这篇准备通过《深度探索C++对象模型》的第四章节Function语意学,简单描述下多重继承下的函数调用,因为最近是在再次阅读此书,顺便记下一些实现并反汇编出具体的实现。
之前面试过一些候选人,对于虚函数的理解,只是停留在一个虚表,但只要一深入问一点就不清楚了。可能背题只是应付面试或者笔试,但如果想要走更深入一些,还是要理解下原理,甚至自己gdb看一下。当然,这些只是个人选择和兴趣。
之前由于阅读STL源码剖析不是特别仔细,有些接口的全名和使用差不多都忘记了。这里也贴上之前分析过的:
类实例的构造
为什么会继续基类虚函数的默认形参?
基类指针指向派生类数组的一些问题
C++对象模型及性能优化杂谈一
C++对象模型及性能优化杂谈二
这里不分析inline/static function的情况,只是单纯分析通过对象和引用/指针调用(非)虚成员函数的表现,及多重继承下的情况。
情况一:
1 #include <iostream>
2 using std::cout;
3 using std::endl;
4
5 class base {
6 public:
7 void show() { cout << "base" << endl;}
8 private:
9 int mvalue;
10 };
11
12 int main() {
13 base obj;
14 obj.show();
15 base* pbase = &obj;
16 pbase->show();
17 pbase = nullptr;
以下经过返回汇编后的结果,由于mac系统反汇编出的结果有点不一样,这里直接gdb进汇编代码看出:
0x0000000100000e30 <+0>: push %rbp
0x0000000100000e31 <+1>: mov %rsp,%rbp
0x0000000100000e34 <+4>: sub $0x10,%rsp //rsp=rsp-16
0x0000000100000e38 <+8>: lea -0xc(%rbp),%rax
=> 0x0000000100000e3c <+12>: mov %rax,%rdi //obj的地址
0x0000000100000e3f <+15>: callq 0x100000ebe
0x0000000100000e44 <+20>: lea -0xc(%rbp),%rax
0x0000000100000e48 <+24>: mov %rax,-0x8(%rbp)
0x0000000100000e4c <+28>: mov -0x8(%rbp),%rax
0x0000000100000e50 <+32>: mov %rax,%rdi //obj的地址
0x0000000100000e53 <+35>: callq 0x100000ebe
(gdb) si
base::show (this=0x0) at struct.cpp:7
从上面反汇编出来的结果看出,没有合成构造函数,所以不会调用,这里并不需要编译器合成一个;再之,成员函数show没有用到类实例中的成员变量,所以这里的this
为null
;这里都到某一跳转callq 0x100000ebe
;成员函数show 被编译器name mangling成__ZN4base4showEv
;所以通过对象或指针调用非表态成员函数(no virtual),最终都会被转化成类似:__ZN4base4showEv(&obj)
,传进对象地址,所以这里的性能其实都是一样的。
带虚函数的单重继承:
1 #include <iostream>
2 using std::cout;
3 using std::endl;
4
5 class base {
6 public:
7 virtual void show() { cout << "base" << endl;}
8 private:
9 int mvalue;
10 };
11
12 class derived : public base {
13 public:
14 void show() {
15 cout << "derived" << endl;
16 }
17 };
18
19 int main() {
20 base obj;
21 obj.show();
22 base* pbase = new derived;//new (std::nothrow) derived; check pbase
23 pbase->show();
24 delete pbase;
25 pbase = nullptr;
这里作为基类的base,并没有声明一个virtual ~base()
,这里测试就省略掉,实际使用中还是要加上。
在起始处:
0x0000000100000cef <+0>: push %rbp
0x0000000100000cf0 <+1>: mov %rsp,%rbp
0x0000000100000cf3 <+4>: push %rbx
0x0000000100000cf4 <+5>: sub $0x28,%rsp
=> 0x0000000100000cf8 <+9>: lea -0x30(%rbp),%rax
0x0000000100000cfc <+13>: mov %rax,%rdi
0x0000000100000cff <+16>: callq 0x100000dac
因为base有虚函数,所以会对base合成一个默认构造函数,此时并不初始化mvalue值(内建类型的整型一般是0),只是设置base实例的vptr:$2 = {_vptr.base = 0x0, mvalue = 0}
Dump of assembler code for function base::base():
0x0000000100000ca4 <+0>: push %rbp
0x0000000100000ca5 <+1>: mov %rsp,%rbp
0x0000000100000ca8 <+4>: mov %rdi,-0x8(%rbp)
0x0000000100000cac <+8>: mov 0x365(%rip),%rax # 0x100001018
0x0000000100000cb3 <+15>: lea 0x10(%rax),%rax
0x0000000100000cb7 <+19>: mov -0x8(%rbp),%rdx
=> 0x0000000100000cbb <+23>: mov %rax,(%rdx)
0x0000000100000cbe <+26>: nop
0x0000000100000cbf <+27>: pop %rbp
0x0000000100000cc0 <+28>: retq
End of assembler dump.
