对于此特定c代码。
#include <stdio.h>
int main()
{
int *p = (int *)20;
int *q = (int *)30;
printf("%d", q-p);
}
这是gdb disass /m main
Dump of assembler code for function main:
0x0000000100000f45 <+0>: push %rbp // save the current base pointer
0x0000000100000f46 <+1>: mov %rsp,%rbp // base pointer reference the highest stack entry
0x0000000100000f49 <+4>: sub $0x10,%rsp // Move stack pointer 0x10/4 = (4) memory entries downwards
0x0000000100000f4d <+8>: movq $0x14,-0x8(%rbp) // save (0x14 = 20), 0x8/4 mem address lower than base pointer ref ( 'q' for quad word )
0x0000000100000f55 <+16>: movq $0x1e,-0x10(%rbp) // save (0x14 = 30), 0x10/4 mem address lower than base pointer ref
0x0000000100000f5d <+24>: mov -0x10(%rbp),%rdx // Save 0x10/4 memory addresses lower than the base pointer is referencing to rdx
0x0000000100000f61 <+28>: mov -0x8(%rbp),%rax // Save 0x8/4 memory addresses lower than the base pointer is referencing to rax
0x0000000100000f65 <+32>: sub %rax,%rdx // subtract rdx from rax and result in rax
0x0000000100000f68 <+35>: mov %rdx,%rax // move value of rax to rdx
0x0000000100000f6b <+38>: sar $0x2,%rax // shift right 2 ( divide the value of rax by 4 )
0x0000000100000f6f <+42>: mov %rax,%rsi // ( move value of rsi to rax )
0x0000000100000f72 <+45>: lea 0x31(%rip),%rdi # 0x100000faa
0x0000000100000f79 <+52>: mov $0x0,%eax
0x0000000100000f7e <+57>: callq 0x100000f8a
0x0000000100000f83 <+62>: mov $0x0,%eax
0x0000000100000f88 <+67>: leaveq
0x0000000100000f89 <+68>: retq
End of assembler dump.
我尝试过，从搜索和理解，我已经评论了每一步。但在此之后我不知道这个程序实际上是如何运行的。
请帮忙。
EDIT。在-O2之后我得到了这个。
Dump of assembler code for function main:
0x0000000100000f70 <+0>: sub $0x8,%rsp
0x0000000100000f74 <+4>: mov $0x2,%esi
0x0000000100000f79 <+9>: xor %eax,%eax
0x0000000100000f7b <+11>: lea 0x2c(%rip),%rdi # 0x100000fae
0x0000000100000f82 <+18>: callq 0x100000f8e
0x0000000100000f87 <+23>: xor %eax,%eax
0x0000000100000f89 <+25>: add $0x8,%rsp
0x0000000100000f8d <+29>: retq
End of assembler dump.