freeing the same memory with distructors - c++

I try to free the memory correctly after the program ends, but I always encounter a problem.
In my code I want to have an array of all the numbers that I allow in my program, and have objects A and B (or more) that each one have some of the numbers that I allowed.
In the end I want to delete 'a' and 'b' only after "ints" getting out of the scope. But A and B calls their distructors to delete some of ints variables.
#define MAX_LEN 255
class IntArray
{
public:
int len;
void add(int* n) {
arr[len] = n; len++;
}
IntArray() : arr(new int* [MAX_LEN]), len(0) {}
~IntArray() {
for (int i = 0; i < len; i++)
delete arr[i];
delete[] arr;
}
private:
int** arr;
};
class Object
{
public:
void add(int* n) {
myIntArr.add(n);
}
private:
IntArray myIntArr;
};
int main(void)
{
int* a = new int(5);
int* b = new int(6);
IntArray ints;
ints.add(a);
ints.add(b);
Object A;
A.add(a);
Object B;
B.add(b);
return 0;
}

If you want to share dynamically allocated ints between multiple objects, use std::shared_ptr<int>.
Also, rather than writing a dynamic array type yourself, use std::vector to do it (correctly) for you.
using int_ptr = std::shared_ptr<int>;
class IntArray
{
public:
void add(int_ptr n) {
arr.push_back(n);
}
private:
std::vector<int_ptr> arr
};
class Object
{
public:
void add(int_ptr n) {
myIntArr.add(n);
}
private:
IntArray myIntArr;
};
int main(void)
{
int_ptr a = std::make_shared<int>(5);
int_ptr b = std::make_shared<int>(6);
IntArray ints;
ints.add(a);
ints.add(b);
Object A;
A.add(a);
Object B;
B.add(b);
return 0;
}
If you just want to have a copyable array of int, use std::vector<int>.

You're deleting a and b twice.
You should only delete something returned by new and exactly once.
But you add them both to IntArray ints; and then one each to Objects A and B and their destructors delete them also. Destructors are called in reverse order to it's when ints is destructed you'll be deleting them again - that's "Undefined Behaviour" but normally a catastrophic failure (crash) either immediately or later during executon.
The shortest fix is:
int* a = new int(5);
int* b = new int(6);
int *ac = new int(*a);//copy of *a
int *ab = new int(*b);//copy of *b
IntArray ints;
ints.add(a);
ints.add(b);
Object A;
A.add(ac);
Object B;
B.add(bc);
But it's not clear what our intention is. IntArray isn't an array of int as it stands, it's an array of pointers to int (which have been allocated by new).
My 'fix' will mean if you modify a (e.g. *a=20) you won't modify the copy (ac) added to the Object A.

Related

return an array from a function for later use [duplicate]

