模板之小试牛刀---我有点明白为什么这么多人骂c++了

有朋友让我写一个模版算素数的程序，我也好久没用模版了，就当是热手。结果让我非常非常郁闷。

另外附带写了算整数平方根，因为素数判断时要用。下面有4个测试宏，分别是打印素数，打印素数个数，打印是否是素数，打印整数平方根。

#include <iostream>using namespace std;template<size_t N, size_t M1 , size_t M2, bool stop >//true? Stop expand template! struct sqrt_check{enum { result = M1 * M1 >= N ?M1 : M2 * M2 <= N ?M2 : (M1 + M2)/2 * (M1 + M2)/2 == N ?(M1 + M2)/2 : (M1 + M2)/2 * (M1 + M2)/2 > N ?//Bsearch begin! sqrt_check<N, M1, (M1 + M2)/2, M2 == (M1 + M2)/2 || M1 * M1 >= N>::result//The 3rd argument is to avoid recursion, when |M1-M2|<=1. //Repeat checking M1*M1>=N is to reduce compiling time!: sqrt_check<N, (M1 + M2)/2, M2, M1 == (M1 + M2)/2 || M2 * M2 <= N>::result//The 3rd argument is to avoid recursion, when |M1-M2|<=1. //Repeat checking M2*M2<=N is to reduce compiling time!};};template<size_t N, size_t M1 , size_t M2>struct sqrt_check<N, M1, M2, true>{enum { result = M1 };//when (M1 * M1)< N < (M2 * M2), the result is M1.};template <size_t N>struct sqrt_nature{enum {result = N <= 5 ?N/2 : sqrt_check < N, 2, N/2, false >::result//begin searching between [2, N/2]};enum {square = result * result == N};};template <>struct sqrt_nature<1>{enum {result = 1};enum {square = true};};template<size_t N, size_t D>struct prime_check{enum {prime = D == 1 ? true : N % D == 0 ? false : prime_check<N , D - 1>::prime };};template<size_t N>struct prime_check<N , 1>{enum {prime = true };};template<size_t N>struct prime_nature{enum {prime = prime_check<N, sqrt_nature<N>::result>::prime};enum {composite = !prime};};template<>struct prime_nature<1>{enum {prime = false};enum {composite = false};};template<size_t N>struct prime_count_nature{enum {prime_count = prime_count_nature<N-1>::prime_count + (prime_nature<N>::prime ? 1 : 0)};enum {composite_count = N - prime_count - 1};};template<>struct prime_count_nature<1>{enum {prime_count = 0};enum {composite_count = 0};};/*#define test(x)/do{/if (prime_nature<x>::prime)/cout<<x<<endl;/}while(false)*///#define test(x) cout<<prime_count_nature<x>::prime_count<<endl//#define test(x) cout<<x<<' '<<(bool(prime_nature<x>::prime) ? "Yes" : "No")<<endl#define test(x) cout<<x<<' ' <<sqrt_nature<x>::result<<endltemplate<size_t N>struct test_nature:test_nature<N-1>{test_nature():test_nature<N-1>(){test(N);}};template<>struct test_nature<1>{test_nature(){test(1);}};int main(){test_nature<100> x;return 0;} 

 

很意外的发现，编译时运算居然不支持逻辑运算短路。也就是说编译时运算结果已经在常数表达式左边算出，逻辑运算依然要强行算完剩下的。

 

以prime_check（检查是否是素数）为例，

prime = D == 1 ? 

true :   N % D == 0 ? 

false : prime_check<N , D - 1>::prime  

 

当D=1时，prime的结果就可以获得。然而编译器会不必要的强行把整个表达式算完。于是如果没有对D=1作偏特化，编译器便会无限制的进行递归展开，直到展开过深报错。我不得不添加了一个其实根本没用的偏特化prime_check<N , 1>，并放弃使用vc6改用其他支持偏特化的编译器。我一时身边能找到的g++, intel, codepad.org等编译器都是完全相同的行为。

 

事实上我并不知道编译器是真的去计算prime_check<N , D - 1>::prime呢，还是仅仅把prime_check<N , D - 1>实例化。总之在逻辑运算之外产生了完全没有必要的副作用。我先后试了&&,||,?:,都是完全相同的结果。

 

最让人不可接受的是：预处理的#if逻辑运算支持短路，运行时的&&,||,?:逻辑运算支持短路，但是在这2个阶段中间的编译阶段却不支持（这算是c++自己搞出来的）。你说凭什么让程序员不骂制定c++标准的人白痴？