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Re: [RFH] Folding of &a[-1]


Sorry, yes:

   &x[-1] == (int *)x + (int *)4*(int *)-1

would be correct/true, and probably
simplest; although equivalent to either:

   &x[-1] == (int *)x + (int *)((int)4*(int)-1)
or:
   &x[-1] == (int *)x + (int *)((size_t)4*(int)-1)
or:
   &x[-1] == (int *)x + (int *)((size_t)4*(size_t)-1)




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