Modelica evaluation of an expression - modelica

Question

I'm new to Modelica, and I'm wondering, what would be the evaluation of rA within model B.
Is the value 5 or 11 ?
model A
Real rA = 1 + rB;
Real rB = 4;
model B
Real rB = 10;
// Case A : Evaluation of rA = 1 + 4 = 5
// Case B : Evaluation of rA = 1 + 10 = 11
end B;
end A;
Edit : what would be the evaluation of rD in model A using the code below ?
Is the value 5 or 11 ?
model A
Real rA = 1 + rB;
Real rB = 4;
model B
Real rB = 10;
Real rC = rA;
end B;
B myB;
Real rD = myB.rC;
// What is the result of the evaluation of rD ?
// Case A : Evaluation of rD = 1 + 4 = 5
// Case B : Evaluation of rD = 1 + 10 = 11
end A;

Answer 1

The rD = 1 + 4. The rB in model B is not involved in the calculation of rC (or rA). The expression is not like a C macro that is evaluated in that context again.
If you want to have the behavior that myB.rB influences the calculation of myB.rC then you need to change your code.
model A
Real rA = 1 + rB;
Real rB = 4;
model B
Real rB = 10;
Real rC = 1 + rB;
end B;
B myB;
Real rD = myB.rC;
// What is the result of the evaluation of rD?
// Case B : Evaluation of rD = 1 + 10 = 11
end A;
or make a function if that expression is complicated:
model A
function f
input Real x;
output Real y;
algorithm
y := 1 + x;
end f;
Real rA = f(rB);
Real rB = 4;
model B
Real rB = 10;
Real rC = f(rB);
end B;
B myB;
Real rD = myB.rC;
// What is the result of the evaluation of rD?
// Case B : Evaluation of rD = 1 + 10 = 11
end A;

Answer 2

Model B is not instantiated so B.rB=10 is not "known" by model A. Thus, rA = 5
If you want the result to be 11 then the code should be something like:
model A
B b;
Real rA = 1 + rB;
Real rB = b.rB;
model B
Real rB = 10;
// Case A : Evaluation of rA = 1 + 4 = 5
// Case B : Evaluation of rA = 1 + 10 = 11
end B;
end A;