Hi! I really want to know how does gcc compile code like *(ptr base +offset), where ptr base is the initial address of a pointer variableand 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 <malloc@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 thenadd the result of "i + j" to the base address of buf to obtain thefinal address. Do I guess right? Is there any exception?ps: My gcc version is 4.3.3. Thank you!
