Related
I need to declare a 2D array as the member variable of a class. I can't use STL (so, no vector), and I've been asked to avoid double/triple pointers. I want to be able to reference elements in this 2D array using subscripting notation, e.g. arr[0][0]. The array also must be declared on the heap due to its size.
Due to the requirements I have, none of the existing answers on StackOverflow meet my needs.
What I've been trying to do is:
class MyClass {
public:
MyClass() : arr(new int[1000][2]) {}
// other stuff here
private:
int arr[1000][2];
};
The error I get after compiling that class is "cannot initialize a parameter of type int * with an lvalue of type int[1000][2]". Clearly, I can solve this by using pointer syntax, but, as mentioned above, I've been asked to use "array syntax" for code clarity. I was hoping someone with a better understanding of C++ could explain how to use "array syntax".
Of course you can do this without double/triple pointers. You can even do this without use of any pointers in the class declaration. But first lets look at the more common approach. A 2D array is a simple extension of a 1D array.
Starting off with the standard way this is done for a 1D array of 1000 ints w/o using vector. The pointer, arr, is on the stack but points to an array of 1000 ints on the heap.
class MyClass {
public:
MyClass() : arr(new int[1000]) {}
private:
int *arr;
};
Elements are accessed the usual way. For instance arr[0]=42;
Extending this to a 2D array in the heap is a simple extension of the above.
You need to declare the member variable as a pointer to a 1D array instead of the basic type.
class MyClass {
public:
MyClass() : arr(new int[1000][2]) {}
private:
int (*arr)[2];
};
Similarly, you can refer to elements of the 2D array the usual way: arr[0][0]=42;
Finally, there is the approach that completely eliminates pointers except the one required for the new. Here we initialize a reference. The trick is to add a third level to new, the [1] so that the *new returns an object that is the actual 2D int array. Structurally, it is no different than what the pointer version above does but lets us directly initialize a reference to a 2D int array. It's certainly not a common idiom so I'd stick with the ptr approach.
class MyClass {
public:
MyClass() : arr(*new int[1][1000][2]) {}
~MyClass() {delete[] arr;}
//private: // to test
int(&arr)[1000][2];
};
int main()
{
MyClass obj;
obj.arr[2][1] = 42;
}
When your class has an array in it, and you use new to create a new instance of that class, that array is on the heap. You can still access the array with arr[i][j].
Why not do something like this?
class myClass {
public:
int arr[1000][2];
};
int main() {
myClass* test = new myClass;
for (int i = 0; i < 1000; i++) {
for (int j = 0; j < 2; j++) {
test->arr[i][j] = 5;
}
}
}
You can use 2 classes to achieve this.
class BaseArray {
public:
int& operator[](int x) { return this->arr[x]; }
int operator[](int index) const { return this->arr[index]; }
int arr[2];
};
class myClass {
public:
myClass() {}
~myClass() {}
BaseArray& operator[](int index) { return this->arr[index]; }
BaseArray operator[](int index) const { return this->arr[index]; }
BaseArray arr[1000];
};
Optionally use can use templates to make this class more dynamic.
template<class TYPE, int arraySize>
class BaseArray {
public:
TYPE& operator[](int x) { return this->arr[x]; }
TYPE operator[](int index) const { return this->arr[index]; }
TYPE arr[arraySize];
};
template<class TYPE, int dim1, int dim2>
class myClass {
public:
myClass() {}
~myClass() {}
BaseArray<TYPE, dim2>& operator[](int index) { return this->arr[index]; }
BaseArray<TYPE, dim2> operator[](int index) const { return this->arr[index]; }
BaseArray<TYPE, dim2> arr[dim1];
};
int main()
{
myClass<int, 1000, 2> myArray;
}
EDIT
When you provide the array dimentions int arr[1000][2]; the variable will automatically be allocated in the stack. If the array needs to be fully dynamic, you can just use a double pointer int** arr = { nullptr }; and initialize it at the constructor as shown below.
class myClass {
public:
myClass()
{
arr = new int* [1000];
for (int i = 0; i < 1000; ++i)
arr[i] = new int[2];
}
~myClass()
{
/* Make sure to delete or else it might flag a memory error. */
for (int i = 0; i < 1000; ++i)
delete[] arr[i];
delete[] arr;
}
int** arr = { nullptr };
};
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 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 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;
I have the following class in C++:
class a {
const int b[2];
// other stuff follows
// and here's the constructor
a(void);
}
The question is, how do I initialize b in the initialization list, given that I can't initialize it inside the body of the function of the constructor, because b is const?
This doesn't work:
a::a(void) :
b([2,3])
{
// other initialization stuff
}
Edit: The case in point is when I can have different values for b for different instances, but the values are known to be constant for the lifetime of the instance.
With C++11 the answer to this question has now changed and you can in fact do:
struct a {
const int b[2];
// other bits follow
// and here's the constructor
a();
};
a::a() :
b{2,3}
{
// other constructor work
}
int main() {
a a;
}
Like the others said, ISO C++ doesn't support that. But you can workaround it. Just use std::vector instead.
int* a = new int[N];
// fill a
class C {
const std::vector<int> v;
public:
C():v(a, a+N) {}
};
It is not possible in the current standard. I believe you'll be able to do this in C++0x using initializer lists (see A Brief Look at C++0x, by Bjarne Stroustrup, for more information about initializer lists and other nice C++0x features).
std::vector uses the heap. Geez, what a waste that would be just for the sake of a const sanity-check. The point of std::vector is dynamic growth at run-time, not any old syntax checking that should be done at compile-time. If you're not going to grow then create a class to wrap a normal array.
