Consider the following situation.
I have a void * first pointing to a certain amount of memory. I have another void * second pointing to something else. I want to save void * second in the memory pointed by the first pointer.
My approach:
*(void**) first = second;
Is this valid? Are there any precautions to consider?
You can write anything you want to the memory location pointed by void* as long as:
It is large enough to hold that data type
It is sufficiently aligned for data type
Reads from that location consistent with writes
So, it's legal to do things like:
void* first = malloc(sizeof(void*));
*(void**)first = second;
....
void* another = *(void**)first; //Now another == second
But anything else, for example:
malloc(sizeof(int*)), or malloc(4)
void* first = &someCharArray[11]
int* another = *(int**)first
Is implementation defined at best. And sometimes undefined behavior.
That is exactly what void* type for - you cast some pointer to void*, then you cast back. You are responsible for types to match.
Try this
#include <string.h>
#include <stdlib.h>
int main()
{
int memory[10];
void *first = ( void * )memory;
void *second = ( void * )&memory[5];
*( int * )first = ( int )second;
return 0;
}
You can validate result in memory[0]
Related
I am learning C++ using C++ Primer 5th edition. In particular, i read about void*. There it is written that:
We cannot use a void* to operate on the object it addresses—we don’t know that object’s type, and the type determines what operations we can perform on that object.
void*: Pointer type that can point to any nonconst type. Such pointers may not
be dereferenced.
My question is that if we're not allowed to use a void* to operate on the object it addressess then why do we need a void*. Also, i am not sure if the above quoted statement from C++ Primer is technically correct because i am not able to understand what it is conveying. Maybe some examples can help me understand what the author meant when he said that "we cannot use a void* to operate on the object it addresses". So can someone please provide some example to clarify what the author meant and whether he is correct or incorrect in saying the above statement.
My question is that if we're not allowed to use a void* to operate on the object it addressess then why do we need a void*
It's indeed quite rare to need void* in C++. It's more common in C.
But where it's useful is type-erasure. For example, try to store an object of any type in a variable, determining the type at runtime. You'll find that hiding the type becomes essential to achieve that task.
What you may be missing is that it is possible to convert the void* back to the typed pointer afterwards (or in special cases, you can reinterpret as another pointer type), which allows you to operate on the object.
Maybe some examples can help me understand what the author meant when he said that "we cannot use a void* to operate on the object it addresses"
Example:
int i;
int* int_ptr = &i;
void* void_ptr = &i;
*int_ptr = 42; // OK
*void_ptr = 42; // ill-formed
As the example demonstrates, we cannot modify the pointed int object through the pointer to void.
so since a void* has no size(as written in the answer by PMF)
Their answer is misleading or you've misunderstood. The pointer has a size. But since there is no information about the type of the pointed object, the size of the pointed object is unknown. In a way, that's part of why it can point to an object of any size.
so how can a int* on the right hand side be implicitly converted to a void*
All pointers to objects can implicitly be converted to void* because the language rules say so.
Yes, the author is right.
A pointer of type void* cannot be dereferenced, because it has no size1. The compiler would not know how much data he needs to get from that address if you try to access it:
void* myData = std::malloc(1000); // Allocate some memory (note that the return type of malloc() is void*)
int value = *myData; // Error, can't dereference
int field = myData->myField; // Error, a void pointer obviously has no fields
The first example fails because the compiler doesn't know how much data to get. We need to tell it the size of the data to get:
int value = *(int*)myData; // Now fine, we have casted the pointer to int*
int value = *(char*)myData; // Fine too, but NOT the same as above!
or, to be more in the C++-world:
int value = *static_cast<int*>(myData);
int value = *static_cast<char*>(myData);
The two examples return a different result, because the first gets an integer (32 bit on most systems) from the target address, while the second only gets a single byte and then moves that to a larger variable.
The reason why the use of void* is sometimes still useful is when the type of data doesn't matter much, like when just copying stuff around. Methods such as memset or memcpy take void* parameters, since they don't care about the actual structure of the data (but they need to be given the size explicitly). When working in C++ (as opposed to C) you'll not use these very often, though.
