I\'ve been messing around with enable_if, and I seem to have stumbled upon some inconsistent behaviour. This is in VS2010. I\'ve reduced it to the following sample.
As question states, is there a reason why people use the struct version over the normal conditionals开发者_JS百科?An excerpt from the Boost Coding Guidelines for Integral Constant Expressions:
I\'m writing a template class, and I want to allow an additional method to exist only for a certain template type.Currently the method exists for all template types, but causes a compilation error for
I\'m trying to write a generic filtering function that performs linear interpolation at a given sampling coordinate in an 开发者_如何学运维multi-dimensional array (arbitrary rank). For this, I need a
Implicit conversion can be really useful when types are semantically equivalent. For example, imagine two libraries that implement a type identically, but in different namespaces. Or just a type that
template<typename T> std::istream & read(std::istream & istr, typename std::enable_if<std::is_pod<T>::value, T>::type & value)
This is more or less copy pasted from boost docs and I keep getting an error (actually alot of errors)
Do I have any way to simplify the following statements? (probably, using boost::enable_if). I have a simple class structure - Base base class, Derived1, Derived2 inherit from Base.
Any chance to use enable_if with a type conversion operator? Seems tricky, since both retur开发者_如何学运维n type and parameters list are implicit.From the little research I did (and ignoring the c++
Consider the following example: struct Scanner { template <typename T> T get(); }; template <> string Scanner::get()
加载中,请稍侯......