How can I return an array from a method, and how must I declare it?
int[] test(void); // ??
int* test();
but it would be "more C++" to use vectors:
std::vector< int > test();
EDIT
I'll clarify some point. Since you mentioned C++, I'll go with new[] and delete[] operators, but it's the same with malloc/free.
In the first case, you'll write something like:
int* test() {
return new int[size_needed];
}
but it's not a nice idea because your function's client doesn't really know the size of the array you are returning, although the client can safely deallocate it with a call to delete[].
int* theArray = test();
for (size_t i; i < ???; ++i) { // I don't know what is the array size!
// ...
}
delete[] theArray; // ok.
A better signature would be this one:
int* test(size_t& arraySize) {
array_size = 10;
return new int[array_size];
}
And your client code would now be:
size_t theSize = 0;
int* theArray = test(theSize);
for (size_t i; i < theSize; ++i) { // now I can safely iterate the array
// ...
}
delete[] theArray; // still ok.
Since this is C++, std::vector<T> is a widely-used solution:
std::vector<int> test() {
std::vector<int> vector(10);
return vector;
}
Now you don't have to call delete[], since it will be handled by the object, and you can safely iterate it with:
std::vector<int> v = test();
std::vector<int>::iterator it = v.begin();
for (; it != v.end(); ++it) {
// do your things
}
which is easier and safer.
how can i return a array in a c++ method and how must i declare it? int[] test(void); ??
This sounds like a simple question, but in C++ you have quite a few options. Firstly, you should prefer...
std::vector<>, which grows dynamically to however many elements you encounter at runtime, or
std::array<> (introduced with C++11), which always stores a number of elements specified at compile time,
...as they manage memory for you, ensuring correct behaviour and simplifying things considerably:
std::vector<int> fn()
{
std::vector<int> x;
x.push_back(10);
return x;
}
std::array<int, 2> fn2() // C++11
{
return {3, 4};
}
void caller()
{
std::vector<int> a = fn();
const std::vector<int>& b = fn(); // extend lifetime but read-only
// b valid until scope exit/return
std::array<int, 2> c = fn2();
const std::array<int, 2>& d = fn2();
}
The practice of creating a const reference to the returned data can sometimes avoid a copy, but normally you can just rely on Return Value Optimisation, or - for vector but not array - move semantics (introduced with C++11).
If you really want to use an inbuilt array (as distinct from the Standard library class called array mentioned above), one way is for the caller to reserve space and tell the function to use it:
void fn(int x[], int n)
{
for (int i = 0; i < n; ++i)
x[i] = n;
}
void caller()
{
// local space on the stack - destroyed when caller() returns
int x[10];
fn(x, sizeof x / sizeof x[0]);
// or, use the heap, lives until delete[](p) called...
int* p = new int[10];
fn(p, 10);
}
Another option is to wrap the array in a structure, which - unlike raw arrays - are legal to return by value from a function:
struct X
{
int x[10];
};
X fn()
{
X x;
x.x[0] = 10;
// ...
return x;
}
void caller()
{
X x = fn();
}
Starting with the above, if you're stuck using C++03 you might want to generalise it into something closer to the C++11 std::array:
template <typename T, size_t N>
struct array
{
T& operator[](size_t n) { return x[n]; }
const T& operator[](size_t n) const { return x[n]; }
size_t size() const { return N; }
// iterators, constructors etc....
private:
T x[N];
};
Another option is to have the called function allocate memory on the heap:
int* fn()
{
int* p = new int[2];
p[0] = 0;
p[1] = 1;
return p;
}
void caller()
{
int* p = fn();
// use p...
delete[] p;
}
To help simplify the management of heap objects, many C++ programmers use "smart pointers" that ensure deletion when the pointer(s) to the object leave their scopes. With C++11:
std::shared_ptr<int> p(new int[2], [](int* p) { delete[] p; } );
std::unique_ptr<int[]> p(new int[3]);
If you're stuck on C++03, the best option is to see if the boost library is available on your machine: it provides boost::shared_array.
Yet another option is to have some static memory reserved by fn(), though this is NOT THREAD SAFE, and means each call to fn() overwrites the data seen by anyone keeping pointers from previous calls. That said, it can be convenient (and fast) for simple single-threaded code.
int* fn(int n)
{
static int x[2]; // clobbered by each call to fn()
x[0] = n;
x[1] = n + 1;
return x; // every call to fn() returns a pointer to the same static x memory
}
void caller()
{
int* p = fn(3);
// use p, hoping no other thread calls fn() meanwhile and clobbers the values...
// no clean up necessary...
}
It is not possible to return an array from a C++ function. 8.3.5[dcl.fct]/6:
Functions shall not have a return type of type array or function[...]
Most commonly chosen alternatives are to return a value of class type where that class contains an array, e.g.
struct ArrayHolder
{
int array[10];
};
ArrayHolder test();
Or to return a pointer to the first element of a statically or dynamically allocated array, the documentation must indicate to the user whether he needs to (and if so how he should) deallocate the array that the returned pointer points to.
E.g.
int* test2()
{
return new int[10];
}
int* test3()
{
static int array[10];
return array;
}
While it is possible to return a reference or a pointer to an array, it's exceedingly rare as it is a more complex syntax with no practical advantage over any of the above methods.
int (&test4())[10]
{
static int array[10];
return array;
}
int (*test5())[10]
{
static int array[10];
return &array;
}
Well if you want to return your array from a function you must make sure that the values are not stored on the stack as they will be gone when you leave the function.
So either make your array static or allocate the memory (or pass it in but your initial attempt is with a void parameter). For your method I would define it like this:
int *gnabber(){
static int foo[] = {1,2,3}
return foo;
}
"how can i return a array in a c++ method and how must i declare it?
int[] test(void); ??"
template <class X>
class Array
{
X *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(X* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new X [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(X));
return m_size;
}
return 0;
}
};
just for int
class IntArray
{
int *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(int* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new int [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(int));
return m_size;
}
return 0;
}
};
example
Array<float> array;
float *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;
example for int
IntArray array;
int *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;