1 "No size" applies to the size of the destination object, not the size of the variable containing the pointer. sizeof(void*) is perfectly valid and returns, the size of a pointer variable. This is always equal to any other pointer size, so sizeof(void*)==sizeof(int*)==sizeof(MyClass*) is always true (for 99% of today's compilers at least). The type of the pointer however defines the size of the element it points to. And that is required for the compiler so he knows how much data he needs to get, or, when used with + or -, how much to add or subtract to get the address of the next or previous elements.
void * is basically a catch-all type. Any pointer type can be implicitly cast to void * without getting any errors. As such, it is mostly used in low level data manipulations, where all that matters is the data that some memory block contains, rather than what the data represents. On the flip side, when you have a void * pointer, it is impossible to determine directly which type it was originally. That's why you can't operate on the object it addresses.
if we try something like
typedef struct foo {
int key;
int value;
} t_foo;
void try_fill_with_zero(void *destination) {
destination->key = 0;
destination->value = 0;
}
int main() {
t_foo *foo_instance = malloc(sizeof(t_foo));
try_fill_with_zero(foo_instance, sizeof(t_foo));
}
we will get a compilation error because it is impossible to determine what type void *destination was, as soon as the address gets into try_fill_with_zero. That's an example of being unable to "use a void* to operate on the object it addresses"
Typically you will see something like this:
typedef struct foo {
int key;
int value;
} t_foo;
void init_with_zero(void *destination, size_t bytes) {
unsigned char *to_fill = (unsigned char *)destination;
for (int i = 0; i < bytes; i++) {
to_fill[i] = 0;
}
}
int main() {
t_foo *foo_instance = malloc(sizeof(t_foo));
int test_int;
init_with_zero(foo_instance, sizeof(t_foo));
init_with_zero(&test_int, sizeof(int));
}
Here we can operate on the memory that we pass to init_with_zero represented as bytes.
You can think of void * as representing missing knowledge about the associated type of the data at this address. You may still cast it to something else and then dereference it, if you know what is behind it. Example:
int n = 5;
void * p = (void *) &n;
At this point, p we have lost the type information for p and thus, the compiler does not know what to do with it. But if you know this p is an address to an integer, then you can use that information:
int * q = (int *) p;
int m = *q;
And m will be equal to n.
void is not a type like any other. There is no object of type void. Hence, there exists no way of operating on such pointers.
This is one of my favourite kind of questions because at first I was also so confused about void pointers.
Like the rest of the Answers above void * refers to a generic type of data.
Being a void pointer you must understand that it only holds the address of some kind of data or object.
No other information about the object itself, at first you are asking yourself why do you even need this if it's only able to hold an address. That's because you can still cast your pointer to a more specific kind of data, and that's the real power.
Making generic functions that works with all kind of data.
And to be more clear let's say you want to implement generic sorting algorithm.
The sorting algorithm has basically 2 steps:
The algorithm itself.
The comparation between the objects.
Here we will also talk about pointer functions.
Let's take for example qsort built in function
void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void*))
We see that it takes the next parameters:
base − This is the pointer to the first element of the array to be sorted.
nitems − This is the number of elements in the array pointed by base.
size − This is the size in bytes of each element in the array.
compar − This is the function that compares two elements.
And based on the article that I referenced above we can do something like this:
int values[] = { 88, 56, 100, 2, 25 };
int cmpfunc (const void * a, const void * b) {
return ( *(int*)a - *(int*)b );
}
int main () {
int n;
printf("Before sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
qsort(values, 5, sizeof(int), cmpfunc);
printf("\nAfter sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
return(0);
}
Where you can define your own custom compare function that can match any kind of data, there can be even a more complex data structure like a class instance of some kind of object you just define. Let's say a Person class, that has a field age and you want to sort all Persons by age.
And that's one example where you can use void * , you can abstract this and create other use cases based on this example.
It is true that is a C example, but I think, being something that appeared in C can make more sense of the real usage of void *. If you can understand what you can do with void * you are good to go.
For C++ you can also check templates, templates can let you achieve a generic type for your functions / objects.
A very general question: I was wondering why we use pointer to pointer?
A pointer to pointer will hold the address of a pointer which in turn will point to another pointer. But, this could be achieved even by using a single pointer.
Consider the following example:
{
int number = 10;
int *a = NULL;
a = &number;
int *b = a;
int *pointer1 = NULL;
pointer1 = b; //pointer1 points to the address of number which has value 10
int **pointer2 = NULL;
pointer2 = &b; //pointer2 points to the address of b which in turn points to the address of number which has value 10. Why **pointer2??
return 0;
}
I think you answered your own question, the code is correct, what you commented isn't.
int number = 10; is the value
int *pointer1 = b; points to the address where int number is kept
int **pointer2 = &b; points to the address where address of int number is kept
Do you see the pattern here??
address = * (single indirection)
address of address = ** (double indirection)
The following expressions are true:
*pointer2 == b
**pointer2 == 10
The following is not!
*pointer2 == 10
Pointer to pointer can be useful when you want to change to what a pointer points to outside of a function. For example
void func(int** ptr)
{
*ptr = new int;
**ptr = 1337;
}
int main()
{
int* p = NULL;
func(&p);
std::cout << *p << std::endl; // writes 1337 to console
delete p;
}
A stupid example to show what can be achieved :) With just a pointer this can not be done.
