SFINAE检查operator + = [英] SFINAE check for operator+=
问题描述
如果运算符+ =
缺失,我试图从重载集中消除重载。
知道如何检查 T + T
是否合法:
typename T,
typename CheckTplusT = decltype(std :: declval< T>()+ std :: declval< T>())>
void foo(T a,T b,...)
{
a = a + b;
}
但不适用于 + = / code>
模板< typename T,
typename CheckTplusT = decltype(std :: declval< T> ;()+ = std :: declval< T>())>
void foo(T a,T b,...)
{
a + = b;这是可以通过使用中的另一个表达式来修复的。
}
或者我需要另一个SFINAE构造?
我需要这个从重载集中消除的原因是它与另一个重载接受函子用作 + =
的替代。编译器是VS2013,gcc4.8
我写第二个形式为:
template< typename T>
auto foo(T a,T b,...) - > decltype(a + = b,void())
{
a + = b;
}
decltype的推导类型(a + = b,如果表达式 a + = b
有效,则void()void
将只是 void
好吧,即使在第一种形式中,我也会使用尾随返回类型方法。
I'm trying to eliminate an overload from an overload set if operator+=
is missing.
I know how to check if T+T
is legal :
template<typename T,
typename CheckTplusT = decltype(std::declval<T>() + std::declval<T>())>
void foo(T a, T b, ...)
{
a = a + b;
}
but this doesn't work for +=
template<typename T,
typename CheckTplusT = decltype(std::declval<T>() += std::declval<T>())>
void foo(T a, T b, ...)
{
a += b;
}
Is this fixable by using another expression inside decltype
or do I need another SFINAE construct?
The reason I need this eliminated from the overload set is that it clashes with another overload that accepts a functor to be used as an alternative to +=
. Compilers are VS2013, gcc4.8
解决方案 I would write the second form as:
template<typename T>
auto foo(T a, T b, ...) -> decltype( a+=b, void() )
{
a += b;
}
The deduced type for decltype(a+=b, void())
would be just void
if the expression a+=b
is valid, else it would result in SFINAE.
Well, even in the first form, I would use the trailing-return type approach.
这篇关于SFINAE检查operator + =的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!