最后构造函数返回时:
(gdb) p *this
$21 = (base) {_vptr.base = 0x100001060 <vtable for base+16>, mvalue = 0}
(gdb) p /a *(void**)0x100001060@1
$22 = {0x100000c12 <base::show()>}
接着调用obj.show:
0x0000000100000d04 <+21>: lea -0x30(%rbp),%rax
0x0000000100000d08 <+25>: mov %rax,%rdi
0x0000000100000d0b <+28>: callq 0x100000da6
接着new derived
分配内存:
0x0000000100000d10 <+33>: mov $0x10,%edi
0x0000000100000d15 <+38>: callq 0x100000dd0
0x0000000100000d1a <+43>: mov %rax,%rbx
=> 0x0000000100000d1d <+46>: mov %rbx,%rdi
此时:
0x0000000100000d1d 22 base* pbase = new derived;
(gdb) p pbase
$26 = (base *) 0x0
这里new derived
分三步走,分配内存,进行构造,再设置pbase
,其他平台可能在后两步顺序对调下:
Dump of assembler code for function derived::derived():
0x0000000100000cc2 <+0>: push %rbp
0x0000000100000cc3 <+1>: mov %rsp,%rbp
0x0000000100000cc6 <+4>: sub $0x10,%rsp
0x0000000100000cca <+8>: mov %rdi,-0x8(%rbp)
0x0000000100000cce <+12>: mov -0x8(%rbp),%rax
0x0000000100000cd2 <+16>: mov %rax,%rdi
=> 0x0000000100000cd5 <+19>: callq 0x100000db2
0x0000000100000cda <+24>: mov 0x33f(%rip),%rax # 0x100001020
0x0000000100000ce1 <+31>: lea 0x10(%rax),%rax
0x0000000100000ce5 <+35>: mov -0x8(%rbp),%rdx
0x0000000100000ce9 <+39>: mov %rax,(%rdx)
0x0000000100000cec <+42>: nop
0x0000000100000ced <+43>: leaveq
0x0000000100000cee <+44>: retq
End of assembler dump.
Dump of assembler code for function base::base():
=> 0x0000000100000c86 <+0>: push %rbp
0x0000000100000c87 <+1>: mov %rsp,%rbp
0x0000000100000c8a <+4>: mov %rdi,-0x8(%rbp)
0x0000000100000c8e <+8>: mov 0x383(%rip),%rax # 0x100001018
0x0000000100000c95 <+15>: lea 0x10(%rax),%rax
0x0000000100000c99 <+19>: mov -0x8(%rbp),%rdx
0x0000000100000c9d <+23>: mov %rax,(%rdx)
0x0000000100000ca0 <+26>: nop
0x0000000100000ca1 <+27>: pop %rbp
0x0000000100000ca2 <+28>: retq
End of assembler dump.
