用汇编的眼光看C++（之拷贝、赋值函数） - c++编程基础

拷贝构造函数和复制函数是类里面比较重要的两个函数。两者有什么区别呢？其实也很简单，我们可以举个例子，加入有这样一个类的定义：

copy to clipboardprint class apple

{

public:

apple() { printf("apple()!\n");}

apple(apple& a) { printf("copy apple()!\n");}

apple& operator=(apple& a) { printf("= apple()\n"); return *this;}

~apple() { printf("~apple()!\n");}

void print() const { return;}

};

class apple

{

public:

apple() { printf("apple()!\n");}

apple(apple& a) { printf("copy apple()!\n");}

apple& operator=(apple& a) { printf("= apple()\n"); return *this;}

~apple() { printf("~apple()!\n");}

void print() const { return;}

};

那么我们在如下的函数里面进行调用的时候，调用的函数分别是哪些呢？

copy to clipboardprint void process()

{

apple a, c;

apple b =a;

c = b;

}

void process()

{

apple a, c;

apple b =a;

c = b;

} 其实汇编的结果是这样的，大家可以一起看一下，自己尝试读一下。如果一次不是很明白，可以多读几次。

copy to clipboardprint 70: apple a, c;

0040127D lea ecx,[ebp-10h]

00401280 call @ILT+70(apple::apple) (0040104b)

00401285 mov dword ptr [ebp-4],0

0040128C lea ecx,[ebp-14h]

0040128F call @ILT+70(apple::apple) (0040104b)

00401294 mov byte ptr [ebp-4],1

71: apple b =a;

00401298 lea eax,[ebp-10h]

0040129B push eax

0040129C lea ecx,[ebp-18h]

0040129F call @ILT+50(apple::apple) (00401037)

004012A4 mov byte ptr [ebp-4],2

72: c = b;

004012A8 lea ecx,[ebp-18h]

004012AB push ecx

004012AC lea ecx,[ebp-14h]

004012AF call @ILT+75(apple::operator=) (00401050)

73: }

004012B4 mov byte ptr [ebp-4],1

004012B8 lea ecx,[ebp-18h]

004012BB call @ILT+0(apple::~apple) (00401005)

004012C0 mov byte ptr [ebp-4],0

004012C4 lea ecx,[ebp-14h]

004012C7 call @ILT+0(apple::~apple) (00401005)

004012CC mov dword ptr [ebp-4],0FFFFFFFFh

004012D3 lea ecx,[ebp-10h]

004012D6 call @ILT+0(apple::~apple) (00401005)

004012DB mov ecx,dword ptr [ebp-0Ch]

004012DE mov dword ptr fs:[0],ecx

004012E5 pop edi

004012E6 pop esi

004012E7 pop ebx

004012E8 add esp,58h

004012EB cmp ebp,esp

004012ED call __chkesp (004087c0)

004012F2 mov esp,ebp

004012F4 pop ebp

004012F5 ret

70: apple a, c;