First of all, a pointer doesn't point to a value. It point to a memory location (that is it contains a memory address) which in turn contains a value. So when you write
pointer1 = b;
pointer1 points to the same memory location as b which is the variable number. Now after that is you execute
pointer2 = &b;
Then pointer2 point to the memory location of b which doesn't contains 10 but the address of the variable number
Your assumption is incorrect. pointer2 does not point to the value 10, but to the (address of the) pointer b. Dereferencing pointer2 with the * operator produces an int *, not an int.
You need pointers to pointers for the same reasons you need pointers in the first place: to implement pass-by-reference parameters in function calls, to effect sharing of data between data structures, and so on.
In c such construction made sense, with bigger data structures. The OOP in C, because of lack of possibility to implement methods withing structures, the methods had c++ this parameter passed explicitly. Also some structures were defined by a pointer to one specially selected element, which was held in the scope global to the methods.
So when you wanted to pass whole stucture, E.g. a tree, and needed to change the root, or 1st element of a list, you passes a pointer-to-a-pointer to this special root/head element, so you could change it.
Note: This is c-style implementation using c++ syntax for convienience.
void add_element_to_list(List** list, Data element){
Data new_el = new Data(element); // this would be malloc and struct copy
*list = new_el; //move the address of list, so it begins at new element
}
In c++ there is reference mechanismm and you generally you can implement nearly anything with it. It basically makes usage of pointers at all obsolete it c++, at least in many, many cases. You also design objects and work on them, and everything is hidden under the hood those two.
There was also a nice question lately "Why do we use pointers in c++?" or something like that.
A simple example is an implementation of a matrix (it's an example, it's not the best way to implement matrices in C++).
int nrows = 10;
int ncols = 15;
double** M = new double*[nrows];
for(unsigned long int i = 0; i < nrows; ++i)
M[i] = new double[ncols];
M[3][7] = 3.1416;
You'll rarely see this construct in normal C++ code, since C++ has references. It's useful in C for "passing by reference:"
int allocate_something(void **p)
{
*p = malloc(whatever);
if (*p)
return 1;
else
return 0;
}
The equivalent C++ code would use void *&p for the parameter.
Still, you could imagine e.g. a resource monitor like this:
struct Resource;
struct Holder
{
Resource *res;
};
struct Monitor
{
Resource **res;
void monitor(const Holder &h) { res = &h.res; }
Resource& getResource() const { return **res; }
}
Yes, it's contrived, but the idea's there - it will keep a pointer to the pointer stored in a holder, and correctly return that resource even when the holder's res pointer changes.
Of course, it's a dangling dereference waiting to happen - normally, you'd avoid code like this.
This question already has answers here:
Need of Pointer to pointer
(5 answers)
Closed 9 years ago.
I am new to C/ C++.
I was going through some of the coding questions related to trees and came across this double pointer notation. Can we do the same things using single pointer as first argument in the below function as we can do with double pointers.
void operate(struct Node *root, struct Node **head_ref){ //do something}
There are two ways of interpreting a pointer; a reference to something, or an array. Considering this is a tree, this is probably the first: a reference to another pointer.
Every argument to a function in C is passed by value, which means that if you change the pointer inside the function, it won't be changed outside. To guarantee it is also changed outside, you can use a reference to the pointer: double pointers. You can consider the following example.
void function(int a) {
a = 5;
}
Even if a is changed above, it is not changed outside of the function. But in this other case,
void function(int * a) {
*a = 5;
}
the value a is changed outside the function as well. The same thought process can be applied to a pointer(which is also a value).
When you want a function to take care of malloc, free is the main reason.
This is useful if you want to encapsulate memory allocation.
For example some init(struct some_struct **), free(struct some_struct **).
And let functions take care of malloc, free. Instead of allocating on stack.
For example a function that packs a string of unknown length.
size_t pack_struct(char** data, const struct some_struct * some_struct)
{
/**
* #brief buffer
* #note verify the needed buffer length
*/
char buffer [256]; // temporary buffer
*data = 0;
//const char* package_pattern = "%cW ;%u.%u;%s%c";
size_t len = sprintf(buffer, weight_package_pattern,
START_CHARACTER,
some_struct->ts.tv_sec,
some_struct->ts.tv_usec,
some_struct->string_of_unknown_length, // but no more then buffer
STOP_CHARACTER);
if(len == 0) {
perror("sprintf failed!\n");
return len;
}
// len++; // for end character if wanna some, see sprintf description
*data = (char*)malloc(len*sizeof(char)); // memory allocation !
strncpy(*data, buffer, len);
return len;
}
However such technic should be avoided when programming in C++.
