When you declare a function parameter as an array, the compiler automatically ignores the array size (if any) and converts it to a pointer. That is, this declaration: int foo(char p[123]); is 100% equivalent to: int foo(char *p); In fact, this isn’t about notation but about the actual type: typedef char array_t[42]; int foo(array_t p); … Read more
I think you’re approaching this from a rather oblique angle, so to speak. According to ยง5.3.1/1: The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or … Read more
In your expression BinaryNode * & t) BinaryNode* & t ————- —– BinaryNode pointer t is reference variable so t is reference to pointer of BinaryNode class. Pointer of the address of t? You are confused ampersand & operator in c++. that give address of an variable. but syntax is different. ampersand & in front … Read more
What is the difference between ptr and rptr? If you do char *world = “World”; rptr = world; and then print str, it will print “World”. If you do ptr = world; and then print str, it will print “Hello”.
To break this down yourself, start from the inner most parentheses and work your way out. (*[10]) <—- Array of 10 pointers (*[10])(int *) <—— Array of 10 pointers to functions which has a pointer to int as its argument (void (*[10])(int *)) <—— Array of 10 pointers to functions which has a pointer to … Read more
Sure, if you don’t mind the linked list having a maximum size, you could statically allocate an array of list nodes, and then use integer indices into the array as your “previous” and “next” values for each node, rather than pointers. I’ve done in this in the past to save a bit of memory (since … Read more
new allocates an amount of memory needed to store the object/array that you request. In this case n numbers of int. The pointer will then store the address to this block of memory. But be careful, this allocated block of memory will not be freed until you tell it so by writing delete [] array;
If you use &hello it prints the address of the pointer, not the address of the string. Cast the pointer to a void* to use the correct overload of operator<<. std::cout << “String address = ” << static_cast<void*>(hello) << std::endl;
A reference is semantically the following: T& <=> *(T * const) const T& <=> *(T const * const) T&& <=> [no C equivalent] (C++11) As with other answers, the following from the C++ FAQ is the one-line answer: references when possible, pointers when needed. An advantage over pointers is that you need explicit casting in … 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