My actual question is it really possible to compare values contained in two void pointers, when you actually know that these values are the same type? For example int.
void compVoids(void *firstVal, void *secondVal){
if (firstVal < secondVal){
cout << "This will not make any sense as this will compare addresses, not values" << endl;
}
}
Actually I need to compare two void pointer values, while outside the function it is known that the type is int. I do not want to use comparison of int inside the function.
So this will not work for me as well: if (*(int*)firstVal > *(int*)secondVal)
Any suggestions?
Thank you very much for help!
In order to compare the data pointed to by a void*, you must know what the type is. If you know what the type is, there is no need for a void*. If you want to write a function that can be used for multiple types, you use templates:
template<typename T>
bool compare(const T& firstVal, const T& secondVal)
{
if (firstVal < secondVal)
{
// do something
}
return something;
}
To illustrate why attempting to compare void pointers blind is not feasible:
bool compare(void* firstVal, void* secondVal)
{
if (*firstVal < *secondVal) // ERROR: cannot dereference a void*
{
// do something
}
return something;
}
So, you need to know the size to compare, which means you either need to pass in a std::size_t parameter, or you need to know the type (and really, in order to pass in the std::size_t parameter, you have to know the type):
bool compare(void* firstVal, void* secondVal, std::size_t size)
{
if (0 > memcmp(firstVal, secondVal, size))
{
// do something
}
return something;
}
int a = 5;
int b = 6;
bool test = compare(&a, &b, sizeof(int)); // you know the type!
This was required in C as templates did not exist. C++ has templates, which make this type of function declaration unnecessary and inferior (templates allow for enforcement of type safety - void pointers do not, as I'll show below).
The problem comes in when you do something (silly) like this:
int a = 5;
short b = 6;
bool test = compare(&a, &b, sizeof(int)); // DOH! this will try to compare memory outside the bounds of the size of b
bool test = compare(&a, &b, sizeof(short)); // DOH! This will compare the first part of a with b. Endianess will be an issue.
As you can see, by doing this, you lose all type safety and have a whole host of other issues you have to deal with.
It is definitely possible, but since they are void pointers you must specify how much data is to be compared and how.
The memcmp function may be what you are looking for. It takes two void pointers and an argument for the number of bytes to be compared and returns a comparison. Some comparisons, however, are not contingent upon all of the data being equal. For example: comparing the direction of two vectors ignoring their length.
This question doesn't have a definite answer unless you specify how you want to compare the data.
You need to dereference them and cast, with
if (*(int*) firstVal < *(int*) secondVal)
Why do you not want to use the int comparison inside the function, if you know that the two values will be int and that you want to compare the int values that they're pointing to?
You mentioned a comparison function for comparing data on inserts; for a comparison function, I recommend this:
int
compareIntValues (void *first, void *second)
{
return (*(int*) first - *(int*) second);
}
It follows the convention of negative if the first is smaller, 0 if they're equal, positive if the first is larger. Simply call this function when you want to compare the int data.
yes. and in fact your code is correct if the type is unsigned int. casting int values to void pointer is often used even not recommended.
Also you could cast the pointers but you have to cast them directly to the int type:
if ((int)firstVal < (int)secondVal)
Note: no * at all.
You may have address model issues doing this though if you build 32 and 64 bits. Check the intptr_t type that you could use to avoid that.
if ((intptr_t)firstVal < (intptr_t)secondVal)
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.
I have such expressions in my code:
QByteArray idx0 = ...
unsigned short ushortIdx0;
if ( idx0.size() >= sizeof(ushortIdx0) ) {
// Do something
}
But I'm getting the warning:
warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
if ( idx0.size() >= sizeof(ushortIdx0) ) {
~~~~~~~~~~~~^~~~~~~~~~
Why size() of QByteArray is returned as int rather than unsigned int? How can I get rid of this warning safely?
Some folk feel that the introduction of unsigned types into C all those years ago was a bad idea. Such types found themselves introduced into C++, where they are deeply embedded in the C++ Standard Library and operator return types.
Yes, sizeof must, by the standard, return an unsigned type.
The Qt developers adopt the modern thinking that the unsigned types were a bad idea, and favour instead making the return type of size a signed type. Personally I find it idiosyncratic.
To solve, you could (i) live with the warning, (ii) switch it off for the duration of the function, or (iii) write something like
(std::size_t)idx0.size() >= sizeof(ushortIdx0)
at the expense of clarity.
Why size() of QByteArray is returned as int rather than unsigned int?
