The default implementation will compare the addresses stored by the pointers, so different objects will be considered as different keys. However, the logical state of the object will not be considered. For example, if you use std::string * as the key, two different std::string objects with the same text of “Hello” would be considered a … Read more
Variable char* matches[1]; is declared on the stack, and it will be automatically released when the current block goes out of scope. This means when you return matches, memory reserved for matches will be freed, and your pointer will point to something that you don’t want to. You can solve this in many ways, and … Read more
Like this: void * p = (void *)0x28ff44; Or if you want it as a char *: char * p = (char *)0x28ff44; …etc. If you’re pointing to something you really, really aren’t meant to change, add a const: const void * p = (const void *)0x28ff44; const char * p = (const char *)0x28ff44; … Read more
As a first remark, I wouldn’t call it ObjectArray if it is a map and not an array. Anyway, you can insert objects this way: ObjectArray myMap; myMap.insert(std::make_pair(0, std::unique_ptr<Class1>(new Class1()))); Or this way: ObjectArray myMap; myMap[0] = std::unique_ptr<Class1>(new Class1()); The difference between the two forms is that the former will fail if the key 0 … Read more
Every variable in Java is a reference. So when you do SomeClass s2 = s1; you just point s2 to the same object as s1 points to. You are actually assigning the value of the reference s1 (which points to an instance of SomeClass) to s2. If you modify s1, s2 will be modified as … Read more
There are many solutions, here’s a few I’ve come up with: int main(int nArgs, char ** vArgs) { vector<int> *v = new vector<int>(10); v->at(2); //Retrieve using pointer to member v->operator[](2); //Retrieve using pointer to operator member v->size(); //Retrieve size vector<int> &vr = *v; //Create a reference vr[2]; //Normal access through reference delete &vr; //Delete the … Read more
Go performs pointer escape analysis. If the pointer escapes the local stack, which it does in this case, the object is allocated on the heap. If it doesn’t escape the local function, the compiler is free to allocate it on the stack (although it makes no guarantees; it depends on whether the pointer escape analysis … Read more
What you get for w is a pointer to the non exported type http.response but as ResponseWriter is an interface, that’s not visible. From server.go: type ResponseWriter interface { … } On the other hand, r is a pointer to a concrete struct, hence the need to pass a reference explicitly. From request.go: type Request … Read more
From the C99 standard (7.21.1/2): Where an argument declared as size_t n specifies the length of the array for a function, n can have the value zero on a call to that function. Unless explicitly stated otherwise in the description of a particular function in this subclause, pointer arguments on such a call shall still … Read more
The reason why the shorthand: int *bar = &foo; in your example can be confusing is that it’s easy to misread it as being equivalent to: int *bar; *bar = &foo; // error: use of uninitialized pointer bar! when it actually means: int *bar; bar = &foo; Written out like this, with the variable declaration … Read more