#include <stdio.h>
template <class Type, size_t MaxLength>
class ConstFixedSizeArrayFiller {
private:
size_t length;
public:
ConstFixedSizeArrayFiller() : length(0) {
}
virtual ~ConstFixedSizeArrayFiller() {
}
virtual void Fill(Type *array) = 0;
protected:
void add_element(Type *array, const Type & element)
{
if(length >= MaxLength) {
// todo: throw more appropriate out-of-bounds exception
throw 0;
}
array[length] = element;
length++;
}
};
template <class Type, size_t Length>
class ConstFixedSizeArray {
private:
Type array[Length];
public:
explicit ConstFixedSizeArray(
ConstFixedSizeArrayFiller<Type, Length> & filler
) {
filler.Fill(array);
}
const Type *Array() const {
return array;
}
size_t ArrayLength() const {
return Length;
}
};
class a {
private:
class b_filler : public ConstFixedSizeArrayFiller<int, 2> {
public:
virtual ~b_filler() {
}
virtual void Fill(int *array) {
add_element(array, 87);
add_element(array, 96);
}
};
const ConstFixedSizeArray<int, 2> b;
public:
a(void) : b(b_filler()) {
}
void print_items() {
size_t i;
for(i = 0; i < b.ArrayLength(); i++)
{
printf("%d\n", b.Array()[i]);
}
}
};
int main()
{
a x;
x.print_items();
return 0;
}
ConstFixedSizeArrayFiller and ConstFixedSizeArray are reusable.
The first allows run-time bounds checking while initializing the array (same as a vector might), which can later become const after this initialization.
The second allows the array to be allocated inside another object, which could be on the heap or simply the stack if that's where the object is. There's no waste of time allocating from the heap. It also performs compile-time const checking on the array.
b_filler is a tiny private class to provide the initialization values. The size of the array is checked at compile-time with the template arguments, so there's no chance of going out of bounds.
I'm sure there are more exotic ways to modify this. This is an initial stab. I think you can pretty much make up for any of the compiler's shortcoming with classes.
ISO standard C++ doesn't let you do this. If it did, the syntax would probably be:
a::a(void) :
b({2,3})
{
// other initialization stuff
}
Or something along those lines. From your question it actually sounds like what you want is a constant class (aka static) member that is the array. C++ does let you do this. Like so:
#include <iostream>
class A
{
public:
A();
static const int a[2];
};
const int A::a[2] = {0, 1};
A::A()
{
}
int main (int argc, char * const argv[])
{
std::cout << "A::a => " << A::a[0] << ", " << A::a[1] << "\n";
return 0;
}
The output being:
A::a => 0, 1
Now of course since this is a static class member it is the same for every instance of class A. If that is not what you want, ie you want each instance of A to have different element values in the array a then you're making the mistake of trying to make the array const to begin with. You should just be doing this:
#include <iostream>
class A
{
public:
A();
int a[2];
};
A::A()
{
a[0] = 9; // or some calculation
a[1] = 10; // or some calculation
}
int main (int argc, char * const argv[])
{
A v;
std::cout << "v.a => " << v.a[0] << ", " << v.a[1] << "\n";
return 0;
}
Where I've a constant array, it's always been done as static. If you can accept that, this code should compile and run.
#include <stdio.h>
#include <stdlib.h>
class a {
static const int b[2];
public:
a(void) {
for(int i = 0; i < 2; i++) {
printf("b[%d] = [%d]\n", i, b[i]);
}
}
};
const int a::b[2] = { 4, 2 };
int main(int argc, char **argv)
{
a foo;
return 0;
}
You can't do that from the initialization list,
Have a look at this:
http://www.cprogramming.com/tutorial/initialization-lists-c++.html
:)
A solution without using the heap with std::vector is to use boost::array, though you can't initialize array members directly in the constructor.
#include <boost/array.hpp>
const boost::array<int, 2> aa={ { 2, 3} };
class A {
const boost::array<int, 2> b;
A():b(aa){};
};
How about emulating a const array via an accessor function? It's non-static (as you requested), and it doesn't require stl or any other library:
class a {
int privateB[2];
public:
a(int b0,b1) { privateB[0]=b0; privateB[1]=b1; }
int b(const int idx) { return privateB[idx]; }
}
Because a::privateB is private, it is effectively constant outside a::, and you can access it similar to an array, e.g.
a aobj(2,3); // initialize "constant array" b[]
n = aobj.b(1); // read b[1] (write impossible from here)
If you are willing to use a pair of classes, you could additionally protect privateB from member functions. This could be done by inheriting a; but I think I prefer John Harrison's comp.lang.c++ post using a const class.
interestingly, in C# you have the keyword const that translates to C++'s static const, as opposed to readonly which can be only set at constructors and initializations, even by non-constants, ex:
readonly DateTime a = DateTime.Now;
I agree, if you have a const pre-defined array you might as well make it static.
At that point you can use this interesting syntax:
//in header file
class a{
static const int SIZE;
static const char array[][10];
};
//in cpp file:
const int a::SIZE = 5;
const char array[SIZE][10] = {"hello", "cruel","world","goodbye", "!"};
however, I did not find a way around the constant '10'. The reason is clear though, it needs it to know how to perform accessing to the array. A possible alternative is to use #define, but I dislike that method and I #undef at the end of the header, with a comment to edit there at CPP as well in case if a change.