void-pointers

What inherent advantages do boost::any and boost::any_cast offer over using void* and dynamic_cast ? The advantage is that boost::any is way more type-safe than void* . E.g. int i = 5; void* p = &i; static_cast<double*>(p); //Compiler doesn't complain. Undefined Behavior. boost::any a; a = i; boost::any_cast<double>(a); //throws, which is good As to your comment, you cannot dynamic_cast from a void* . You can dynamic_cast only from pointers and references to class types which have at least one virtual function (aka polymorphic types) boost::any calls destructors: { boost::any x = std::string(

Why is it impossible to have a reference to void? The only thing I found in the C++ Standard is this line, at 8.3.2.1 A declarator that specifies the type "reference to cv void" is ill-formed. Why is it that way? Why can't I write a "generic" function that accept a void& ? Just to be clear, I have no useful application in mind where using a reference-to-void could be better than using templates, but I'm just curious about the rationale for forbidding this construct. To clarify a little, I understand that using a reference-to-void "as is" would be as meaningless as dereferencing a pointer-to

How to free a void pointer. struct vStruct { void *vPtr; struct vStruct *next; }; struct vStruct sObj; struct vStruct *sObjNew = sObj; delete sObjNew->vPtr; -----------> Is this correct way to delete void pointer delete sObjNew; Showing error Operator 'delete', applied to void* argument having has undefined behavior, and most likely does not invoke the object's destructor. You should not delete a void pointer. delete works for specific types (such that compiler knows, which destructor should be called - as stated in error message). If you want to hold unspecified type in your structure, you

Here's my code: #include <iostream> using namespace std; int main() { void *x; int arr[10]; x = arr; *x = 23; //This is where I get the error } As you can see, the code is very simple. It just creates a void pointer x which points to the memory address of the array 'arr' and puts the integer 23 into that memory address. But when I compile it, I get the error message "'void*' is not a pointer-to-object type". When I use an 'int' pointer instead of a void pointer and then compile it, I don't get any errors or warnings. I wanna know why I get this error. Thank you. As the compiler message says,

I am trying to make void* to hold a value (to avoid default constructor calling). I want to:- copy K to void* e.g. K k1; --> void* raw=k1; copy void* to K e.g. void* raw; --> K k2=raw; try not to break destructor and causes memory leak don't use any dynamic allocation (heap, performance reason) Here is what I tried:- class K{ public: std::string yes="yes" ; }; int main() { //objective: k1->raw->k2 , then delete "raw" void* raw[sizeof(K)]; //<--- try to avoid heap allocation K k1; static_cast<K>( raw)=k1; //<--- compile fail K k2= static_cast<K>( raw); std::cout<<k2.yes; //test static_cast<K&>

If I create a void pointer, and malloc a section of memory to that void pointer, how can I print out the individual bits that I just allocated? For example: void * p; p = malloc(24); printf("0x%x\n", (int *)p); I would like the above to print the 24 bits that I just allocated. size_t size = 24; void *p = malloc(size); for (int i = 0; i < size; i++) { printf("%02x", ((unsigned char *) p) [i]); } Of course it invokes undefined behavior (the value of an object allocated by malloc has an indeterminate value). You can't - reading those bytes before initializing their contents leads to undefined

I'm trying to make an universal bubble sort function. It allow to user to write its own compare and swap function. I implemented a swap and compare function for int type, but when I run the code for next array: {3, 5, 8, 9, 1, 2, 4, 7, 6, 0} , I get: 0 0 84214528 2312 1 2 4 7 6 0. Why this is happening? #include <stdio.h> #include <stdlib.h> #define true 1 #define false 0 int compInt(void *a, void *b) // FUNCTION FOR COMPARE INT { if (*(int*)(a) > *(int*)(b)) { return false; } // IF FIRST INT > SECOND INT (WRONG ORDER) RETURN FALSE return true; // RIGHT ORDER -> RETURN TRUE } I think the

Consider the following C# function: void DoWork() { ... } The C# documentation states: When used as the return type for a method, void specifies that the method does not return a value . That seems straight-forward enough and, in most cases, that suits as a fair definition. However, for many lower-level languages (i.e. C) "void" has a slightly different meaning. Specifically, all blocks of code must return something . Therefore, void is a representation of the null pointer which is a representation of "nothing". In such a case, you do not need to have a return statement within your code