I literally have no idea why Qt chose a signed return for size(). However, there are good reasons to use a signed instead of an unsigned.
One infamous example where a unsigned size() fails miserably is this quite innocent looking loop:
for (int i = 0; i < some_container.size() - 1; ++i) {
do_somehting(some_container[i] , some_container[i+1] );
}
Its not too uncommon to make the loop body operate on two elements and in that case its seems to be a valid choice to iterate only till some_container.size() - 1.
However, if the container is empty some_container.size() - 1 will silently (unsigned overflow is well defined) turn into the biggest value for the unsigned type. Hence, instead of avoiding the out-of-bounds access it leads to the maximum out of bounds you can get.
Note that there are easy fixes for this problem, but if size() does return a signed value, then there is no issue that needs to be fixed in the first place.
Because in Qt containers (like: QByteArray, QVector, ...) there are functions which can return a negative number, like: indexOf, lastIndexOf, contains, ... and some can accept negative numbers, like: mid, ...; So to be class-compatible or even framework-compatibe, developers use a signed type (int).
You can use standard c++ casting:
if ( static_cast<size_t>(idx0.size()) >= sizeof(ushortIdx0) )
The reason why is a duplicate part of the question, but the solution to the type mismatch is a valid problem to solve. For the comparisons of the kind you're doing, it'd probably be useful to factor them out, as they have a certain reusable meaning:
template <typename T> bool fitsIn(const QByteArray &a) {
return static_cast<int>(sizeof(T)) <= a.size();
}
template <typename T> bool fitsIn(T, const QByteArray &a) {
return fitsIn<T>(a);
}
if (fitsIn(ushortIdx0, idx0)) ...
Hopefully you'll have just a few kinds of such comparisons, and it'd make most sense to DRY (do not repeat yourself) and instead of a copypasta of casts, use functions dedicated to the task - functions that also express the intent of the original comparison. It then becomes easy to centralize handling of any corner cases you might wish to handle, i.e. when sizeof(T) > INT_MAX.
Another approach would be to define a new type to wrap size_t and adapt it to the types you need to use it with:
class size_of {
size_t val;
template <typename T> static typename std::enable_if<std::is_signed<T>::value, size_t>::type fromSigned(T sVal) {
return (sVal > 0) ? static_cast<size_t>(sVal) : 0;
}
public:
template <typename T, typename U = std::enable_if<std::is_scalar<T>::value>::type>
size_of(const T&) : val(sizeof(T)) {}
size_of(const QByteArray &a) : val(fromSigned(a.size())) {}
...
bool operator>=(size_of o) const { return value >= o.value; }
};
if (size_of(idx0) >= size_of(ushortIdx0)) ...
This would conceptually extend sizeof and specialize it for comparison(s) and nothing else.
I am a bit confused. There are two ways to return an array from a method. The first suggests the following:
typedef int arrT[10];
arrT *func(int i);
However, how do I capture the return which is an int (*)[]?
Another way is through a reference or pointer:
int (*func(int i)[10];
or
int (&func(int i)[10];
The return types are either int (*)[] or int (&)[].
The trouble I am having is how I can assign a variable to accept the point and I continue to get errors such as:
can't convert int* to int (*)[]
Any idea what I am doing wrong or what is lacking in my knowledge?
If you want to return an array by value, put it in a structure.
The Standard committee already did that, and thus you can use std::array<int,10>.
std::array<int,10> func(int i);
std::array<int,10> x = func(77);
This makes it very straightforward to return by reference also:
std::array<int,10>& func2(int i);
std::array<int,10>& y = func2(5);
First, the information you give is incorrect.
You write,
“There are two ways to return an array from a method”
and then you give as examples of the ways
typedef int arrT[10];
arrT *func(int i);
and
int (*func(int i))[10];
(I’ve added the missing right parenthesis), where you say that this latter way, in contrast to the first, is an example of
“through a reference or pointer”
Well, these two declarations mean exactly the same, to wit:
typedef int A[10];
A* fp1( int i ) { return 0; }
int (*fp2( int i ))[10] { return 0; }
int main()
{
int (*p1)[10] = fp1( 100 );
int (*p2)[10] = fp2( 200 );
}
In both cases a pointer to the array is returned, and this pointer is typed as "pointer to array". Dereferencing that pointer yields the array itself, which decays to a pointer to itself again, but now typed as "pointer to item". It’s a pointer to the first item of the array. At the machine code level these two pointers are, in practice, exactly the same. Coming from a Pascal background that confused me for a long time, but the upshot is, since it’s generally impractical to carry the array size along in the type (which precludes dealing with arrays of different runtime sizes), most array handling code deals with the pointer-to-first-item instead of the pointer-to-the-whole-array.