(gdb) p *this
$3 = {_vptr.base = 0x100001060 <vtable for base+16>, mvalue = 0}
当执行完base的构造函数时,回到derived的构造函数那边,会重新设置派生类derived自己的虚表:
$5 = {<base> = {_vptr.base = 0x100001048 <vtable for derived+16>, mvalue = 0}, <No data fields>}
(gdb) p pbase //调用derived 构造函数前
$7 = (base *) 0x0
(gdb) i r rbx
rbx 0x100600440 4301259840
(gdb) p pbase//调用derived 构造函数后
$8 = (base *) 0x100600440
最后几行中:
0x0000000100000d25 <+54>: mov %rbx,-0x18(%rbp)
0x0000000100000d29 <+58>: mov -0x18(%rbp),%rax//pbase地址
0x0000000100000d2d <+62>: mov (%rax),%rax//虚表地址
0x0000000100000d30 <+65>: mov (%rax),%rdx//show虚函数地址
=> 0x0000000100000d33 <+68>: mov -0x18(%rbp),%rax
0x0000000100000d37 <+72>: mov %rax,%rdi//准备this参数
0x0000000100000d3a <+75>: callq *%rdx//调用show虚函数
(gdb) p /a *(void**)0x100001060@1
$15 = {0x100000c12 <base::show()>}
(gdb) p /a *(void**)0x100001048@1
$16 = {0x100000c4c <derived::show()>}
所以通过上面的调试,可以看出通过对象调用的虚函数,并不会引发虚机制,和调用普通函数一样__ZN4base4showEv(&obj)
;而通过指向派生类的基类指针,则会转化成类似:(* pbase->vptr[1])(pbase)
,所以从汇编代码上看,是多了一些指令。
如果在base的show下面再加个hello的虚函数,并且derived类重写那反汇编后面一段:
0x0000000100000cb9 <+62>: mov (%rax),%rax
=> 0x0000000100000cbc <+65>: add $0x8,%rax//hello的偏移量
0x0000000100000cc0 <+69>: mov (%rax),%rdx
0x0000000100000cc3 <+72>: mov -0x18(%rbp),%rax
0x0000000100000cc7 <+76>: mov %rax,%rdi //hello函数地址
0x0000000100000cca <+79>: callq *%rdx
$2 = {_vptr.base = 0x100001048 <vtable for derived+16>, mvalue = 0}
(gdb) p /a *(void**)0x100001048@1
$3 = {0x100000b9e <derived::show()>}
(gdb) p /a *(void**)0x100001048@2
$4 = {0x100000b9e <derived::show()>, 0x100000bd8 <derived::hello()>}
所以在获取到hello函数在虚表中的位置时,是需要调整位置add $0x8,%rax
,类似(* pbase->vptr[2])(pbase)
;因为base是没有声明并定义virtual ~base
所以没在虚表中,当然虚表中也包含其他执行期需要的信息,比如type_info for base
等。
多重继承下的virtual functions问题,因为这里涉及到指针的调整,所以当使用第一个base class指针时和第二及后面的base class指针时,是有些区别,后者需要对this指针作一定的偏移量,这里写上virtual析构可以查看虚表中的内容:
5 class base1 {
6 public:
7 virtual void show() {}
8 virtual ~base1() {}
9 private:
10 int mvalue;
11 };
12
13 class base2 {
14 public:
15 virtual void show() {}
16 virtual ~base2() {}
17 private:
18 int mvalue;
19 };
20
21 class derived : public base1, public base2 {
22 public:
23 void show() {}
24 virtual ~derived() {}
25 };
26
27 int main() {
28 base1* pbase1 = new derived;
29 pbase1->show();
30 delete pbase1;
31 pbase1 = nullptr;
32
33 base2* pbase2 = new derived;
34 pbase2->show();
35 delete pbase2;
36 pbase2 = nullptr;
37 return 0;
38 }
因为之前分析过相关的多重继承,这里只是重点关注下class derived
的虚表内容,和main中指针的调整:
(gdb) p *(class derived*)pbase1
$25 = {<base1> = {_vptr.base1 = 0x100001040 <vtable for derived+16>, mvalue = 0}, <base2> = {
_vptr.base2 = 0x100001068 <vtable for derived+56>, mvalue = 0}, <No data fields>}
(gdb) p /a *(void**)0x100001040@2
$26 = {0x1000008ee <derived::show()>, 0x100000902 <derived::~derived()>}
(gdb) p /a *(void**)0x100001068@2
$27 = {0x1000008f9 <_ZThn16_N7derived4showEv>, 0x100000952 <_ZThn16_N7derivedD1Ev>}
当执行完base1* pbase1 = new derived
后的derived对象实例的内容为上面的情况,但查看pbase1只能看到class base1
的那一部分:
