以二进制格式输出对象基本思路是直接将对象的内存数值以二进制的格式输出,但是如何获取对象内存的二进制结构 Xo*DvD
是关键。解决方法是通过位操作找出对象内存的每一个bit的值。并把该方法封装成一个迭代器。 4BwQA#zE
要注意的问题是位操作符只能对整数类型进行操作,为了对所有类型的对象都有效,必须将对 t5lO'Ll*Q]
象当作char数组来处理。 $Q*<96M
template<class Container> />j';6vi
class bit_iterator : public std::iterator<bidirectional_iterator_tag, eW>3XD4
void, void> XerbUkZ
{ 95<EN(oUD
public: %2V-~.Ro6
explicit bit_iterator(Container& c) : m_container(&c) Rml2"9"`
{
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m_size = 8*sizeof(Container); Q#IG;
m_index = 0; nQGQWg`
} F V,4pi
explicit bit_iterator() : m_container(0) ,y%3mR_~
{ _Ob@`
m_size = 8*sizeof(Container); `|Or{ih
m_index = m_size; !!o8N<NU
} 1 n%?l[o
bool operator* () b]a@
{ "U\JV)N
char mask = 1; p^iRPI
char* pc = (char*)m_container; RQFI'@Ks
int i = (m_size-m_index-1)/8; +<prgP`v
int off = (m_size-m_index-1)%8; ;us%/kOR
mask <<=off; eX_D/25 $
return pc & mask; jV8q)=}*)
} hkOsm6
bit_iterator<Container>& operator++() jP~Z`yf
{ rS1fK1dys
m_index++; *Y@nVi
return *this; RyRpl*^
} b$eXFi/
bit_iterator<Container>& operator++(int) t^ZV|s 1
{ }y%oT
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m_index++; [{r}u
return *this; &gI ~LP
} Ssk}e=]
bool operator==(bit_iterator<Container>& bitIt) V
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{ 7$rjlVe
return m_index == bitIt.m_index; |X`/
} +78CvjG
bool operator!=(bit_iterator<Container>& bitIt) !pJeA)W;
{ *9p |HX=
return !(*this == bitIt); VAC iVKk
} +1~Z#^{&
protected: 2!Bd2
Container* m_container; n$[f94d=
private: DD44"w_9
int m_size; s[gKc '
int m_index; XW?b\!@ $
}; (Y^X0yA/
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用该迭代器可以将任意类型对象以二进制格式输出: RaTH\>n
double a = 10; z]3 `*/B
copy(bit_iterator<double>(a), bit_iterator<double> (), ostream_iterator<bool>(cout, ""));