std::function is a type erasure object. That means it erases the details of how some operations happen, and provides a uniform run time interface to them. For std::function, the primary1 operations are copy/move, destruction, and ‘invocation’ with operator() — the ‘function like call operator’. In less abstruse English, it means that std::function can contain almost … Read more
The syntax is wrong. A member pointer is a different type category from a ordinary pointer. The member pointer will have to be used together with an object of its class: class A { public: int f(); int (A::*x)(); // <- declare by saying what class it is a pointer to }; int A::f() { … Read more
Read this for detail : // 1 define a function pointer and initialize to NULL int (TMyClass::*pt2ConstMember)(float, char, char) const = NULL; // C++ class TMyClass { public: int DoIt(float a, char b, char c){ cout << “TMyClass::DoIt”<< endl; return a+b+c;}; int DoMore(float a, char b, char c) const { cout << “TMyClass::DoMore” << endl; … Read more
Both foo() and bar() can only be assigned to func_ptr2. ยง8.3.6/2: A default argument is not part of the type of a function. [Example: int f(int = 0); void h() { int j = f(1); int k = f(); // OK, means f(0) } int (*p1)(int) = &f; int (*p2)() = &f; // error: type … Read more
Funcp func = std::bind(&A::func, &a, std::placeholders::_1, std::placeholders::_2);
As others have pointed out, this is undefined behavior, so all bets are off about what may in principle happen. But assuming that you’re on an x86 machine, there’s a plausible explanation as to why you’re seeing this. On x86, the g++ compiler doesn’t always pass arguments by pushing them onto the stack. Instead, it … Read more
Yes, it is correct. I find that not very readable, so I suggest declaring the signature of the function to be pointed: typedef void sigrout_t(void*); I also have the coding convention that types ending with rout_t are such types for functions signatures. You might name it otherwise, since _t is a suffix reserved by POSIX. … Read more
Yes, you can export static functions by handing out pointers to them. This is a common means of implementing the Factory pattern in C, where you can hide the implementations of a whole bunch of functions from the modules that use them, and have a FuncPtr_t GetFunction( enum whichFunctionIWant) that hands them out to consumers. … Read more
You cannot, at least it won’t be only a pointer to a function. Member functions are common for all instances of this class. All member functions have the implicit (first) parameter, this. In order to call a member function for a specific instance you need a pointer to this member function and this instance. class … Read more
One of the standard conversions, in both C and C++, is the function-to-pointer conversion; when a function name appears in an expression, it can be converted into a pointer to that function. So: foo is equivalent to &foo *foo is equivalent to *(&foo), or foo **foo is eqivalent to **(&foo), or *foo, or foo and … Read more