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$ o, U- t& N5 P" N$ B* T5 V/ m可以求得成员函数,成员变量的地址 - p# U; e5 J0 m- d" F% X' z" j
也可以得到普通变量和函数的地址或者值
$ K9 Z' v. M* ~* W6 T# h% o- `4 F换句话,就是又一个版本的AnythingToLong
& Q; M9 X! _, k3 a优化过后,毫无性能损失
" O4 z* e" e& [& g8 {9 z+ v0 |$ ?7 h所以我觉得是一个较好的解决办法 ' O. C2 I0 R5 Z* _/ V' N
0 ]/ D, W4 ^* n# @9 b// test.cpp : Defines the entry point for the console application.
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#include "stdafx.h"
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template<class T> % }7 U$ M& [, d' }) }+ y* o
inline unsigned long ValueOf(const T &value)
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return *(unsigned long*)(&value); l2 h! x7 n. Z( @* W+ i
}
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P @* m, |. S. [" B* ]template<class T> # v- Y; q H* ^8 B
inline int SizeOfArray(const T &array)
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return sizeof(array)/sizeof(array[0]); ) \2 k7 Q% E: ^
}
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. `4 u2 O- X+ |2 {; w9 lclass Test " a, _( Z; h, `/ q0 ^8 {* J
{
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int a,b; 4 [# ~/ _# B/ u& F& r
void f(){} 3 J( X4 f4 R' J0 N: }0 ^2 I
};
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+ V9 d u7 r- u tint main(int argc, char* argv[]) : w. l& W( w0 k3 N% G9 f
{
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3 K% [! f9 n, F8 T // cout<<SizeOfArray(t)<<endl; # T9 V# {4 l- c/ L0 a
cout<<ValueOf(&Test::a)<<endl;
- I2 n8 J) u7 o9 b- S+ | cout<<ValueOf(&Test::b)<<endl; 8 p2 N6 v8 p$ o. m2 @5 ~
cout<<ValueOf(&Test::f)<<endl; * x/ D" R& I0 v
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cout<<ValueOf(&t)<<endl; - o- n6 ]/ b; M1 K
cout<<(long)&t<<endl;
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cout<<ValueOf(n)<<endl; ! Q% _+ t" `, U# p+ T! s
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return 0; " J; H( E* N& E' X7 v1 \4 ]
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//运行结果
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7 y$ e) a8 s8 F$ Y4198425
9 _! E' S* h/ T# J8 v* Q6684088 6 C! ^1 r& q+ \& r e
6684088 " k9 T# q2 g3 a8 Z- v; Y0 G$ ^
1000
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