function-pointers

问题 I understand what function pointers are in C as well as how to use them. However, I do not know how to have one function pointer that can point to functions with different return types. Is this possible? I know how to use an array of function pointers, but I have only found examples of different parameters, not return types. (C: How can I use a single function pointer array for functions with variable parameter counts?) If an array would make more sense, can you provide an example of how to

问题 I was just going through some code on the Internet and found this: float * (*(*foo())[SIZE][SIZE])() How do I read this declaration? Is there a specific set of rules for reading such complex declarations? 回答1: I haven't done this in a while! Start with foo and go right. float * (*(* foo() )[SIZE][SIZE])() foo is a function with no arguments... Can't go right since there's a closing parenthesis. Go left: float * (*( * foo() )[SIZE][SIZE])() foo is a function with no arguments returning a

问题 example struct B1{int x; void f(){x = 1;}}; struct D : B1{int x; void f(){B1::x = 2;}}; using Dmp = void(D::*)(); using B1mp = void(B1::*)(); int main() { Dmp dmp = &D::f; D d; (d.*dmp)(); // ok B1mp b1mp = static_cast<B1mp>(dmp); // hm, well that's weird B1 b1; (b1.*b1mp)(); dmp = &B1::f; // ok } And this example will compile and run just fine, and no problem will arise. But wait, now I'm going to use D::x in D::f , and now -- anything can happen at runtime. Yes, you can also static_cast a

问题 This question already has answers here : How to print function pointers with cout? (7 answers) Closed 2 years ago . Check following code: #include <iostream> using namespace std; int& foo() { static int i = 0; return i; } int main() { cout << &foo() << endl; cout << &foo << endl; return 0; } As you see, the first cout prints address of return value of foo() which will be static variable i inside foo() . For 2nd cout I was expecting that &foo returns address of foo() function, as stated here:

问题 This question already has answers here : How to print function pointers with cout? (7 answers) Closed 2 years ago . Check following code: #include <iostream> using namespace std; int& foo() { static int i = 0; return i; } int main() { cout << &foo() << endl; cout << &foo << endl; return 0; } As you see, the first cout prints address of return value of foo() which will be static variable i inside foo() . For 2nd cout I was expecting that &foo returns address of foo() function, as stated here:

问题 I've got a question about how to properly use the new C++11 std::function variable. I've seen several examples from searching the Internet, but they don't seem to cover the usage case I'm considering. Take this minimum example, where the function fdiff is an implementation of the finite forward differencing algorithm defined in numerical.hxx (which isn't the problem, I just wanted to give a contextual reason why I'd want to take an arbitrary function and pass it around). #include <functional>

问题 I've been experimenting with function pointers and found the behavior of the following program rather misterious: void foo(int(*p)()) { std::cout << p << std::endl; } int alwaysReturns6() { return 6; } int main() { foo(alwaysReturns6); return 0; } The above code prints the number '1' on the screen. I know I should access the function pointer like this: p() (and then 6 gets printed), but I still don't get what the plain p or *p means when used in the foo function. 回答1: std::cout << p << std:

问题 I've been experimenting with function pointers and found the behavior of the following program rather misterious: void foo(int(*p)()) { std::cout << p << std::endl; } int alwaysReturns6() { return 6; } int main() { foo(alwaysReturns6); return 0; } The above code prints the number '1' on the screen. I know I should access the function pointer like this: p() (and then 6 gets printed), but I still don't get what the plain p or *p means when used in the foo function. 回答1: std::cout << p << std:

问题 Usually, a C++ lambda without a capture should be convertable to a c-style function pointer. Somehow, converting it using std::function::target does not work (i.e. returns a nullptr), also the target_type does not match the signature type even though it seems to be the same. Tested on VC13 and GCC 5.3 / 5.2.0 / 4.8 Minimal testing example: #include <functional> #include <iostream> void Maybe() { } void callMe(std::function<void()> callback) { typedef void (*ftype)(); std::cout << (callback

问题 I need to pass class function as argument for glutIdleFunc. This construction doesn't work: void (*func)(void) = this->step(); glutIdleFunc(func); and this too: void (*func)(void) = Core::step(); glutIdleFunc(func); How can I do it? Sorry for my stupidity... :) 回答1: In C++, callbacks such as the one accepted by glutIdleFunc() have to be static methods. Try: class Core { public: static void step() { // ... } }; glutIdleFunc(Core::step); 回答2: glutIdleFunc simply does not support calling a non