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 f2 end C1 redefine f1, g1 select g1 end feature f1 do ... end g1 do ... end end

This system is defined by:

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

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 ${\beta }_{S_3}$.

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:

1. $\text{id}\left(\text{dom}\left({\tau }_1\right)\right)$
   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:

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

And hence:

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

Calculating ${\beta }_{S_2}$

We show the calculation of the different (four) parts of the function in separate sections:

Calculate ${\beta }_{S_2}(C_1\mapsto C_1)$

According to the definition: ${\beta }_{S_2}(C_1\mapsto C_1)={\beta }_{S_1}(C_1\mapsto C_1)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$

Calculate ${\beta }_{S_2}(C_2\mapsto C_1)$

According to the definition: ${\beta }_{S_2}(C_2\mapsto C_1)=\mathrm{\varnothing }$

Calculate ${\beta }_{S_2}(C_2\mapsto C_2)$

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

Lets calculate the left part first:

1. ${\tau }_2=\tau (C_2)=\{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}$
2. $\text{id}\left(\text{dom}\left({\tau }_2\right)\right)=\{f_2\mapsto f_2,g_2\mapsto g_2\}$

And now the right part:

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

So we get:

${\beta }_{S_2}(C_2\mapsto C_2)=\{f_2\mapsto f_2,g_2\mapsto g_2\}$

Calculate ${\beta }_{S_2}(C_1\mapsto C_2)$

According to the definition: ${\beta }_{S_2}(C_1\mapsto C_2)={{\beta }_{S_2}}^{\prime }(C_1\mapsto C_2)\ll {\theta }_{S_2}(C_1\mapsto C_2)$

Where: ${{\beta }_{S_2}}^{\prime }(C_1\mapsto C_2)=\text{UNION}b\cdot (b\in {\alpha }_2|{\beta }_{S_1}(C_1\mapsto b);(\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto b)\right)\right)\ll {\eta }_{2,b}))$

And: ${\theta }_{S_2}(C_1\mapsto C_2)=\text{UNION}b\cdot (b\in {\alpha }_2|{\beta }_{S_1}(C_1\mapsto b);(\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto b)\right)\right)\ll {\eta }_{2,b})▹{\sigma }_{2,b})$

Here are the steps taken to compute the part:

1. ${\beta }_{S_1}(C_1\mapsto C_1);(\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto C_1)\right)\right)\ll {\eta }_{2,C_1})$
   1. $\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto C_1)\right)\right)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$
   2. $\{f_1\mapsto f_1,g_1\mapsto g_1\}\ll \{f_1\mapsto f_2,g_1\mapsto g_2\}=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
   3. $\{f_1\mapsto f_1,g_1\mapsto g_1\};\{f_1\mapsto f_2,g_1\mapsto g_2\}=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
   4. ${{\beta }_{S_2}}^{\prime }(C_1\mapsto C_2)=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
2. ${\beta }_{S_1}(C_1\mapsto C_1);(\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto C_1)\right)\right)\ll {\eta }_{2,C_1})▹{\sigma }_{2,C_1}$
   1. ${\sigma }_{2,C_1}=\mathrm{\varnothing }$
   2. $\text{id}\left(\text{ran}\left({\beta }_{S_1}(C_1\mapsto C_1)\right)\right)\ll \mathrm{\varnothing }=\mathrm{\varnothing }$
   3. ${\beta }_{S_1}(C_1\mapsto C_1);\mathrm{\varnothing }=\mathrm{\varnothing }$
   4. ${\theta }_{S_2}(C_1\mapsto C_2)=\mathrm{\varnothing }$
3. $\{f_1\mapsto f_2,g_1\mapsto g_2\}\ll \mathrm{\varnothing }=\{f_1\mapsto f_2,g_1\mapsto g_2\}$

So we get:

${\beta }_{S_2}(C_1\mapsto C_2)=\{f_1\mapsto f_2,g_1\mapsto g_2\}$

Putting the parts together:

We can know show the complete total function:

${\beta }_{S_2}=\{$	$(C_1\mapsto C_1)\mapsto \{f_1\mapsto f_1,g_1\mapsto g_1\},$
	$(C_1\mapsto C_2)\mapsto \{f_1\mapsto f_2,g_1\mapsto g_2\},$
	$(C_2\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_2\mapsto C_2)\mapsto \{f_2\mapsto f_2,g_2\mapsto g_2\}\}$

Calculating ${\beta }_{S_3}$

We will not show the complete construction of ${\beta }_{S_3}$ but restrict our self to the most interesting parts.

Calculate ${\beta }_{S_3}(C_1\mapsto C_3)$

According to the definition: ${\beta }_{S_3}(C_1\mapsto C_3)={{\beta }_{S_3}}^{\prime }(C_1\mapsto C_3)\ll {\theta }_{S_3}(C_1\mapsto C_3)$

Where: ${{\beta }_{S_3}}^{\prime }(C_1\mapsto C_3)=\text{UNION}b\cdot (b\in {\alpha }_3|{\beta }_{S_2}(C_1\mapsto b);(\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto b)\right)\right)\ll {\eta }_{3,b}))$

And: ${\theta }_{S_3}(C_1\mapsto C_3)=\text{UNION}b\cdot (b\in {\alpha }_3|{\beta }_{S_2}(C_1\mapsto b);(\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto b)\right)\right)\ll {\eta }_{3,b})▹{\sigma }_{3,b})$

Note that we can ommit ${\eta }_{3,b}$ from the formulas since it is always empty.

