A (somewhat) outdated article explores ways to use decltype along with SFINAE to detect if a type supports certain operators, such as == or <.
With a class such as Foo: struct Foo { static const int i = 9; }; 开发者_如何学Go I find that GCC 4.5 will reject the following
As an exercise in understanding C++0x, I am trying to create a C++ class that wraps a pointer of some template-ized type:
#include <iostream> #include <map> using namespace std; int main() { int x = 5; decltype(x) y = 10;
While reading Wikipedia\'s page on decltype, I was curious about the statement, Its [decltype\'s] primary intended use is in generic
The following code: struct开发者_如何转开发 A { int f(int); auto g(int x) -> decltype(f(x)); };
template<class IntT, IntT low = IntT(), IntT high = IntT()> struct X { static_assert(std::is_same<decltype(low),decltype(high)>::value,\"Different types not allowed\");//this should give
Why this (static_assert) in a definition of a class doesn\'t work? template<class IntT, IntT low = IntT(), IntT high = IntT()>
I am trying to use declt开发者_如何学JAVAype inside a template class as follows: #include <functional>
I\'m trying to use decltype in the late specified return of a member function in a CRTP base class and it\'s erroring with: invalid use of incomplete type const struct AnyOp<main()::<lambda(int)
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