房陈 <fancyfl...@gmail.com> writes: > I really want to how does gcc compile code like *(ptr base + > offset), where ptr base is the initial address of a pointer variable > and offset is any legal integer expression. There is a example here: > > int i = 1; > int j = 1; > int *buf = (int*)malloc(10 *sizeof(int)); > *(buf + i + j) = 7; > > And the correspondent assembly code is : > ...... > int i = 1; > 80483b5: c7 45 f0 01 00 00 00 movl $0x1,-0x10(%ebp) > int j = 1; > 80483bc: c7 45 f4 01 00 00 00 movl $0x1,-0xc(%ebp) > int *buf = (int*)malloc(10 * sizeof(int)); > 80483c3: c7 04 24 28 00 00 00 movl $0x28,(%esp) > 80483ca: e8 09 ff ff ff call 80482d8 <mal...@plt> > 80483cf: 89 45 f8 mov %eax,-0x8(%ebp) > > *(buf + i + j) = 7; > 80483d2: 8b 55 f0 mov -0x10(%ebp),%edx > 80483d5: 8b 45 f4 mov -0xc(%ebp),%eax > 80483d8: 8d 04 02 lea (%edx,%eax,1),%eax > 80483db: c1 e0 02 shl $0x2,%eax > 80483de: 03 45 f8 add -0x8(%ebp),%eax > 80483e1: c7 00 07 00 00 00 movl $0x7,(%eax) > ...... > So I guess that gcc would always compute offset "i+j" first, and then > add the result of "i + j" to the base address of buf to obtain the > final address. Do I guess right? Is there any exception? > ps: My gcc version is 4.3.3.
Unless you plan to modify gcc itself, this question would be more appropriate for the gcc-help mailing list. Please take any followups there. If you compile at -O0, gcc will probably generate code more or less as you describe. However, there is no guarantee of that. If you compile with optimization, then the instructions can and will be completely changed. In particular, for your example, gcc will most likely forward propagate i + j, fold the constant, and simply use buf + 2. One way to see what gcc does is to use -fdump-tree-all and examine the generated dump files. Ian
