DynBindModelExamples

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Example 1

The system

In this example we are studying a symple system with three classes that contains both renaming and selects:

class C1 feature f1 do ... end g1 do ... end end

class C2 inherit B rename f1 as f2, g1 as g2 redefine f2, g2 end feature f2 do ... end g2 do ... end end

class C3 inherit C2 select f1 end C1 redefine f1, g1 select g1 end feature f1 do ... end g1 do ... end end

This system is defined by:

• $S=\{C_1,C_2,C_3\}$
• $S_1=\left\{C_1\right\},S_2=\{C_1,C_2\},S_3=\{C_1,C_2,C_3\}$
• $\tau =\{C_1\mapsto \{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\},C_2\mapsto \{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\},C_3\mapsto \{f_1\mapsto \underset{\_}{C_3.f_1},g_1\mapsto \underset{\_}{C_3.g_1}\}\}$
• $\eta =\{C_2\mapsto \{C_1\mapsto \{f_1\mapsto f_2,g_1\mapsto g_2\}\}\}$
• $\alpha =\{C_2\mapsto \left\{C_1\right\},C_3\mapsto \{C_2,C_1\}\}$
• $\sigma =\{C_3\mapsto \{C_2\mapsto \left\{f_1\right\},C_2\mapsto \left\{g_1\right\}\}\}$

The naming function

We proceed by calculating the naming function. According to its definition we have to do that incrementaly. Meaning that we first calculate ${\beta }_{S_1}$ then ${\beta }_{S_2}$ and finally ${\beta }_{S_3}$.

Some operators

We will give some rather informal definitions for the set-theoretic operators used here.

Operator: identity

$\text{id}\left(\{e_1,\dots ,e_n\}\right)=\{e_1\mapsto e_1,\dots ,e_n\mapsto e_n\}$

Operator: domain

$\text{dom}\left(\{l_1\mapsto r_1,\dots ,l_n\mapsto r_n\}\right)=\{l_1,\dots ,l_n\}$

Operator: range

$\text{ran}\left(\{l_1\mapsto r_1,\dots ,l_n\mapsto r_n\}\right)=\{r_1,\dots ,r_n\}$

Operator: override

$E\mapsto F\in p\ll q$ is equivalent to: $E\mapsto F\in q\vee (E\mapsto F\in p\wedge E\in ̸\text{dom}\left(p\right))$

Example:

$\{1\mapsto 2,2\mapsto 3\}\ll \{1\mapsto 4,3\mapsto 6\}=\{1\mapsto 4,2\mapsto 3,3\mapsto 6\}$

Operator: union

$E\in \text{UNION}x\cdot (x\in S|T)$ is equivalent to: $\mathrm{\exists }x\cdot (x\in S\wedge E\in T)$

Example:

$\text{UNION}x\cdot (x\in \{1,2,3\}|\{x,x^2\})$ is equivalent to: $\{1,2,3,4,9\}$

Calculating ${\beta }_{S_1}$

Since ${\beta }_{S_1}$ contains only one class it is sufficient to define ${\beta }_{S_1}(C_1\mapsto C_1)$. According to the definition:

${\beta }_{S_1}(C_1\mapsto C_1)=\text{id}\left(\text{dom}\left({\tau }_1\right)\right)\cup \text{UNION}b\cdot (b\in {\alpha }_1|\text{id}\left(\text{ran}({\beta }_{S_0}(b\mapsto b)\ll {\eta }_{1,b})\right))$

We can forget about the right part of the union since $C_1$ has no base classes. So we need to calculate: $\text{id}\left(\text{dom}\left({\tau }_1\right)\right)$. To do so we give a little reminder on the identity and domain operators:

This done we calculate:

1. ${\tau }_1=\tau (C_1)=\{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\}$
2. $\text{dom}\left({\tau }_1\right)=\{f_1,g_1\}$
3. $\text{id}\left(\text{dom}\left({\tau }_1\right)\right)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$

So we get:

${\beta }_{S_1}(C_1\mapsto C_1)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$

Calculating ${\beta }_{S_2}$

Lets calculate the easy part first:

${\beta }_{S_2}(C_1\mapsto C_1)={\beta }_{S_1}(C_1\mapsto C_1)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$

${\beta }_{S_2}(C_2\mapsto C_1)=\mathrm{\varnothing }$

The next least difficult thing we can calculate is:

${\beta }_{S_2}(C_2\mapsto C_2)=\text{id}\left(\text{dom}\left({\tau }_2\right)\right)\cup \text{UNION}b\cdot (b\in {\alpha }_2|{\beta }_{S_1}(b\mapsto b)\ll {\eta }_{2,b})$

We now calculate:

1. ${\tau }_2=\tau (C_2)=\{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}$
2. $\text{id}\left(\text{dom}\left({\tau }_2\right)\right)=\{f_2\mapsto f_2,g_2\mapsto g_2\}$
3. $\text{UNION}b\cdot (b\in {\alpha }_2|{\beta }_{S_1}(b\mapsto b)\ll {\eta }_{2,b})=\text{UNION}b\cdot (b\in \left\{C_1\right\}|{\beta }_{S_1}(b\mapsto b)\ll {\eta }_{2,b})={\beta }_{S_1}(C_1\mapsto C_1)\ll {\eta }_{2,C_1}$
4. ${\beta }_{S_1}(C_1\mapsto C_1)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$
5. ${\eta }_{2,C_1}=\eta (C_2)(C_1)=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
6. $\{f_1\mapsto f_1,g_1\mapsto g_1\}\ll \{f_1\mapsto f_2,g_1\mapsto g_2\}=\{f_1\mapsto f_2,g_1\mapsto g_2\}$