What exactly is the ->* operator?

Question

I\'ve never used it before and just stumbled upon it in an article... I thought it would be the equivalent to *x->y but apparently it isn\'t.

Here\'s

Answer 1

The .* and ->* operators will point to member functions of a class or structure. The code below will show a simple example of how to use the .* operator, if you change the line: Value funcPtr = &Foo::One; to Value funcPtr = &Foo::Two; the result displayed will change to 1000 since that function is inValue*2

for example Taken From Here:

#include 
#include 

class Foo { 
  public: 
    double One( long inVal ) { return inVal*1; }
    double Two( long inVal ) { return inVal*2; }
}; 

typedef double (Foo::*Value)(long inVal); 

int main( int argc, char **argv ) { 
  Value funcPtr = &Foo::One; 

  Foo foo;

  double result = (foo.*funcPtr)(500); 
  std::cout << result << std::endl;
  system("pause");
  return 0; 
}