Destructor call in a class constructor with a vector

I have the following class which simply wraps an array and adds with the constructor some elements to it:
class myArray {
public:
myArray();
myArray(int a, int b);
myArray(int a, int b, int c);
myArray(myArray&& emplace);
~myArray();
int& operator [](int id);
private:
int *data;
};
myArray::myArray() {
data = new int[1];
data[0] = 0;
}
myArray::myArray(int a, int b) {
data = new int[3];
data[0] = 0;
data[1] = a;
data[2] = b;
}
myArray::myArray(int a, int b, int c) {
data = new int[4];
data[0] = 0;
data[1] = a;
data[2] = b;
data[3] = c;
}
myArray::~myArray() {
std::cout << "Destructor"<<std::endl;
delete[] data;
}
int& myArray::operator [](int id) {
return data[id];
}
myArray::myArray(myArray&& emplace) : data(std::move(emplace.data)) {}
Furthermore I have a second class which contains a vector of elements of the first class (myArray).
class Queue {
public:
Queue();
private:
std::vector<myArray> _queue;
};
Queue::Queue {
_queue.reserve(1000);
for(int a = 0; a < 10; a++)
for(int b = 0; b < 10; b++)
for(int c = 0; c < 10; c++)
_queue.emplace_back(a,b,c);
}
My question here is: Why is the destructor called for the myArray elements at the end of the Queue constructor? The Queue object is still alive in my main program but the destructor of myArray frees the allocated memory and I consequently get a segmentation fault.
Is there a way to avoid the call of the destructor or rather call it not until at the end of the Queue objects lifetime?
Your move constructor doesn't set data to null on the moved from object so when the moved from object is destructed it will try to free data.
If you have c++14 you can use std::exchange to implement this:
myArray::myArray(myArray&& emplace)
: data{std::exchange(emplace.data, nullptr)})
{}
Otherwise you need to do:
myArray::myArray(myArray&& emplace)
: data{emplace.data)
{
emplace.data = nullptr;
}
The move constructor will be invoked by std::vector as it reallocates to increase its capacity when you call emplace_back. Something like the following steps are performed:
Allocate new memory to hold the elements
Move construct using placement new elements in the new memory from the elements in the previous memory
Destruct the elements in the previous memory
Deallocate the previous memory

How to delete dynamic allocated array in c++?

I got stuck with deleting an dynamically allocated array of int.
I've got a destructor, where I'm trying to use a loop for to delete all elements of array and finally delete it.
I have code on http://rextester.com/OTPPRQ8349
Thanks!
class MyClass
{
public:
int _a;
int* c;
int fRozmiar;
static int fIlosc;
MyClass() //default constructor
{
_a=0;
c = new int [9];
for(int i = 0; i<=9; i++)
{
c[i] = 1;
}
fIlosc++;
}
MyClass(int a1, int c1) // parametrized constructor
{
_a=a1;
c = new int [c1];
for(int i = 0; i<=c1; i++)
{
c[i] = rand();
}
fIlosc++;
}
MyClass(const MyClass &p2) // copy constructor
{
_a =p2._a;
c = p2.c;
fRozmiar = p2.fRozmiar;
fIlosc = fIlosc;
fIlosc++;
}
~MyClass(); // destructor
static int getCount() {
return fIlosc;
}
};
//Initialize static member of class
int MyClass::fIlosc = 0;
MyClass::~MyClass()
{
for(int i = 0; i<sizeof(c); ++i)
{
delete[] c[i];
}
delete[] c;
fIlosc--;
}
int main()
{
}
Remove the for-loop, but keep the delete[] c after it.
Each int doesn't need to be deleted because they're not dynamically allocated. If you needed to delete them, then the for-loop wouldn't work becuase: sizeof(c) is not the size of the array, and delete[] should have been delete instead.
There are several problems in the code.
First, that loop in the destructor must go. If you didn’t new it, don’t delete it.
Second, a loop through an array of N elements should be for (int i = 0; i < N; ++i). Note that the test is i < N, not i <= N. The loops as currently written go off the end of the array. That’s not good.
Third, the copy constructor copies the pointer. When the first object goes out of scope its destructor deletes the array; when the copy goes out of scope its destructor also deletes the array. Again, not good. The copy constructor has to make a copy of the array. In order to do that the class needs to also store the number of elements the array.

