C programming: casting a void pointer to an int?, You're casting 5 to be a void pointer and assigning it to ptr . 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. https://docs.microsoft.com/en-us/dotnet/csharp/language-reference/unsafe-code Maybe something is going wrong prior to the cast. The void pointer can point to which type of objects? However, be aware that if this code is inside a function and. A char pointer pointer can also be looked at as a pointer to a string. Since the array (aiData) is the collection of integer element so the type of &aiData[0] would be a pointer to int (int*). It is also called general purpose pointer. this way you won't get any warning. 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. Answer: Maybe your first thought was, well reinterpret_cast(&f) of course. 1. It helps in implementing two types of pointers namely you can pass the int value as void pointer like (void *)&n where n is integer, and in the function accept void pointer as parameter like void foo(void *n);and finally inside the function convert void pointer to int like, int num = *(int *)n;. Syntax. (Note that in C++ casting any pointer to void* is also done implicitly (except for function pointers and function-member / method pointers which cannot be cast to void*), but casting back requires an explicit cast.) Generally, you cannot convert between ints and pointers in this way. Pa is declared as a pointer to int variables, Pd is declared as a pointer to double type variables, and Pc is declared as pointer to character type variables. Well, it turns out, C++ wants to support architectures, where data and instruction pointers are physically distinct things. Now ptr points at the memory address 0x5. In C, malloc () and calloc () functions return void * or generic pointers… Although you can cast a pointer to a structure to a void *, since both are addresses, you cannot cast a structure to a void *, since a structure is not an address.The only thing you can do is cast a pointer to a structure from a void *:. In C++, a pointer to a specific type (primitive or user-defined) can be assigned to a void* without an explicit typecast. In the following code lines, A is an int type variable, D is variable of type double, and ch is a variable of type char. The syntax is as follows − * ( (type cast) void pointer) For example, int i=10; void *vp; vp = &i; printf ("%d", * ((int*) vp)); type cast Example When does the void pointer can be dereferenced? QAxBase::dynamicCall () is documented as returning an invalid QVariant if the call fails, or if the called method doesn't return a value. a is of type int[4], which can be implicitly cast to int* (ie, a pointer to an int) &a is of type int(*)[4] (ie: a pointer to an array of 4 ints). Finally, cast it to a void pointer, and inspect it again. As a result, it is possible to silently convert from one pointer type to another without the compiler diagnosing the problem by storing or casting a pointer to void * and then storing or casting it to the final type. You need to cast the void* pointer to a char* pointer - and then dereference that char* pointer to give you the char that it points to! So we have to typecast the void pointer pvData from the pointer to int (int*) before performing an arithmetic operation. The void pointer in C is a pointer which is not associated with any data types. If the value in a pointer is cast to a different type and it does not have the correct alignment for the new type, the behavior is undefined. So the cast in your mallo statement is pointless; the cast should come when you reference the pointer: ptr = malloc( sizeof(testStructure) ); ((testStructure*)ptr)->a = 5; … 5.3.2 Struct Pointer Cast of Void Pointer. void *pointername; For example, void *vp; Accessing − Type cast operator is for accessing the value of a variable through its pointer. The most general answer is — in no way. You could use this code, and it would do the same thing as casting ptr to (char*) returnPtr [j] = ( void *) ( (size_t) (ptr) + i); Do not cast pointers to int, long, ULONG, or DWORD. Now the type of vptr temporarily changes from void pointer to pointer to int or (int*) , and we already know how to dereference a pointer to int, just precede it with indirection operator ( *) *(int *)vptr. That was mine, at least. Cast void pointer to int. GCC does not warn on casts from pointers to enumerators, while clang currently does. 2,149. On many systems, an 8-bit unsigned int can be stored at any address while an unsigned 32-bit int must be aligned on an address that is a multiple of 4. This causes a bunch of extra warnings in the Linux kernel, where certain structs contain a void pointer to avoid using a gigantic union for all of the various types of driver data, such as version. This property of void* makes it quite useful as a generic or opaque handle. Following is the declaration for the void pointer −. you want to return that pointer from the function, you will. The pointer concept in C is very useful as it helps in memory