so you'll receive a similar type of error as the original poster. To fix it, add code to handle the case where x is exactly equal to 0 in fun343107; that's the code that's commented out in the function.
Look for this same type of situation in your code.
y = fun343107(1)
y = fun343107(-1)
y = fun343107(0)
function y = fun343107(x)
if x > 0
y = 'positive';
elseif x < 0
y = 'negative';
% else
% y = 'zero';
end
end
Categorías
Más información sobre Audio I/O and Waveform Generation en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!