On Thu, Mar 31, 2016 at 03:18:17PM +0100, KatolaZ wrote: > On Thu, Mar 31, 2016 at 03:12:32PM +0200, Edward Bartolo wrote: > > Hi, thanks for taking time to reply, > > > > KatolaZ wrote: > > >> c) type **ss; // declare a pointer to pointer. System only allocates > > >> space for one address > > > > > C pointers are *always* one variable, precisely a variable large > > > enough to store a memory address on the current architecture. > > > > I think, I did not understand what you want to say. As far as I can > > imagine a pointer to pointer is something like this: > > [pointer1]------>[pointer2]------->[data] int the case of data** dt. > > > > OR > > > > [pointer1]------>[pointer2] in the case of void** ptr. > >
Your example would be correct *only if* and pointer1, pointer2 actually contained proper addresses of memory instead of containing uninitialized junk or NULL. And that, I think, is what KatolaZ is tryng to point out. > > Nope. This is totally wrong. What you have in memory after a > declaration: > > type *a; > > is just *one variable*, namely a variable able to contain a memory > address. Yes. It is only when you actually assgn a pointer to a that your pointer----->data diagram becomes valied. -- hendrik > When you declare: > > type **b; > > you still have exactly *one variable* allocated in memory, namely one > variable able to contain a memory address, and not *two variables* as > in your example above. This does not depend on what "type" is, so > each of the following declarations: > > int *a; > double ***b; > void ****c; > myowntype **d; > > will result in the allocation of exactly *one variable*, namely a > variable large enough to contain a pointer (i.e., a memory address). I > don't want to confuse you, but in my example myowntype might also be > declared as: > > typedef char*** myowntype; > > and still the declaration: > > myowntype **d; > > will reserve exactly one variable in memory, namely a variable large > enough to contain a memory address. > > The declaration is used only by the compiler to understand what is the > implicit semantic of the pointer arithmetic to be used with that > pointer, and to check that you are not palying nastly with it, so that > > int *p; > ... > p += 1; > > assigns to p the address of the memory location which is 1*sizeof(int) > bytes after the old address stored into p, while: > > double *p; > ... > p +=2; > > will assign to p the address of the memory location which is > 2*sizeof(double) bytes after the old address stored in p. > > You have not broken the pointers spell, yet. I warmly suggst you to > read a good introduction to C pointers, but the only thing that comes > to my mind is the Kernighan and Ritchie, which I admit is not the > easiest book around (although it is certainly the best on the > subject). > > My2Cents > > KatolaZ > > -- > [ Enzo Nicosia aka KatolaZ --- GLUG Catania -- Freaknet Medialab ] > [ me [at] katolaz.homeunix.net -- http://katolaz.homeunix.net -- ] > [ GNU/Linux User:#325780/ICQ UIN: #258332181/GPG key ID 0B5F062F ] > [ Fingerprint: 8E59 D6AA 445E FDB4 A153 3D5A 5F20 B3AE 0B5F 062F ] > _______________________________________________ > Dng mailing list > Dng@lists.dyne.org > https://mailinglists.dyne.org/cgi-bin/mailman/listinfo/dng _______________________________________________ Dng mailing list Dng@lists.dyne.org https://mailinglists.dyne.org/cgi-bin/mailman/listinfo/dng