1. ${\beta }_{S_2}(C_1\mapsto C_2);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_2)\right)\right)\cup {\beta }_{S_2}(C_1\mapsto C_1);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_1)\right)\right)$
   1. ${\beta }_{S_2}(C_1\mapsto C_2)=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
   2. $\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_2)\right)\right)=\{f_2\mapsto f_2,g_2\mapsto g_2\}$
   3. ${\beta }_{S_2}(C_1\mapsto C_2);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_2)\right)\right)=\{f_1\mapsto f_2,g_1\mapsto g_2\}$
   4. ${\beta }_{S_2}(C_1\mapsto C_1)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$
   5. $\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_1)\right)\right)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$
   6. ${\beta }_{S_2}(C_1\mapsto C_1);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_1)\right)\right)=\{f_1\mapsto f_1,g_1\mapsto g_1\}$
   7. $\{f_1\mapsto f_2,g_1\mapsto g_2\}\cup \{f_1\mapsto f_1,g_1\mapsto g_1\}=\{f_1\mapsto f_1,f_1\mapsto f_2,g_1\mapsto g_1,g_1\mapsto g_2\}$
2. ${\beta }_{S_2}(C_1\mapsto C_2);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_2)\right)\right)▹{\sigma }_{3,C_2}\cup {\beta }_{S_2}(C_1\mapsto C_1);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_1)\right)\right)▹{\sigma }_{3,C_1}$
   1. ${\sigma }_{3,C_2}=\sigma \left(C_3\right)\left(C_2\right)=\left\{f_2\right\}$
   2. ${\sigma }_{3,C_1}=\sigma \left(C_3\right)\left(C_1\right)=\left\{g_1\right\}$
   3. ${\beta }_{S_2}(C_1\mapsto C_2);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_2)\right)\right)▹{\sigma }_{3,C_2}=\{f_1\mapsto f_2\}$
   4. ${\beta }_{S_2}(C_1\mapsto C_1);\text{id}\left(\text{ran}\left({\beta }_{S_2}(C_1\mapsto C_1)\right)\right)▹{\sigma }_{3,C_1}=\{g_1\mapsto g_1\}$
   5. $\{f_1\mapsto f_2\}\cup \{g_1\mapsto g_1\}=\{f_1\mapsto f_2,g_1\mapsto g_1\}$
   6. $\{f_1\mapsto f_1,f_1\mapsto f_2,g_1\mapsto g_1,g_1\mapsto g_2\}\ll \{f_1\mapsto f_2,g_1\mapsto g_1\}=\{f_1\mapsto f_2,g_1\mapsto g_1\}$

So we get:

${\beta }_{S_3}(C_1\mapsto C_3)=\{f_1\mapsto f_2,g_1\mapsto g_1\}$

The complete naming function

${\beta }_{S_3}=\{$	$(C_1\mapsto C_1)\mapsto \{f_1\mapsto f_1,g_1\mapsto g_1\},$
	$(C_1\mapsto C_2)\mapsto \{f_1\mapsto f_2,g_1\mapsto g_2\},$
	$(C_2\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_2\mapsto C_2)\mapsto \{f_2\mapsto f_2,g_2\mapsto g_2\},$
	$(C_1\mapsto C_3)\mapsto \{f_1\mapsto f_2,g_1\mapsto g_1\},$
	$(C_2\mapsto C_3)\mapsto \{f_2\mapsto f_2,g_2\mapsto g_2\},$
	$(C_3\mapsto C_3)\mapsto \{f_1\mapsto f_1,g_1\mapsto g_1,f_2\mapsto f_2,g_2\mapsto g_2\},$
	$(C_3\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_3\mapsto C_2)\mapsto \mathrm{\varnothing }\}$

The dynamic binding function

To calculate the dynamic binding function, we need to get ${\delta }_{S_3}$. We will again calculate it incrementaly:

Calculating ${\delta }_{S_1}$

According to the definition: ${\delta }_{S_1}(C_1\mapsto C_1)=(\text{UNION}b\cdot (b\in {\alpha }_1|{({\delta }_{S_0}{(b\mapsto b)}^{-1};{\beta }_S(b\mapsto C_1))}^{-1}))\ll {\tau }_1$

Since ${\alpha }_1$ is empty this simplifies to: ${\delta }_{S_1}(C_1\mapsto C_1)={\tau }_1=\tau \left(C_1\right)=\{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\}$

So we get: ${\delta }_{S_1}=\{(C_1\mapsto C_1)\mapsto \{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\}\}$

Calculating ${\delta }_{S_2}$

Again we calculate the different parts of the function in separate sections:

