64-bit lessons. A void pointer can point to a variable of any data type. Or does it actually do anything? int MyFunction (double money); // Function prototype const void* arg = (const void*)MyFunction; // type casting function pointer to const void* in C-style int , char or float etc. Featured on Meta Stopping the compiler from complaining about unused variables. Cast from void* to int; Andrey Karpov, Evgeniy Ryzhkov. Sometimes it's necessary to use void* pointers, for example when passing between C++ code and C functions.

We have learned in chapter Pointer Basics in C that if a pointer is of type pointer to int or (int *) then it can hold the address of the variable of type int only. The type given for a variable in its declation or definition is fixed; if you declare ptr as a pointer to void, then it will always be a pointer to void. your coworkers to find and share information.

site design / logo © 2020 Stack Exchange Inc; user contributions licensed under Or does it actually do anything?It doesn't do anything, except one thing. Not possible in standard C++. Therefore, you can use any particular pointer to function type as a generic pointer to function type, in the same sense that void* is a generic pointer to object type. Stack Overflow works best with JavaScript enabled Pattern 7.

In this chapter we will be looking special type of pointer called void pointer or general purpose pointer. ) Let me show you what I mean. If r is empty, so is the new shared_ptr (but its stored pointer is not necessarily null). If you want to use it as a pointer to something else, then you have to cast it at the point that you use it.

A pointer to void simply points to a raw memory location. What is the POINTER_32 macro?

That's literally it.

It's not calling a function and casting its return value as void - that would require a couple more parentheses. Is this just a no-op to prevent the function from being optimized away? When I cast a void * vPointer to a unigned int - like (uint32_t)vPointer - the compiler is happy - but when I cast to a signed char - like (int8_t)vPointer the compiler complains and says: Source_App\TerminalDrv.c(56): warning: #767-D: conversion from pointer to smaller integer Unlike the pointer-to-pointer conversions discussed in Par.A.6.6, which generally require an explicit cast, pointers may be assigned to and from pointers of type void*, and may be compared with them. Note the to and from part of the above quote. We have learned in chapter Pointer Basics in C that if a pointer is of type pointer to We can't just dereference a void pointer using indirection (It simply doesn't work that way!. (Steps 1 and 2 are often combined to cast and dereference in one expression. The following program demonstrates pointer arithmetic in void pointers.The void pointers are used extensively in dynamic memory allocation which we will discuss next.// wrong since type of fp is pointer to float How to typecast a "function pointer" to "const void*" type in C++ way?

By using our site, you acknowledge that you have read and understand our About size_t and ptrdiff_t. Discussion at bytes.com. void * Sometimes we know we want a pointer, but we don't necessarily know or care what it points to. Do not cast pointers to int, long, ULONG, or DWORD. (void)(foo); It's not a void pointer cast - that would be straightforward. If you must cast a pointer to test some bits, set or clear bits, or otherwise manipulate its contents, use the UINT_PTR or INT_PTR type.

Knowledge Base. While perusing some STM32 middleware code, I came across this very odd line and can't parse it. Before you apply pointer arithmetic in void pointers make sure to provide a proper typecast first otherwise you may get unexcepted results. Before you dereference a void pointer it must be typecasted to appropriate pointer type. It would be incorrect, if we assign an address of a float variable to a pointer of type pointer to int.But void pointer is an exception to this rule. Discussion at stackoverflow.com. When a typed pointer is cast to a void pointer, the contents of the memory location are unchanged. The following program demonstrates how to dereference a Another important point I want to mention is about pointer arithmetic with void pointer. Private self-hosted questions and answers for your enterpriseProgramming and related technical career opportunitiesMy guess is that (1) the line itself is there to suppress a warning about Creates a new instance of std::shared_ptr whose stored pointer is obtained from r's stored pointer using a cast expression.. It's basically,It's not a void pointer cast - that would be straightforward. Otherwise, the new shared_ptr will share ownership with the initial value of r, except that it is empty if the dynamic_cast performed by dynamic_pointer_cast returns a null pointer. Andrey Karpov. Pointer packing. You can only cast object pointers to void*, not function pointers.

Error: cast from void* to int loses precision.

By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader. It looks like an rvalue without an lvalue.

The void pointer can then be cast back to the appropriate type and used when required. It's not calling a function and casting its return value as void - that would require a couple more parentheses. It is actually being recommended here, on SO, as a portable way of doing it: It's to prevent the compiler from complaining about foo being unused.

The Overflow Blog Is this just a no-op to prevent the function from being optimized away? Free 30 Day Trial Summary Void Pointers & Casting.