用模板方法来判断数据类型的符号

 comp.lang.c++.moderated 上面有一篇帖子，说的是下面一段程序

</p> <p>#include <iostream></p> <p>#define OK</p> <p>template<typename TIntegral></p> <p>struct is_signed</p> <p>{</p> <p>#ifdef OK</p> <p>static const bool value = TIntegral(TIntegral() – 1) > 0 ? false :</p> <p>true;</p> <p>#else</p> <p>static const bool value = ((TIntegral)0) – ((TIntegral)1) > ((TIntegral)</p> <p>0) ? false : true;</p> <p>#endif</p> <p>};</p> <p>template<typename TIntegral></p> <p>struct the_no</p> <p>{</p> <p>static const TIntegral value = TIntegral() – 1;</p> <p>};</p> <p>using std::cout;</p> <p>template<typename T></p> <p>void display(void)</p> <p>{</p> <p>bool b = is_signed<T>::value;</p> <p>cout << typeid(T).name();</p> <p>if (b)</p> <p>cout << ” is signed”;</p> <p>else</p> <p>cout << ” is UNsigned”;</p> <p>cout  << std::endl;</p> <p>}</p> <p>int main(void)</p> <p>{</p> <p>bool is_unsigned_char_signed = the_no<unsigned char>::value > 0 ?false : true;</p> <p>unsigned char value = the_no<unsigned char>::value;</p> <p>cout << “is_unsigned_char_signed is : “<< is_unsigned_char_signed << std::endl;</p> <p>cout << “value is: ” << (short) value << std::endl;</p> <p> typedef unsigned char UCHAR;</p> <p>unsigned char n = UCHAR() -1;</p> <p>cout << “UNSigned types:” << std::endl;</p> <p>display<unsigned long long>();</p> <p>display<unsigned long>();</p> <p>display<unsigned int>();</p> <p>display<unsigned short>();</p> <p>display<unsigned char>();</p> <p>cout << std::endl;</p> <p> cout << “Signed types:” << std::endl;</p> <p>display<signed long long>();</p> <p>display<signed long>();</p> <p>display<signed int>();</p> <p>display<signed short>();</p> <p>display<signed char>();</p> <p>return 0;</p> <p>}</p> <p>

为什么当OK宏未定义的时候，显示的结果如下：

UNSigned types:

unsigned __int64 is UNsigned

unsigned long is UNsigned

unsigned int is UNsigned

unsigned short is signed

unsigned char is signed

Signed types:

__int64 is signed

long is signed

int is signed

short is signed

signed char is signed

其中无符号的16位整型和无符号字符型都显示为有符号的了。而OK宏被定义的时候，却是正常的。作者很纳闷。为啥呢？

其实，作者此处绕了很多弯子，去用0和1做减法，看结果的符号来判断结果的类型，其实完全没有必要这么做，直接吧-1转换为目标类型，然后和零比较即可。而他认为能正常工作的，其实也只不过是在减法之后，又转换为目标类型了罢了。

现在来讨论 static const bool value = ((TIntegral)0) – ((TIntegral)1) > ((TIntegral)

0) ? false : true; 不能正常工作的原因吧，因为任何小于int的类型在计算时都已经提升为int了。呜呜。2008年过完了~