Calculate ${\delta }_{S_2}(C_1\mapsto C_1)$

According to the definition: ${\delta }_{S_2}(C_1\mapsto C_1)={\delta }_{S_1}(C_1\mapsto C_1)=\{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\}$

Calculate ${\delta }_{S_2}(C_2\mapsto C_1)$

According to the definition: ${\delta }_{S_2}(C_1\mapsto C_1)=\mathrm{\varnothing }$

Calculate ${\delta }_{S_2}(C_2\mapsto C_2)$

According to the definition: ${\delta }_{S_2}(C_2\mapsto C_2)=(\text{UNION}b\cdot (b\in {\alpha }_2|{({\delta }_{S_1}{(b\mapsto b)}^{-1};{\beta }_S(b\mapsto C_2))}^{-1}))\ll {\tau }_2$

1. $(\text{UNION}b\cdot (b\in \left\{C_1\right\}|{({\delta }_{S_1}{(b\mapsto b)}^{-1};{\beta }_S(b\mapsto C_2))}^{-1}))={({\delta }_{S_1}{(C_1\mapsto C_1)}^{-1};{\beta }_S(C_1\mapsto C_2))}^{-1}$
   1. ${\delta }_{S_1}{(C_1\mapsto C_1)}^{-1}=\{\underset{\_}{C_1.f_1}\mapsto f_1,\underset{\_}{C_1.g_1}\mapsto g_1\}$
   2. ${\delta }_{S_1}{(C_1\mapsto C_1)}^{-1};{\beta }_S(C_1\mapsto C_2)=\{\underset{\_}{C_1.f_1}\mapsto f_2,\underset{\_}{C_1.g_1}\mapsto g_2\}$
   3. ${({\delta }_{S_1}{(C_1\mapsto C_1)}^{-1};{\beta }_S(C_1\mapsto C_2))}^{-1}=\{f_2\mapsto \underset{\_}{C_1.f_1},g_2\mapsto \underset{\_}{C_1.g_1}\}$
2. ${\tau }_2=\tau \left(C_2\right)=\{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}$
3. $\{f_2\mapsto \underset{\_}{C_1.f_1},g_2\mapsto \underset{\_}{C_1.g_1}\}\ll \{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}=\{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}$

So we get: ${\delta }_{S_2}(C_2\mapsto C_2)=\{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}$

Calculate ${\delta }_{S_2}(C_1\mapsto C_2)$

According to the definition:

${\delta }_{S_2}(C_1\mapsto C_2)=\text{UNION}b\cdot (b\in {\alpha }_m|{\beta }_S(C_1\mapsto C_2);({({\delta }_{S_1}{(C_1\mapsto b)}^{-1};{\beta }_S(C_1\mapsto C_2))}^{-1}\ll {\tau }_2))$

At this stage it should be clear how to calculate this, so we just give the result:

${\delta }_{S_2}(C_1\mapsto C_2)=\{f_1\mapsto \underset{\_}{C_2.f_2},g_1\mapsto \underset{\_}{C_2.g_2}\}$

Putting the parts together:

We can know show the complete total function:

${\delta }_{S_2}=\{$	$(C_1\mapsto C_1)\mapsto \{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\},$
	$(C_1\mapsto C_2)\mapsto \{f_1\mapsto \underset{\_}{C_2.f_2},g_1\mapsto \underset{\_}{C_2.g_2}\},$
	$(C_2\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_2\mapsto C_2)\mapsto \{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\}\}$

Calculating ${\delta }_{S_3}$

There is nothing new for this calculation so we just present the result:

${\delta }_{S_3}=\{$	$(C_1\mapsto C_1)\mapsto \{f_1\mapsto \underset{\_}{C_1.f_1},g_1\mapsto \underset{\_}{C_1.g_1}\},$
	$(C_1\mapsto C_2)\mapsto \{f_1\mapsto \underset{\_}{C_2.f_2},g_1\mapsto \underset{\_}{C_2.g_2}\},$
	$(C_2\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_2\mapsto C_2)\mapsto \{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\},$
	$(C_1\mapsto C_3)\mapsto \{f_1\mapsto \underset{\_}{C_2.f_2},g_1\mapsto \underset{\_}{C_3.g_1}\},$
	$(C_2\mapsto C_3)\mapsto \{f_2\mapsto \underset{\_}{C_2.f_2},g_2\mapsto \underset{\_}{C_2.g_2}\},$
	Failed to parse (syntax error): {\displaystyle \left( C_3 \mapsto C_3 \right) \mapsto \left\{ f_1 \mapsto f_1, g_1 \mapsto \left\{ f_1 \mapsto \underline{C_3.f_1}, g_1 \mapsto \underline{C_3.g_1}, f_2 \mapsto \underline{C_2.f_2}, g_2 \mapsto \underline{C_2.g_2} \right\}, }
	$(C_3\mapsto C_1)\mapsto \mathrm{\varnothing },$
	$(C_3\mapsto C_2)\mapsto \mathrm{\varnothing }\}$