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