I.e., normally such a low level C language like function would be declared as just
int* func()
return a pointer to the first item of an array of size established at run time.
Now, if you want to return an array by value then you have two choices:
Returning a fixed size array by value: put it in a struct.
The standard already provides a templated class that does this, std::array.
Returning a variable size array by value: use a class that deals with copying.
The standard already provides a templated class that does this, std::vector.
For example,
#include <vector>
using namespace std;
vector<int> foo() { return vector<int>( 10 ); }
int main()
{
vector<int> const v = foo();
// ...
}
This is the most general. Using std::array is more of an optimization for special cases. As a beginner, keep in mind Donald Knuth’s advice: “Premature optimization is the root of all evil.” I.e., just use std::vector unless there is a really really good reason to use std::array.
using arrT10 = int[10]; // Or you can use typedef if you want
arrT10 * func(int i)
{
arrT10 a10;
return &a10;
// int a[10];
// return a; // ERROR: can't convert int* to int (*)[]
}
This will give you a warning because func returns an address of a local variable so we should NEVER code like this but I'm sure this code can help you.
I declare the following array:
char* array [2] = { "One", "Two"};
I pass this array to a function. How can I find the length of this array in the function?
You can't find the length of an array after you pass it to a function without extra effort. You'll need to:
Use a container that stores the size, such as vector (recommended).
Pass the size along with it. This will probably require the least modification to your existing code and be the quickest fix.
Use a sentinel value, like C strings do1. This makes finding the length of the array a linear time operation and if you forget the sentinel value your program will likely crash. This is the worst way to do it for most situations.
Use templating to deduct the size of the array as you pass it. You can read about it here: How does this Array Size Template Work?
1 In case you were wondering, most people regret the fact that C strings work this way.
When you pass an array there is NOT an easy way to determine the size within the function.
You can either pass the array size as a parameter
or
use std::vector<std::string>
If you are feeling particularly adventurous you can use some advanced template techniques
In a nutshell it looks something like
template <typename T, size_t N>
void YourFunction( T (&array)[N] )
{
size_t myarraysize = N;
}
C is doing some trickery behind your back.
void foo(int array[]) {
/* ... */
}
void bar(int *array) {
/* ... */
}
Both of these are identical:
6.3.2.1.3: Except when it is the operand of the sizeof operator or the unary & operator,
or is a string literal used to initialize an array, an expression that has type
‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’
that points to the initial element of the array object and is not an lvalue. If
the array object has register storage class, the behavior is undefined.
As a result, you don't know, inside foo() or bar(), if you were
called with an array, a portion of an array, or a pointer to a single
integer:
int a[10];
int b[10];
int c;
foo(a);
foo(&b[1]);
foo(&c);
Some people like to write their functions like: void foo(int *array)
just to remind themselves that they weren't really passed an array,
but rather a pointer to an integer and there may or may not be more
integers elsewhere nearby. Some people like to write their functions
like: void foo(int array[]), to better remind themselves of what the
function expects to be passed to it.
Regardless of which way you like to do it, if you want to know how long
your array is, you've got a few options:
Pass along a length paramenter too. (Think int main(int argc, char
*argv)).
Design your array so every element is non-NULL, except the last
element. (Think char *s="almost a string"; or execve(2).)
Design your function so it takes some other descriptor of the
arguments. (Think printf("%s%i", "hello", 10); -- the string describes
the other arguments. printf(3) uses stdarg(3) argument handling, but
it could just as easily be an array.)
Getting the array-size from the pointer isn't possible. You could just terminate the array with a NULL-pointer. That way your function can search for the NULL-pointer to know the size, or simply just stop processing input once it hits the NULL...
If you mean how long are all the strings added togather.
int n=2;
int size=0;
char* array [n] = { "One", "Two"};
for (int i=0;i<n;++i)
size += strlen(array[i];
Added:
yes thats what im currently doing but i wanted to remove that extra
paramater. oh well –
Probably going to get a bad response for this, but you could always use the first pointer to store the size, as long as you don't deference it or mistake it for actually being a pointer.
char* array [] = { (char*)2,"One", "Two"};
long size=(long)array[0];
for(int i=1; i<= size;++i)
printf("%s",array[i]);
Or you could NULL terminate your array
char* array [] = { "One", "Two", (char*)0 };
for(int i=0;array[i]!=0;++i)
{
printf("%s",array[i]);
}
Use the new C++11 std::array
http://www.cplusplus.com/reference/stl/array/
the standard array has the size method your looking for
//Prints out a given array
template <typename T>
void print(T t)
{
for(int i = 0; i < t.size(); i++)
{
cout << t[i] << " ";
}
cout << endl;
}
I have an idea but it includes passing the size of the array. Is it possible to avoid this?
