#include using namespace std; struct cheese { string type; cheese (string s) : type(s) { } }; void int_val(int a) { a = 4; } void int_ref(int& a) { a = 4; } void cheese_val(cheese c) { cheese brie("brie"); c = brie; } void cheese_ref(cheese& c) { cheese camembert("camembert"); c = camembert; } void cheeseptr_val(cheese* c) { c = new cheese("emmentaler"); } void cheeseptr_ref(cheese*& c) { c = new cheese("parmesan"); } int main() { int x = 5; // Passing to `int_val` obviously // does nothing as `x` will be // copied when calling it. int_val(x); cout << "int_val: " << x << endl; // Passing to `int_ref` will affect // `x`, because it is not `x` that // is passed in, but rather a reference // to x. A reference is similar to // a pointer. int_ref(x); cout << "int_ref: " << x << endl; ////////////////////// cheese c("moelltaler"); // The same goes for objects. // This will have no effect, because // c will be copied: cheese_val(c); cout << "cheese_val: " << c.type << endl; // And if we pass a reference, we will // see change: cheese_ref(c); cout << "cheese_ref: " << c.type << endl; /////////////////////// // This is a heap-allocated cheese; // a heap of cheese if you will. cheese* d = new cheese("gouda"); // It helps to realize that a pointer // is just a number, much like an int. // Assigning to a pointer means switching // these numbers. // Naturally, this will have no effect: // The pointer is copied as it is // passed in. So assining to the // pointer will only affect the // function's copy of the pointer! // (Though the function leaks memory) cheeseptr_val(d); cout << "cheeseptr_val: " << d->type << endl; // So, to get around that, we pass // the pointer by reference. cheeseptr_ref(d); cout << "cheeseptr_ref: " << d->type << endl; // It's the same concept but we just // dressed up an integer and called // it cheese* :) delete d; // Let's try to be safe even though we are leaking everywhere return 0; }

[Q/A] Reference to Pointer