allocation and address management. However even though their types are different, the address is going to be the same. int * intPtr {static_cast < int * > (voidPtr)}; // however, if we cast our void pointer to an int pointer... std :: cout << * intPtr << '\n' ; // then we can use indirection through it like normal This prints: In this noncompliant code example, loop_function() is passed the char pointer char_ptr but returns an object of type int pointer: a) int b) float c) double d) all of the mentioned Answer: d Clarification: Because it doesn’t know the type of object it is pointing to, So it can point to all objects. For example − void *vp; Accessing − Type cast operator is used for accessing the value of a variable through its pointer. BulldogLowell: Sorry, is your intention to dereference a pointer to an int, float or double and assign to a single class data member? Void Pointers in C. 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. 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. That’s why the standard forbids casting of function pointers to data pointers. )can be assigned to a void pointer … How to correctly cast a pointer to int in a 64-bit application? 2. Well the void pointer called value IS a member of the menu class. void f(void) { char *ptr; /* ... */ unsigned int number = (unsigned int)ptr; /* ... Compliant Solution Any valid pointer to void can be converted to intptr_t or … (int *)vptr. A "void pointer" (or more accurately, a pointer-to-void) is a pointer where you don't know the type of what it's pointing to. I'm doing some changes in the Linux kernel code and have noticed a pointer being cast into integer. Then verify that the QVariant is valid, at least for development purposes. Yes, but you'll need a location to store the float: float my_float ; void *tmp ; my_float = atof ("123.123") ; tmp = (void*) &my_float ; You can now probably do what you want with your void* pointer. When a pointer variable is declared using keyword void – it becomes a general purpose pointer variable. Hence the proper typecast in this case is (int*). We saw a conversion from a void pointer above. Address of any variable of any data type (char, int, float etc. The compiler is perfectly correct & accurate! A void pointer can hold address of any type and can be typcasted to any type. A void pointer is nothing but a pointer variable declared using the reserved word in C ‘void’. you could have used int pointers or char pointers and got the same problems. It points to some data location in the storage means points to the address of variables. 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. 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. The syntax for void pointer is given below − * ( (type cast) void pointer) Example 1 int i=10; void *vp; vp = &i; printf ("%d", * ((int*) vp)); // int * type cast Example. 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 Alternatively, if you choose to cast the ptr variable to (size_t) instead, then you don't need to worry about the pointer type anymore. There's no need for the cast in the second example. Check out buf below (full code): snd_pcm_sframes_t snd_pcm_lib_read(struct snd_pcm_substream * Its casting a void pointer to a char pointer pointer (pointer to pointer to char), then dereferencing that to get a char pointer that represents a string. A structure is a bunch of data, of known size. However, pointers may be type cast from one type to another type. A conversion to a void pointer happens in the following code: void foo ( void * vptr) { } int main () { int * p = ... /* some initialization */ ; foo (p); return 0 ; } Note that foo expects a void pointer, but we pass it int*. So it's casting void to string pointer, and dereferencing that to get the actual string to compare. But because its a void pointer, another pointer of the correct type must be cast to change the value that its POINTING to, e.g. A pointer is an address. And when assigning to a void pointer, all type information is lost. In the following example, the void pointer vp, is cast as a struct pointer. 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. Also, a void* can be typecasted back to a pointer of any type: void* vp = new int(); // OK int* ip = static_cast(vp); //OK with typecast. With lint -Xalias_level=weak (or higher), this generates a warning. int a = 10; char b = 'x'; void *p = &a; p = &b; Advantages of void pointers: 1) malloc () and calloc () return void * type and this allows these functions to be used to allocate memory of any data type (just because of void *) int main (void) Similarly, Pc may be type cast to type int and assigned the value Pa. After the execution of the above code all the three pointers, i.e., Pa, Pd, and Pc, point to the value 150. In C, casting to void* from any pointer type and vice-versa is done implicitly. Offline ImPer Westermark over 10 years ago in reply to Andy Neil Except that you sometimes stores an integer in the pointer itself.

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