(gdb) p pbase1
$28 = (base1 *) 0x100700020
(gdb) p *pbase1
$29 = {_vptr.base1 = 0x100001040 <vtable for derived+16>, mvalue = 0}
因为pbase1声明时的静态类型为class base1
,告诉编译器他的寻址范围是sizeof(class base1)
这么大;接着调用derived::show
。
=> 0x0000000100000a55 <+57>: test %rax,%rax
0x0000000100000a58 <+60>: je 0x100000a69 <main()+77>
0x0000000100000a5a <+62>: mov (%rax),%rdx
0x0000000100000a5d <+65>: add $0x10,%rdx
0x0000000100000a61 <+69>: mov (%rdx),%rdx
0x0000000100000a64 <+72>: mov %rax,%rdi
0x0000000100000a67 <+75>: callq *%rdx
0x0000000100000a69 <+77>: movq $0x0,-0x18(%rbp)
上面几行对应delete pbase1; pbase1 = nullptr;
,会先判断pbase1是否为空,否则取class derived
的析构函数,并准备this
参数,调用~derived()
:
Dump of assembler code for function derived::~derived():
=> 0x000000010000095c <+0>: push %rbp
0x000000010000095d <+1>: mov %rsp,%rbp
0x0000000100000960 <+4>: sub $0x10,%rsp
0x0000000100000964 <+8>: mov %rdi,-0x8(%rbp)
0x0000000100000968 <+12>: mov -0x8(%rbp),%rax
0x000000010000096c <+16>: mov %rax,%rdi
0x000000010000096f <+19>: callq 0x100000b6e
0x0000000100000974 <+24>: mov -0x8(%rbp),%rax
0x0000000100000978 <+28>: mov $0x20,%esi
0x000000010000097d <+33>: mov %rax,%rdi
0x0000000100000980 <+36>: callq 0x100000b7a
0x0000000100000985 <+41>: leaveq
0x0000000100000986 <+42>: retq
End of assembler dump.
这里发现一个现像,如果pbase1的内容为0x100700020,那么delete pbase1后不执行pbase1 = nullptr
,然后再执行base2* pbase2 = new derived时,new derived
的地址也是0x100700020,如果这里手误delete pbase1,那么会出现问题,比如破坏pbase2内存,double delete的情况。
0x0000000100000a81 <+101>: callq 0x100000b62
0x0000000100000a86 <+106>: test %rbx,%rbx
0x0000000100000a89 <+109>: je 0x100000a91 <main()+117>
0x0000000100000a8b <+111>: lea 0x10(%rbx),%rax
0x0000000100000a8f <+115>: jmp 0x100000a96 <main()+122>
0x0000000100000a91 <+117>: mov $0x0,%eax
=> 0x0000000100000a96 <+122>: mov %rax,-0x20(%rbp)
0x0000000100000a9a <+126>: mov -0x20(%rbp),%rax
0x0000000100000a9e <+130>: mov (%rax),%rax
0x0000000100000aa1 <+133>: mov (%rax),%rdx
0x0000000100000aa4 <+136>: mov -0x20(%rbp),%rax
0x0000000100000aa8 <+140>: mov %rax,%rdi
0x0000000100000aab <+143>: callq *%rdx
(gdb) p pbase2
$35 = (base2 *) 0x100700030
(gdb) p *pbase2
$36 = {_vptr.base2 = 0x100001068 <vtable for derived+56>, mvalue = 0}
上面是base2* pbase2 = new derived
的汇编,这里new一段内存后,会对返回的地址进行判断,不为0则会加上sizeof(class base1)
的偏移量作为pbase2内容;接着准备参数调用虚函数。
这句base2* pbase2 = new derived
,差不多会编译成这样的语句:
derived* temp = new derived;
base2* pbase2 = temp ? temp + sizeof(base1) : 0;
最后delete pbase2
时需要调整指针位置:
0x0000000100000aad <+145>: mov -0x20(%rbp),%rax
0x0000000100000ab1 <+149>: test %rax,%rax
0x0000000100000ab4 <+152>: je 0x100000ac5 <main()+169>
0x0000000100000ab6 <+154>: mov (%rax),%rdx
0x0000000100000ab9 <+157>: add $0x10,%rdx
0x0000000100000abd <+161>: mov (%rdx),%rdx
0x0000000100000ac0 <+164>: mov %rax,%rdi
0x0000000100000ac3 <+167>: callq *%rdx
以上的程序只是测试使用,实际工程项目中的代码要更严谨些,有良好的编码习惯,比如作为base class的要带个virtual destructor,delete后要置null,对于new的使用要加std::nothrow并判断空,并作后续处理等。
最后关于虚拟继承这个,其实算是比较复杂的,之前经历中也很少见到此类的使用场景,有兴趣自行参考书《深度探索C++对象模型》。