How to return a pointer to an array for use in a separate function [duplicate]

How can I return an array from a method, and how must I declare it?
int[] test(void); // ??
int* test();
but it would be "more C++" to use vectors:
std::vector< int > test();
EDIT
I'll clarify some point. Since you mentioned C++, I'll go with new[] and delete[] operators, but it's the same with malloc/free.
In the first case, you'll write something like:
int* test() {
return new int[size_needed];
}
but it's not a nice idea because your function's client doesn't really know the size of the array you are returning, although the client can safely deallocate it with a call to delete[].
int* theArray = test();
for (size_t i; i < ???; ++i) { // I don't know what is the array size!
// ...
}
delete[] theArray; // ok.
A better signature would be this one:
int* test(size_t& arraySize) {
array_size = 10;
return new int[array_size];
}
And your client code would now be:
size_t theSize = 0;
int* theArray = test(theSize);
for (size_t i; i < theSize; ++i) { // now I can safely iterate the array
// ...
}
delete[] theArray; // still ok.
Since this is C++, std::vector<T> is a widely-used solution:
std::vector<int> test() {
std::vector<int> vector(10);
return vector;
}
Now you don't have to call delete[], since it will be handled by the object, and you can safely iterate it with:
std::vector<int> v = test();
std::vector<int>::iterator it = v.begin();
for (; it != v.end(); ++it) {
// do your things
}
which is easier and safer.
how can i return a array in a c++ method and how must i declare it? int[] test(void); ??
This sounds like a simple question, but in C++ you have quite a few options. Firstly, you should prefer...
std::vector<>, which grows dynamically to however many elements you encounter at runtime, or
std::array<> (introduced with C++11), which always stores a number of elements specified at compile time,
...as they manage memory for you, ensuring correct behaviour and simplifying things considerably:
std::vector<int> fn()
{
std::vector<int> x;
x.push_back(10);
return x;
}
std::array<int, 2> fn2() // C++11
{
return {3, 4};
}
void caller()
{
std::vector<int> a = fn();
const std::vector<int>& b = fn(); // extend lifetime but read-only
// b valid until scope exit/return
std::array<int, 2> c = fn2();
const std::array<int, 2>& d = fn2();
}
The practice of creating a const reference to the returned data can sometimes avoid a copy, but normally you can just rely on Return Value Optimisation, or - for vector but not array - move semantics (introduced with C++11).
If you really want to use an inbuilt array (as distinct from the Standard library class called array mentioned above), one way is for the caller to reserve space and tell the function to use it:
void fn(int x[], int n)
{
for (int i = 0; i < n; ++i)
x[i] = n;
}
void caller()
{
// local space on the stack - destroyed when caller() returns
int x[10];
fn(x, sizeof x / sizeof x[0]);
// or, use the heap, lives until delete[](p) called...
int* p = new int[10];
fn(p, 10);
}
Another option is to wrap the array in a structure, which - unlike raw arrays - are legal to return by value from a function:
struct X
{
int x[10];
};
X fn()
{
X x;
x.x[0] = 10;
// ...
return x;
}
void caller()
{
X x = fn();
}
Starting with the above, if you're stuck using C++03 you might want to generalise it into something closer to the C++11 std::array:
template <typename T, size_t N>
struct array
{
T& operator[](size_t n) { return x[n]; }
const T& operator[](size_t n) const { return x[n]; }
size_t size() const { return N; }
// iterators, constructors etc....
private:
T x[N];
};
Another option is to have the called function allocate memory on the heap:
int* fn()
{
int* p = new int[2];
p[0] = 0;
p[1] = 1;
return p;
}
void caller()
{
int* p = fn();
// use p...
delete[] p;
}
To help simplify the management of heap objects, many C++ programmers use "smart pointers" that ensure deletion when the pointer(s) to the object leave their scopes. With C++11:
std::shared_ptr<int> p(new int[2], [](int* p) { delete[] p; } );
std::unique_ptr<int[]> p(new int[3]);
If you're stuck on C++03, the best option is to see if the boost library is available on your machine: it provides boost::shared_array.
Yet another option is to have some static memory reserved by fn(), though this is NOT THREAD SAFE, and means each call to fn() overwrites the data seen by anyone keeping pointers from previous calls. That said, it can be convenient (and fast) for simple single-threaded code.
int* fn(int n)
{
static int x[2]; // clobbered by each call to fn()
x[0] = n;
x[1] = n + 1;
return x; // every call to fn() returns a pointer to the same static x memory
}
void caller()
{
int* p = fn(3);
// use p, hoping no other thread calls fn() meanwhile and clobbers the values...
// no clean up necessary...
}
It is not possible to return an array from a C++ function. 8.3.5[dcl.fct]/6:
Functions shall not have a return type of type array or function[...]
Most commonly chosen alternatives are to return a value of class type where that class contains an array, e.g.
struct ArrayHolder
{
int array[10];
};
ArrayHolder test();
Or to return a pointer to the first element of a statically or dynamically allocated array, the documentation must indicate to the user whether he needs to (and if so how he should) deallocate the array that the returned pointer points to.
E.g.
int* test2()
{
return new int[10];
}
int* test3()
{
static int array[10];
return array;
}
While it is possible to return a reference or a pointer to an array, it's exceedingly rare as it is a more complex syntax with no practical advantage over any of the above methods.
int (&test4())[10]
{
static int array[10];
return array;
}
int (*test5())[10]
{
static int array[10];
return &array;
}
Well if you want to return your array from a function you must make sure that the values are not stored on the stack as they will be gone when you leave the function.
So either make your array static or allocate the memory (or pass it in but your initial attempt is with a void parameter). For your method I would define it like this:
int *gnabber(){
static int foo[] = {1,2,3}
return foo;
}
"how can i return a array in a c++ method and how must i declare it?
int[] test(void); ??"
template <class X>
class Array
{
X *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(X* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new X [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(X));
return m_size;
}
return 0;
}
};
just for int
class IntArray
{
int *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(int* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new int [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(int));
return m_size;
}
return 0;
}
};
example
Array<float> array;
float *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;
example for int
IntArray array;
int *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;