Double pointer is normally used when allocating memory.
#include <stdlib.h>
void new_malloc(void **p, size_t s) {
*p = malloc(s);
/* do something */
}
int main() {
int *p;
new_malloc((void **)&p, sizeof(int) * 10);
}
I have a void pointer called ptr. I want to increment this value by a number of bytes. Is there a way to do this?
Please note that I want to do this in-place without creating any more variables.
Could I do something like ptr = (void *)(++((char *) ptr)); ?
You cannot perform arithmetic on a void pointer because pointer arithmetic is defined in terms of the size of the pointed-to object.
You can, however, cast the pointer to a char*, do arithmetic on that pointer, and then convert it back to a void*:
void* p = /* get a pointer somehow */;
// In C++:
p = static_cast<char*>(p) + 1;
// In C:
p = (char*)p + 1;
No arithmeatic operations can be done on void pointer.
The compiler doesn't know the size of the item(s) the void pointer is pointing to. You can cast the pointer to (char *) to do so.
In gcc there is an extension which treats the size of a void as 1. so one can use arithematic on a void* to add an offset in bytes, but using it would yield non-portable code.
Just incrementing the void* does happen to work in gcc:
#include <stdlib.h>
#include <stdio.h>
int main() {
int i[] = { 23, 42 };
void* a = &i;
void* b = a + 4;
printf("%i\n", *((int*)b));
return 0;
}
It's conceptually (and officially) wrong though, so you want to make it explicit: cast it to char* and then back.
void* a = get_me_a_pointer();
void* b = (void*)((char*)a + some_number);
This makes it obvious that you're increasing by a number of bytes.
You can do:
++(*((char **)(&ptr)));
How can we access variables of a structure? I have a struct:
typedef struct {
unsigned short a;
unsigned shout b;
} Display;
and in my other class I have a method:
int NewMethod(Display **display)
{
Display *disp=new Display();
*display = disp;
disp->a=11;
}
What does **display mean? To access variables of struct I have used ->, are there other methods too?
As Taylor said, the double asterisk is "pointer to pointer", you can have as many levels of pointers as you need.
As I'm sure you know, the arrow operator (a->b) is a shortcut for the asterisk that dereferences a pointer, and the dot that accesses a field, i.e.
a->b = (*a).b;
The parentheses are necessary since the dot binds tighter. There is no such operator for double asterisks, you have to first de-reference to get to the required level, before accessing the fields:
Display **dpl = ...;
(*dpl)->a = 42;
or
(**dpl).a = 42;
Think of it as *(*display). When you want to pass the address of an integer to a function so that you can set the integer, you use:
void setTo7 (int *x) {
*x = 7;
}
: : :
int a = 4;
setTo7 (&a);
// a is now 7.
It's no different from what you have except that you want to set the value of a pointer so you need to pass the pointer to that pointer. Simple, no?
Try this out:
#include <stdio.h>
#include <string.h>
static void setTo7 (int *x) { *x = 7; }
void appendToStr (char **str, char *app) {
// Allocate enough space for bigger string and NUL.
char *newstr = malloc (strlen(*str) + strlen (app) + 1);
// Only copy/append if malloc worked.
if (newstr != 0) {
strcpy (newstr, *str);
strcat (newstr, app);
}
// Free old string.
free (*str);
// Set string to new string with the magic of double pointers.
*str = newstr;
}
int main (void) {
int i = 2;
char *s = malloc(6); strcpy (s, "Hello");
setTo7 (&i); appendToStr (&s, ", world");
printf ("%d [%s]\n",i,s);
return 0;
}
The output is:
7 [Hello, world]
This will safely append one string value to another, allocating enough space. Double pointers are often used in intelligent memory allocation functions, less so in C++ since you have a native string type, but it's still useful for other pointers.
**display is just a double pointer (a pointer to a pointer of type Display).
The ** means that its a pointer-to-a-pointer. Basically it points to another pointer that then points to something else, in your case a Display structure.
If you called the function with only the object you can access the members with the . operator.
int NewMethod(Display display)
{
Display disp = display;
disp.a=11;
}
But this way you are not modifying directly the Display display object but a local copy. Your code suggests that the changes to the object are needed outside of the function so your only option is the one you described (well, maybe passing the argument by refference but the syntax then would more or less the same (->)).
Since disp is a Pointer you have to use ->
If you just have a "normal" variable (i.e. on the stack)
Display d;
you can write d.a
A struct is the same as a class. The only difference (I am aware of) is that all members are public by default.
You can do (*disp).a=11;
it is called dereferencing