在 C++ 编译时计算和打印阶乘

问题描述

template<unsigned int n>
struct Factorial {
    enum { value = n * Factorial<n-1>::value};
};

template<>
struct Factorial<0> {
    enum {value = 1};
};

int main() {
    std::cout << Factorial<5>::value;
    std::cout << Factorial<10>::value;
}

上面的程序在编译时计算阶乘值.我想在编译时而不是在运行时使用 cout 打印阶乘值.我们如何才能在编译时打印阶乘值?

above program computes factorial value during compile time. I want to print factorial value at compile time rather than at runtime using cout. How can we achive printing the factorial value at compile time?

我使用的是 VS2009.

I am using VS2009.

谢谢！

推荐答案

阶乘可以在编译器生成的消息中打印为:

The factorial can be printed in compiler-generated message as:

template<int x> struct _;
int main() {
        _<Factorial<10>::value> __;
        return 0;
}

错误信息:

prog.cpp:14:32: 错误:聚合 ‘_<3628800> __’ 类型不完整，无法定义_::值> __;^

prog.cpp:14:32: error: aggregate ‘_<3628800> __’ has incomplete type and cannot be defined _::value> __; ^

这里3628800是10的阶乘.

在 ideone 上查看:http://ideone.com/094SJz

See it at ideone : http://ideone.com/094SJz

所以你在找这个吗?

Matthieu 需要一个聪明的技巧来打印阶乘并让编译继续.这是一种尝试.它没有给出任何错误，因此编译成功并发出一个警告.

Matthieu asked for a clever trick to both print the factorial AND let the compilation continue. Here is one attempt. It doesn't give any error, hence the compilation succeeds with one warning.

template<int factorial> 
struct _{ operator char() { return factorial + 256; } }; //always overflow
int main() {
        char(_<Factorial<5>::value>());
        return 0;
}

编译时带有此警告:

main.cpp: 在实例化 '_::operator char() [with intfactorial = 120]': main.cpp:16:39: 从这里需要main.cpp:13:48: 警告:隐式常量转换溢出[-Woverflow] struct _{ operator char() { return factorial + 256;} };//总是溢出

main.cpp: In instantiation of '_::operator char() [with int factorial = 120]': main.cpp:16:39: required from here main.cpp:13:48: warning: overflow in implicit constant conversion [-Woverflow] struct _{ operator char() { return factorial + 256; } }; //always overflow

这里120是5的阶乘.

ideone 上的演示:http://coliru.stacked-crooked.com/a/c4d703a670060545

Demo at ideone : http://coliru.stacked-crooked.com/a/c4d703a670060545

你可以写一个很好的宏，然后用它来代替:

You could just write a nice macro, and use it instead as:

#define PRINT_AS_WARNING(constant) char(_<constant>())    

int main() 
{
         PRINT_AS_WARNING(Factorial<5>::value);
         return 0;
}

那个看起来很棒.

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