How can I return this 2d array? [duplicate]

How can I return an array from a method, and how must I declare it?
int[] test(void); // ??
int* test();
but it would be "more C++" to use vectors:
std::vector< int > test();
EDIT
I'll clarify some point. Since you mentioned C++, I'll go with new[] and delete[] operators, but it's the same with malloc/free.
In the first case, you'll write something like:
int* test() {
return new int[size_needed];
}
but it's not a nice idea because your function's client doesn't really know the size of the array you are returning, although the client can safely deallocate it with a call to delete[].
int* theArray = test();
for (size_t i; i < ???; ++i) { // I don't know what is the array size!
// ...
}
delete[] theArray; // ok.
A better signature would be this one:
int* test(size_t& arraySize) {
array_size = 10;
return new int[array_size];
}
And your client code would now be:
size_t theSize = 0;
int* theArray = test(theSize);
for (size_t i; i < theSize; ++i) { // now I can safely iterate the array
// ...
}
delete[] theArray; // still ok.
Since this is C++, std::vector<T> is a widely-used solution:
std::vector<int> test() {
std::vector<int> vector(10);
return vector;
}
Now you don't have to call delete[], since it will be handled by the object, and you can safely iterate it with:
std::vector<int> v = test();
std::vector<int>::iterator it = v.begin();
for (; it != v.end(); ++it) {
// do your things
}
which is easier and safer.
how can i return a array in a c++ method and how must i declare it? int[] test(void); ??
This sounds like a simple question, but in C++ you have quite a few options. Firstly, you should prefer...
std::vector<>, which grows dynamically to however many elements you encounter at runtime, or
std::array<> (introduced with C++11), which always stores a number of elements specified at compile time,
...as they manage memory for you, ensuring correct behaviour and simplifying things considerably:
std::vector<int> fn()
{
std::vector<int> x;
x.push_back(10);
return x;
}
std::array<int, 2> fn2() // C++11
{
return {3, 4};
}
void caller()
{
std::vector<int> a = fn();
const std::vector<int>& b = fn(); // extend lifetime but read-only
// b valid until scope exit/return
std::array<int, 2> c = fn2();
const std::array<int, 2>& d = fn2();
}
The practice of creating a const reference to the returned data can sometimes avoid a copy, but normally you can just rely on Return Value Optimisation, or - for vector but not array - move semantics (introduced with C++11).
If you really want to use an inbuilt array (as distinct from the Standard library class called array mentioned above), one way is for the caller to reserve space and tell the function to use it:
void fn(int x[], int n)
{
for (int i = 0; i < n; ++i)
x[i] = n;
}
void caller()
{
// local space on the stack - destroyed when caller() returns
int x[10];
fn(x, sizeof x / sizeof x[0]);
// or, use the heap, lives until delete[](p) called...
int* p = new int[10];
fn(p, 10);
}
Another option is to wrap the array in a structure, which - unlike raw arrays - are legal to return by value from a function:
struct X
{
int x[10];
};
X fn()
{
X x;
x.x[0] = 10;
// ...
return x;
}
void caller()
{
X x = fn();
}
Starting with the above, if you're stuck using C++03 you might want to generalise it into something closer to the C++11 std::array:
template <typename T, size_t N>
struct array
{
T& operator[](size_t n) { return x[n]; }
const T& operator[](size_t n) const { return x[n]; }