*Update
Thanks for all of the answers/ideas but this problem is getting way deeper than my snorkeler can handle. I wanted to rewrite my C++ code in C because it was horribly written and slow. I see now that I have an opportunity to make it even worse in C. I'll rewrite it from the ground up in Python(performance be damned). Thanks again
If you don't have ELEMENTS, it's
#define ELEMENTS(a) (sizeof(a)/sizeof(*a))
Then,
#define print_array(a, specifier) print_array_impl(a, specifier, ELEMENTS(a), sizeof(*a))
void print_array_impl(void* a, char* specifier, size_t asize, size_t elsize)
{
for(int i = 0; i < asize; i++)
{
// corrected based on comment -- unfortunately, not as general
if (strcmp(specifier, "%d") == 0)
printf(specifier, ((int*)a)[i]);
// else if ... // check other specifiers
printf(" ");
}
printf("\n");
}
Use like this
print_array(a, "%d") // if a is a int[]
and, a needs to be an array name, not a pointer (or else ELEMENTS won't work)
You cannot know what is the size of an array without passing the size of that array (except operating with sizeof in static arrays). This is because the a pointer to a block of memory will only point to the base of the block of memory, from which you can know where the array/block of memory starts, but as there is no end defined you cannot determine where it will end.
You either need to set your own length per array and preserve it, and use it with the array like as described:
You can make a new type like:
struct _my_array {
typename arr[MAX];
int n;
} my_array;
OR
struct _my_array {
typename *arr;
int n;
} my_array;
In this case you need to allocate the a block of memory dynamically with new or malloc , and when finished free the memory with delete or free (respectively).
Or you can simply pass the array number of elements through the function.
Another way is to use a special terminator value of your array type which if encountered will be determined as the end of the array. In this case you need not preserve the size. For example a string is '\0' terminated, so all the string functions know that when a '\0' character is encounter in the char array it will consider that the string has end.
UPDATE
Because this is a generic function and the array can be of any type, one thing which you can do is like this:
struct _my_generic_arr {
void *arr;
int n;
int type;
} my_generic_arr;
When populating this array you can use any type. To identify which type, pass an identified in the type component. Each unique value will determine which type does the arr pointer actually points to (was actually the intended type to be pointed). The n will define the length. Now, depending on different values of type make a switch - case or an if - else ladder or nest, and process the array as you need.
It is impossible in c to track the size of an array in other block,,
It would be a better option to pass the size of the array along..
The other option would be to declare a global variable that has the size and using that variable inside the function
Eg,,
int size=<some value>
void main()
{
int arr[<same value>];
}
void print(T t)
{
for(int i = 0; i < size; i++)
{
printf("%d ",t[i]) //assuming T as int
}
printf("\n");
}
In C, you would need to pass two additional parameters: the size of the array (as you mentioned), and some way of indicating how to convert t[i] into a string. To convert t[i] to a string, you could create a custom switch statement to decode possible types, pass a pointer to a function that will return the string pointer, or you could pass the printf format specifier (e.g. "%d" for integer).
The problem is larger than you think. If you have an array of size 12, how do you know what data is held in that array? It could be 3 char*'s (on 32 bit system), 3 int32_t's, or even 12 chars. You have no way of knowing how to interpret the data. The best you could do is to implement your own version of a v-table and putting a print or toString function into it.
typedef struct {
void *array;
size_t length;
int element_width;
printer_t to_string;
} container;
printer_t is a type that describes a function pointer that takes an element pointer and returns a string (or prints it, if you don't want to free the string). This is almost never worth doing in C. That doesn't mean it can't be done. I would emphasize, though, that none of this is intended to imply that it should be done.
The function itself would look something like this:
void print(container *thing)
{
size_t offset;
int width;
char *stringified;
width = thing->element_width;
for (offset = 0; offset * width < thing->length; offset += width)
{
stringified = thing->to_string(thing->array + offset);
printf("%s ", stringified);
free(stringified);
}
}
What this does is essentially turn a struct into a faux class with a function pointer for a method. You could be more object-oriented and put the method in the type being printed and make it an array of those instead. Either way, it's not a good idea. C is for writing C. If you try to write in a different language, you'll end up with all sorts of terrible stuff like this.