How can I understand what is declared here: (this is taken from another post on this forum)
template<typename C> static char (&f(ChT<int Fallback::*, &C::x>*))[1];
Here's how I read:
template of static function f called with (ChT<int Fallback::*, &C::x>*), but then I can't make sense why is there an address-of operator and why is there an array?
I'm still learning how to understand 开发者_开发知识库C++ declarations, so please explain this slowly and carefully.
Using some typedefs:
typedef char (&arrayref_t)[1];
This is a reference to an array of characters. The array has one element.
typedef ChT<int Fallback::*, &C::x> tmpl_t;
This is a template class, instantiated with the type "pointer to an int member of the Fallback class", and a member pointer to x in class C.
static arrayref_t f(tmpl_t*);
The function now takes a pointer to a tmpl_t and returns an arrayref_t.
It's important to see return type.
So, return type of this function is reference to char[1];
Imagine that f returns something like reference to the following:
char ret[1];
For example
template<typename C> static char (&f(ChT<int Fallback::*, &C::x>*))[1]
{
static char xx[1] = {'F'};
return xx;
}
加载中,请稍侯......
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