From 8715805071829158251 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 1149ec,c0f8f764e8b7d30f X-Google-Attributes: gid1149ec,public X-Google-Thread: f78e5,c0f8f764e8b7d30f X-Google-Attributes: gidf78e5,public X-Google-ArrivalTime: 1993-01-01 08:35:18 PST Xref: sparky comp.std.c++:1946 comp.std.c:3311 Path: sparky!uunet!think.com!sdd.hp.com!cs.utexas.edu!rutgers!twwells!bill From: bill@twwells.com (T. William Wells) Newsgroups: comp.std.c++,comp.std.c Subject: Re: Question about sizeof in C/C++. Message-ID: Date: 1 Jan 93 13:54:52 GMT References: <1992Dec24.001540.30000@microsoft.com> Organization: None, Mt. Laurel, NJ Lines: 30 In article jamesc@swapsdev.state.state.COM.AU (James Cribb) writes: : jimad@microsoft.com (Jim Adcock) writes: : >jamesc@swapsdev.state.state.COM.AU (James Cribb) writes: : >| Given : >| struct Something* ps; : >| unsigned k; : >| do the C or C++ standards guarantee that : >| &ps[k] == ((char*) ps) + k * sizeof(struct Something) : > : >No. The relationship holds only if ... k is less than or equal : >to the number of elements in the array pointed-to by ps : : My code meets those conditions, but just out of interest, : if k is too big, what could cause an inequality (of the addresses)? In the compiler I've written, if an object is known to be less than 64K bytes, its index arithmetic is done with 16 bits but all pointer arithmetic is done with 32 bits. So, if sizeof(*ps) * k is over 64K, the pointer arithmetic will create a different result from the index arithmetic. (Well OK, for this particular example, the compiler is actually smart enough to see the equivalence between the index expression and the pointer addition and so the equality would hold. But if the (char*)ps weren't detectable as an object pointer, they wouldn't be equal.) --- Bill { rutgers | decwrl | telesci }!twwells!bill bill@twwells.com