编译时间递归和条件 [英] Compile time recursion and conditionals

问题描述

我正在阅读对打印1到1000无回路或条件，我想知道为什么有必要在NumberGeneration< 1>的特殊情况在顶部的答案。

I was reading the responses to "Printing 1 to 1000 without loop or conditionals" and I am wondering why it is necessary to have the special case for NumberGeneration<1> in the top answer.

如果我删除，添加一个检查N == 1在模板（代码下面），代码失败编译与模板实例化深度超过最大，但我不知道为什么。是在编译时处理不同的条件？

If I remove that and add a check for N == 1 in the template (code below), the code fails compilation with "template instantiation depth exceeds maximum" but I'm not sure why. Are conditionals handled differently in compile-time?

#include <iostream>

template<int N>
struct NumberGeneration
{
    static void out(std::ostream& os)
    {
        if (N == 1)
        {
            os << 1 << std::endl;
        }
        else
        {
            NumberGeneration<N-1>::out(os);
            os << N << std::endl;
        }
    }
};

int main()
{
    NumberGeneration<1000>::out(std::cout);
}

推荐答案

代码生成和编译t分支取决于条件！考虑这个：

Code generation and compilation doesn't branch depending on conditionals! Consider this:

// don't declare bar()!

void foo()
{
     if (false) { bar(); }
}

如果您从未声明 bar code>，这是一个编译错误，即使内部作用域永远无法达到。出于同样的原因，不管是否可以到达该分支， NumberGeneration< N-1> 总是实例化，并且你有无限递归。

If you never declare bar(), this is a compilation error even though the inner scope can never be reached. For the same reason, NumberGeneration<N-1> is always instantiated, no matter whether that branch can be reached or not, and you have infinite recursion.

事实上，条件语句的静态模拟正是模板特化：

Indeed, the static analogue of conditionals is precisely template specialization:

template <> struct NumberGeneration<0> { /* no more recursion here */ };

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