元整数平方根中的无限递归 [英] Infinite Recursion in Meta Integer Square Root

问题描述

美好的一天，

我的一个朋友正在询问有关将整数平方根函数转换为元函数的问题.这是原始功能:

A friend of mine is asking about transforming an integer square root function into a meta-function. Here is the original function:

unsigned isqrt(unsigned value)
{
    unsigned sq = 1, dlt = 3;
    while(sq<=value)
    {
        sq  += dlt;
        dlt += 2;
    }
    return (dlt>>1) - 1;
}

我使用constexpr编写了元版本，但他说由于某些原因他不能使用该新功能:

I wrote a meta version using constexpr, but he said he can't use the new feature for some reason:

constexpr std::size_t isqrt_impl
    (std::size_t sq, std::size_t dlt, std::size_t value){
    return sq <= value ?
        isqrt_impl(sq+dlt, dlt+2, value) : (dlt >> 1) - 1;
}

constexpr std::size_t isqrt(std::size_t value){
    return isqrt_impl(1, 3, value);
}

因此，我认为将其转换为可自我递归调用的模板结构并不难:

So I thought it shouldn't be that hard to transform that into template struct that calls it self recursively:

template <std::size_t value, std::size_t sq, std::size_t dlt>
struct isqrt_impl{
    static const std::size_t square_root = 
        sq <= value ?
        isqrt_impl<value, sq+dlt, dlt+2>::square_root :
        (dlt >> 1) - 1;
};

template <std::size_t value>
struct isqrt{
    static const std::size_t square_root = 
        isqrt_impl<value, 1, 3>::square_root;
};

不幸的是，这导致了无限递归(在GCC 4.6.1上)，我无法弄清楚代码出了什么问题.这是错误:

Unfortunately, this is causing infinite recursion(on GCC 4.6.1) and I am unable to figure out what is wrong with the code. Here is the error:

 C:\test>g++ -Wall test.cpp
test.cpp:6:119: error: template instantiation depth exceeds maximum of 1024 (use
 -ftemplate-depth= to increase the maximum) instantiating 'struct isqrt_impl<25u
, 1048576u, 2049u>'
test.cpp:6:119:   recursively instantiated from 'const size_t isqrt_impl<25u, 4u
, 5u>::square_root'
test.cpp:6:119:   instantiated from 'const size_t isqrt_impl<25u, 1u, 3u>::squar
e_root'
test.cpp:11:69:   instantiated from 'const size_t isqrt<25u>::square_root'
test.cpp:15:29:   instantiated from here

test.cpp:6:119: error: incomplete type 'isqrt_impl<25u, 1048576u, 2049u>' used i
n nested name specifier

谢谢，

推荐答案

默认情况下，模板评估不是惰性的.

Template evaluation isn't lazy by default.

static const std::size_t square_root = 
    sq <= value ?
    isqrt_impl<value, sq+dlt, dlt+2>::square_root :
    (dlt >> 1) - 1;

无论条件如何，总是会实例化模板.您需要boost::mpl::eval_if或类似的东西才能使该解决方案生效.

will always instantiate the template, no matter what the condition. You need boost::mpl::eval_if or something equivalent to get that solution to work.

或者，您可以有一个基本情况下的局部模板专门化，如果满足条件，则可以停止递归，就像在K-ballos答案中一样.

Alternatively you can have a base case partial template specialization that stops the recursion if the condition is met, like in K-ballos answer.

实际上，与部分专业化相比，我更喜欢使用某种形式的惰性求值的代码，因为我觉得它更易于理解，并且可以降低模板所带来的噪音.

I'd actually prefer code that uses some form of lazy evaluation over partial specialization because I feel it is easier to comprehend and keeps the noise that comes with templates lower.

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