Saturday, 4 January 2020
c++ - C++11: How to alias a function?
Answer
Answer
If I have a class Foo in namespace bar:
namespace bar
{
class Foo { ... }
};
I can then:
using Baz = bar::Foo;
and now it is just like I defined the class in my namespace with the name Baz.
Is it possible to do the same for functions?
namespace bar
{
void f();
}
And then:
using g = bar::f; // error: ‘f’ in namespace ‘bar’ does not name a type
What is the cleanest way to do this?
The solution should also hold for template functions.
Definition: If some entity B is an alias of A, than if any or all usages (not declarations or definitions of course) of A are replaced by B in the source code than the (stripped) generated code remains the same. For example typedef A B is an alias. #define B A is an alias (at least). T& B = A is not an alias, B can effectively implemented as an indirect pointer, wheres an "unaliased" A can use "immediate semantics".
Answer
You can define a function alias (with some work) using perfect forwarding:
template
auto g(Args&&... args) -> decltype(f(std::forward(args)...)) {
return f(std::forward(args)...);
}
This solution does apply even if f is overloaded and/or a function template.
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