function-pointers

Is it possible to pass a function pointer as an argument to a function in C? If so, how would I declare and define a function which takes a function pointer as an argument? Mehrdad Afshari Definitely. void f(void (*a)()) { a(); } void test() { printf("hello world\n"); } int main() { f(&test); return 0; } Michal Sznajder Let say you have function int func(int a, float b); So pointer to it will be int (*func_pointer)(int, float); So than you could use it like this func_pointer = func; (*func_pointer)(1, 1.0); /*below also works*/ func_pointer(1, 1.0); To avoid specifying full pointer type every

Consider this javascript code: var bar = function () { alert("A"); } var foo = bar; bar = function () { alert("B"); }; foo(); When running this code I get "A". Is this behavior a part of javascript specification and can I rely on it? Yes that is expected and by design. Your question is basically: does foo reference bar as a pointer or reference would in another language? The answer is no: the value of bar at the time of assignment is assigned to foo . siingCoder In other examples, nothing was passed by value; everything was passed by reference. bar and foo are BOTH pointers All vars/handles to

I came across this piece of code: char code[] = "\xb0\x01\x31\xdb\xcd\x80"; int main(int argc, char **argv) { int (*func)(); func = (int (*)()) code; (int)(*func)(); } It is copied from Writing Shellcode for Linux and Windows Tutorial . Could someone explain that what this function invocation (int)(*func)(); is doing? Jean-Baptiste Yunès It calls a function whose machine code is in the array code . The string contains some machine-level instructions ((three I think, have a look at x86 instruction set). func is declared as a pointer to a function that takes no argument and returns an int . func

What are all operations supported by function pointer differs from raw pointer? Is > , < , <= , >=operators supported by raw pointers if so what is the use? For both function and object pointers, they compile but their result is only guaranteed to be consistent for addresses to sub-objects of the same complete object (you may compare the addresses of two members of a class or array) and if you compare a function or object against itself. Using std::less<> , std::greater<> and so on will work with any pointer type, and will give consistent results, even if the result of the respective built-in

I am developing a C++ application using a C library. I have to send a pointer to function to the C library. This is my class: class MainWindow : public QMainWindow { Q_OBJECT public: explicit MainWindow(QWidget *parent = 0); private: Ui::MainWindow *ui; void f(int*); private slots: void on_btn_clicked(); }; This is my on_btn_clicked function: void MainWindow::on_btn_clicked() { void (MainWindow::* ptfptr) (int*) = &MainWindow::f; c_library_function(static_cast<void()(int*)>(ptfptr), NULL); } The C function should get a pointer to a such function : void f(int*). But the code above doesn't work,