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尤怨

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尤怨 2025-02-07 13:34:47

sqrt＆lt; x，n＆gt; instantiates sqrt＆lt; x+1，n＆gt; and instantiates sqrt＆lt; x+2，n＆gt;等永不停止。

即使仅采用一个分支机，也通过编译器评估两个分支。编译器还不够聪明，无法看到该条件在某个时候是错误的，然后sqrt＆lt; value+1，n＆gt;不需要实例化。您必须更加解释。（实际上更正确：编译器确实需要知道的两侧：以确定其常见类型，因为这就是条件运算符类型的类型）。

由于C ++ 17，您可以使用constexpr，如果丢弃false-branch：

#include <iostream>


template<int value, int N>
int sqrt(){
    if constexpr (value*value == N) { 
        return value;
    } else {
        return sqrt<value+1,N>();
    }
    
};

int main()
{
    std::cout << sqrt<1,4>();
}

在C ++ 17之前，您可以将模板专业化作为递归的停止条件：

#include <iostream>


template<int value, int N>
struct sqrt
{
    enum { val = ((value*value == N) ? value : (sqrt<(value+1), N>::val)) };
};

template <int value>
struct sqrt<value,value*value> {
    enum { val = value };
};


int main()
{
    std::cout << sqrt<1,4>::val;
}

但是，当<时，两者都会失败。代码> n 不是平方数。您应该将条件更改为value*value＆gt; = n以确保递归始终停止（并在addionionally中检查n == value*value*value）。

此外，我建议交换参数的顺序，以便您可以将默认的1用于value。另外，enum东西看起来有些过时。我不记得要克服什么限制。无论如何，您可以简单地使用static const val = ...。

sqrt<X,N> instantiates sqrt<X+1,N> and that instantiates sqrt<X+2,N> etc. it never stops.

Both branches are evaluated by the compiler, even if only one of them is taken. The compiler is not clever enough to see that the condition is false at some point and then sqrt<value+1,N> does not need to be instantiated. You have to tell it more explictly. (Actually more correct: The compiler does need to know both sides of the : to determine their common type, because thats what the conditional operators type is).

Since C++17 you can use constexpr if to get the false-branch discarded:

#include <iostream>


template<int value, int N>
int sqrt(){
    if constexpr (value*value == N) { 
        return value;
    } else {
        return sqrt<value+1,N>();
    }
    
};

int main()
{
    std::cout << sqrt<1,4>();
}

Before C++17 you could use template specialization as the stop condition for the recursion:

#include <iostream>


template<int value, int N>
struct sqrt
{
    enum { val = ((value*value == N) ? value : (sqrt<(value+1), N>::val)) };
};

template <int value>
struct sqrt<value,value*value> {
    enum { val = value };
};


int main()
{
    std::cout << sqrt<1,4>::val;
}

However, both will fail when N is not a square number. You should change the condition to value*value >= N to make sure the recursion always stops (and addtionally check if N == value*value).

Further I suggest to swap the order of the arguments so you can use a default 1 for value. Also the enum thingy looks a little outdated. I don't remember what restriction is was meant to overcome. Anyhow, you can simply use a static const val = ... instead.

此模板元图有什么问题找到平方根？

情感失落者 2025-01-31 13:34:47 浏览 4 评论 0 收藏 0

尤怨

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