size_t size() const { return N; }
// iterators, constructors etc....
private:
T x[N];
};
Another option is to have the called function allocate memory on the heap:
int* fn()
{
int* p = new int[2];
p[0] = 0;
p[1] = 1;
return p;
}
void caller()
{
int* p = fn();
// use p...
delete[] p;
}
To help simplify the management of heap objects, many C++ programmers use "smart pointers" that ensure deletion when the pointer(s) to the object leave their scopes. With C++11:
std::shared_ptr<int> p(new int[2], [](int* p) { delete[] p; } );
std::unique_ptr<int[]> p(new int[3]);
If you're stuck on C++03, the best option is to see if the boost library is available on your machine: it provides boost::shared_array.
Yet another option is to have some static memory reserved by fn(), though this is NOT THREAD SAFE, and means each call to fn() overwrites the data seen by anyone keeping pointers from previous calls. That said, it can be convenient (and fast) for simple single-threaded code.
int* fn(int n)
{
static int x[2]; // clobbered by each call to fn()
x[0] = n;
x[1] = n + 1;
return x; // every call to fn() returns a pointer to the same static x memory
}
void caller()
{
int* p = fn(3);
// use p, hoping no other thread calls fn() meanwhile and clobbers the values...
// no clean up necessary...
}
It is not possible to return an array from a C++ function. 8.3.5[dcl.fct]/6:
Functions shall not have a return type of type array or function[...]
Most commonly chosen alternatives are to return a value of class type where that class contains an array, e.g.
struct ArrayHolder
{
int array[10];
};
ArrayHolder test();
Or to return a pointer to the first element of a statically or dynamically allocated array, the documentation must indicate to the user whether he needs to (and if so how he should) deallocate the array that the returned pointer points to.
E.g.
int* test2()
{
return new int[10];
}
int* test3()
{
static int array[10];
return array;
}
While it is possible to return a reference or a pointer to an array, it's exceedingly rare as it is a more complex syntax with no practical advantage over any of the above methods.
int (&test4())[10]
{
static int array[10];
return array;
}
int (*test5())[10]
{
static int array[10];
return &array;
}
Well if you want to return your array from a function you must make sure that the values are not stored on the stack as they will be gone when you leave the function.
So either make your array static or allocate the memory (or pass it in but your initial attempt is with a void parameter). For your method I would define it like this:
int *gnabber(){
static int foo[] = {1,2,3}
return foo;
}
"how can i return a array in a c++ method and how must i declare it?
int[] test(void); ??"
template <class X>
class Array
{
X *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(X* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new X [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(X));
return m_size;
}
return 0;
}
};
just for int
class IntArray
{
int *m_data;
int m_size;
public:
// there constructor, destructor, some methods
int Get(int* &_null_pointer)
{
if(!_null_pointer)
{
_null_pointer = new int [m_size];
memcpy(_null_pointer, m_data, m_size * sizeof(int));
return m_size;
}
return 0;
}
};
example
Array<float> array;
float *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;
example for int
IntArray array;
int *n_data = NULL;
int data_size;
if(data_size = array.Get(n_data))
{ // work with array }
delete [] n_data;