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On 12/05/2013 23:37, David Relson wrote: |
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>> [1] The logic goes something like this: it's a compiler, so the code |
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>> > it produces must be consistently identical for identical inputs. So, |
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>> > the current compiler builds gcc, giving version Y built by version X. |
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>> > That instance of gcc in turn builds a gcc, giving version Y built by |
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>> > version Y. |
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> Haven't you left out the third compile? |
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> |
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> Let me rephrase the 3 builds. |
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> |
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> 1) gcc-X builds gcc-Y giving gcc-Y1 |
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> 2) gcc-Y1 builds gcc-Y giving gcc-Y2 |
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> 3) gcc-Y2 builds gcc-Y giving gcc-Y3 |
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> |
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> gcc-Y1 and gcc-Y2 are likely to be different (since they were build by |
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> gcc-X and gcc-Y which are likely to have optimizations). |
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> |
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> gcc-Y2 and gcc-Y3 should be identical (since both were built by gcc-Y) |
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> |
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>> > |
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Yeah, I think you're right. Thanks for that. |
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Like I said, I'm fuzzy on the details after all these years. My |
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intention was not to be completely accurate and correct, it was more to |
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get the general idea across to Dale |
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-- |
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Alan McKinnon |
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alan.mckinnon@×××××.com |