I am trying to recreate the original function func4(), in C program from the following Assembly code.
func4() takes 3 arguments.
...
40104e: ba 0e 00 00 00 mov $0xe,%edx
401053: be 00 00 00 00 mov $0x0,%esi
401058: 8b 3c 24 mov (%rsp),%edi
40105b: e8 85 ff ff ff call 400fe5 <func4>
...
0000000000400fe5 <func4>:
400fe5: 53 push %rbx
400fe6: 89 d0 mov %edx,%eax
400fe8: 29 f0 sub %esi,%eax
400fea: 89 c3 mov %eax,%ebx
400fec: c1 eb 1f shr $0x1f,%ebx
400fef: 01 d8 add %ebx,%eax
400ff1: d1 f8 sar %eax
400ff3: 8d 1c 30 lea (%rax,%rsi,1),%ebx
400ff6: 39 fb cmp %edi,%ebx
400ff8: 7e 0c jle 401006 <func4+0x21>
400ffa: 8d 53 ff lea -0x1(%rbx),%edx
400ffd: e8 e3 ff ff ff call 400fe5 <func4> // fist recursion call
401002: 01 d8 add %ebx,%eax
401004: eb 10 jmp 401016 <func4+0x31>
401006: 89 d8 mov %ebx,%eax
401008: 39 fb cmp %edi,%ebx
40100a: 7d 0a jge 401016 <func4+0x31>
40100c: 8d 73 01 lea 0x1(%rbx),%esi
40100f: e8 d1 ff ff ff call 400fe5 <func4> // second recursion call
401014: 01 d8 add %ebx,%eax
401016: 5b pop %rbx
401017: c3 ret
I am unable to understand 400ff1: d1 f8 sar %eax instruction. How it carries out the arithmetic right shift operation and from where it takes the shift amount ?
The shift amount should come from the lower %cl part of %rcx register, but the during that instruction execution the %rcx register contains value 0x0, but instead of that the value in the %rax register is shifted some amount.
Detailed technical explanation is appreciated.
