Visual Studios 2012中的默认模板参数

This question is a followup after this one. The actual problem is that default template parameters for function templates are not supported by Visual Studios 2012 as indicated by this list.

这个问题是在这个问题之后的后续问题。实际问题是Visual Studio 2012不支持函数模板的默认模板参数，如此列表所示。

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Since default template parameters are not supported by Visual Studios 2012, is there any workaround to have the same result without it? So is it possible to define a template function such as

由于Visual Studios 2012不支持默认模板参数，是否有任何解决方法可以在没有它的情况下获得相同的结果？因此可以定义模板函数，例如

template <typename T, typename Ret = T>
Ret round(T val, Ret ret = Ret()) {
    return static_cast<Ret>(
        (val >= 0) ?
        floor(val + (T)(.5)) :
        ceil( val - (T)(.5))
    );
}

without the use of default template arguments? The function works as

不使用默认模板参数？该功能的作用如下

auto a = round(5.5, int()); // int a = 6
auto b = round(5.5); // double b = 6.0

1 个解决方案

#1

Like this, also, passing a value to force a return type is not really a nice way to do it, use the template argument instead :

同样，传递一个值来强制返回类型也不是一个很好的方法，而是使用模板参数：

#include <iostream>
#include <cmath>

template <typename Ret, typename T>
Ret round( T val ) {
    return static_cast<Ret>(
        ( val >= 0 ) ?
        std::floor( val + (T) ( .5 ) ) :
        std::ceil( val - (T) ( .5 ) )
        );
}

template <typename T>
T round( T val ) {
    return round<T,T>( val );
}

auto a = round<int>( 5.5 ); // int a = 6
auto b = round( 5.5 ); // double b = 6.0

static_assert( std::is_same<decltype(a), int>::value, "a must be int" );
static_assert( std::is_same<decltype(b), double>::value, "b must be double" );

int main() {
    std::cout << a << " " << b; 
}