register allocation

Question

Comments

This is a snippet from the stanford slide

You can access the slides from this link : http://openclassroom.stanford.edu/MainFolder/DocumentPage.php?course=Compilers&doc=docs/slides.html

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Coming back to the question

The variable p is live at Basic block 1

1. as statement 2 uses p
2. There is a direct path from 1→ 2
3. No intervening assignment to p is there

Hence option D is eliminated

The variable p is not live at BB 2 as there is a assignment to p in BB4

The variable p is live at Basic block 4

1. as statement 5 uses p
2. There is a direct path from 4→ 5
3. No intervening assignment to p is there

The variable p is live at Basic block 5

1. as statement 2 uses p
2. There is a direct path from 5→2
3. No intervening assignment to p is there

Hence option (C) is the answer

You can also use USE(),IN(),OUT() sets, but i find this way more easy to analyse

 

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But ,at statement 1 there is a definition of P ,i.e there is a new definition of p from 1>2.So ,how can you say that p is live at statement 1?

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I hope your doubt is clear why p is live from 1→ 2