有这样的东西吗?这是我第一次遇到对它的实际需求，但我在Stroustrup中没有看到它的列表。我打算写:
// Detect when exactly one of A,B is equal to five.
return (A==5) ^^ (B==5);
但是没有^^操作符。我可以使用这里的位^并得到正确的答案(不管机器表示的是真还是假)吗?我从不把&和&&混在一起，或者|和||混在一起，所以我不太愿意把^和^^混在一起。
我更愿意写我自己的bool XOR(bool,bool)函数。
当前回答
#if defined(__OBJC__)
#define __bool BOOL
#include <stdbool.h>
#define __bool bool
#endif
static inline __bool xor(__bool a, __bool b)
{
return (!a && b) || (a && !b);
}
它按定义工作。条件是用来检测你是否在使用Objective-C，它要求BOOL而不是BOOL(长度不同!)
其他回答
#if defined(__OBJC__)
#define __bool BOOL
#include <stdbool.h>
#define __bool bool
#endif
static inline __bool xor(__bool a, __bool b)
{
return (!a && b) || (a && !b);
}
它按定义工作。条件是用来检测你是否在使用Objective-C，它要求BOOL而不是BOOL(长度不同!)
异或运算符不能短路;也就是说，你不能仅仅通过计算一个XOR表达式的左操作数来预测它的结果。因此，没有理由提供^^版本。
有一些好的代码比!a != !b更好地解决了这个问题
注意，我必须添加BOOL_DETAIL_OPEN/CLOSE，这样它才能在MSVC 2010上工作
/* From: http://groups.google.com/group/comp.std.c++/msg/2ff60fa87e8b6aeb
Proposed code left-to-right? sequence point? bool args? bool result? ICE result? Singular 'b'?
-------------- -------------- --------------- ---------- ------------ ----------- -------------
a ^ b no no no no yes yes
a != b no no no no yes yes
(!a)!=(!b) no no no no yes yes
my_xor_func(a,b) no no yes yes no yes
a ? !b : b yes yes no no yes no
a ? !b : !!b yes yes no no yes no
[* see below] yes yes yes yes yes no
(( a bool_xor b )) yes yes yes yes yes yes
[* = a ? !static_cast<bool>(b) : static_cast<bool>(b)]
But what is this funny "(( a bool_xor b ))"? Well, you can create some
macros that allow you such a strange syntax. Note that the
double-brackets are part of the syntax and cannot be removed! The set of
three macros (plus two internal helper macros) also provides bool_and
and bool_or. That given, what is it good for? We have && and || already,
why do we need such a stupid syntax? Well, && and || can't guarantee
that the arguments are converted to bool and that you get a bool result.
Think "operator overloads". Here's how the macros look like:
Note: BOOL_DETAIL_OPEN/CLOSE added to make it work on MSVC 2010
*/
#define BOOL_DETAIL_AND_HELPER(x) static_cast<bool>(x):false
#define BOOL_DETAIL_XOR_HELPER(x) !static_cast<bool>(x):static_cast<bool>(x)
#define BOOL_DETAIL_OPEN (
#define BOOL_DETAIL_CLOSE )
#define bool_and BOOL_DETAIL_CLOSE ? BOOL_DETAIL_AND_HELPER BOOL_DETAIL_OPEN
#define bool_or BOOL_DETAIL_CLOSE ? true:static_cast<bool> BOOL_DETAIL_OPEN
#define bool_xor BOOL_DETAIL_CLOSE ? BOOL_DETAIL_XOR_HELPER BOOL_DETAIL_OPEN
异或还有另一种方式:
bool XOR(bool a, bool b)
{
return (a + b) % 2;
}
这显然可以通过以下方式来证明:
#include <iostream>
bool XOR(bool a, bool b)
{
return (a + b) % 2;
}
int main()
{
using namespace std;
cout << "XOR(true, true):\t" << XOR(true, true) << endl
<< "XOR(true, false):\t" << XOR(true, false) << endl
<< "XOR(false, true):\t" << XOR(false, true) << endl
<< "XOR(false, false):\t" << XOR(false, false) << endl
<< "XOR(0, 0):\t\t" << XOR(0, 0) << endl
<< "XOR(1, 0):\t\t" << XOR(1, 0) << endl
<< "XOR(5, 0):\t\t" << XOR(5, 0) << endl
<< "XOR(20, 0):\t\t" << XOR(20, 0) << endl
<< "XOR(6, 6):\t\t" << XOR(5, 5) << endl
<< "XOR(5, 6):\t\t" << XOR(5, 6) << endl
<< "XOR(1, 1):\t\t" << XOR(1, 1) << endl;
return 0;
}
使用简单的:
return ((op1 ? 1 : 0) ^ (op2 ? 1 : 0));