I'm trying to create my own version of an array called a safearray, to test my knowledge of operator overloading and creating proper class's and such.
I'm encountering two errors.
SafeArray.h:11:15: error: ‘const int SafeArray::operator’ cannot be overloaded
SafeArray.h:10:10: error: with ‘int& SafeArray::operator’
My code is split between three files.
Main.cpp
#include <cstdlib>
#include <iostream>
#include "SafeArray.h"
using namespace std;
int main(int argc, char** argv) {
SafeArray a(10); // 10 integer elements
for (int i = 0; i < a.length(); i++) {
cout << i << " " << a[i] << "s" << endl; // values initialise to 0
}
cout << endl << a[1]; // Program exits here.
a[3] = 42;
cout << a[3];
a[10] = 10;
cout << a[10];
a[-1] = -1; // out-of-bounds is "safe"?
SafeArray b(20); // another array
b = a; // array assignment
for (int i = 0; i < b.length(); i++) {
cout << b[i] << endl; // values copied from a
}
return 0;
}
SafeArray.h
#ifndef SAFEARRAY_H
#define SAFEARRAY_H
class SafeArray {
public:
SafeArray(int); // int variable will be the array size
int length();
int boundsCheck(int y); // constructor will call this function
// const SafeArray operator= (const SafeArray&);
int& operator[] (int y);
const int operator [] (const int y); // you need this one too.
SafeArray &operator=(SafeArray rhs) {
std::swap(array, rhs.array);
std::swap(length_, rhs.length_);
}
SafeArray(SafeArray const &other);
~SafeArray();
private:
int length_;
int *array;
//int array[];
};
#endif /* SAFEARRAY_H */
SafeArray.cpp
#include "SafeArray.h"
#include <iostream>
SafeArray::SafeArray(int x) {
length_ = x;
array = new int[length];
for (int i = 0; i < length_; i++) {
array[i] = 0;
}
}
int SafeArray::length() {
return this->length_;
}
int SafeArray::boundsCheck(int y) {
}
int& SafeArray::operator[] (int y) {
return array[y];
}
SafeArray::~SafeArray() {
delete [] array;
}
SafeArray::SafeArray(SafeArray const &other) {
int *temp = new int[rhs.size_];
for (int i=0; i<rhs.size_; i++)
temp[i] = rhs.array[i];
std::swap(temp, array);
delete [] temp;
return *this;
}
Your class definition isn't valid. int array[] is an incomplete type, which must not appear as a (non-static) class member. Some compilers accept this as a synonym for int array[0], but zero-sized arrays are not valid in C++, either (only in C99).
In short, you cannot write your code the way you do. You need to learn about dynamic allocation and manage your own memory. Check out how std::vector is implemented.
In C++11, I might recommend a std::unique_ptr<int[]> array as a quick-fix approach, to be initialized as array(new int[x]).
Actually int array[] is valid, and may appear as a class member. The following compiles with strict C++11 conformance:
class foo
{
public:
foo() {}
int length;
int A[];
};
void ralph()
{
foo *bar = (foo *)new int[ 21 ];
bar->length = 20;
bar->A[0] = 1;
}
This is legal, and has its advantages (occasionally). Although it is not commonly used.
However, I suspect that the OP wanted something more along the lines of
class SafeArray {
public:
SafeArray(int); // int variable will be the array size
int length();
int boundsCheck(int y); // constructor will call this function
int& operator[] (int y);
const int operator [] (const int y) // you need this one too.
private:
int length_;
int *array;
};
along with
SafeArray::SafeArray(int x) {
length_ = x;
array = new int[length];
for (int i = 0; i < length_; i++) {
array[i] = 0;
}
}
As #Kerrek already pointed out, your class definition is clearly wrong (shouldn't compile).
To fix it, you want to change the definition to something like:
int *array;
Then in your default ctor you could use something like this:
SafeArray::SafeArray(unsigned size = 0)
: array(new int[size])
{
for (unsigned i=0; i<size; i++)
array[i] = 0;
}
Then, yes, you'll need to write an assignment operator. The usual way is called the copy and swap idiom. You create a copy, then swap the contents of the current one with those of the copy:
SafeArray &operator=(SafeArray rhs) {
std::swap(array, rhs.array);
std::swap(length_, rhs.length_);
}
Along with that, you'll need a copy constructor that makes a copy of the data as well:
SafeArray::SafeArray(SafeArray const &other) {
int *temp = new int[rhs.size_];
for (int i=0; i<rhs.size_; i++)
temp[i] = rhs.array[i];
std::swap(temp, array);
delete [] temp;
return *this;
}
Finally, you'll need a destructor to destroy an object and (particularly) delete the memory it holds:
SafeArray::~SafeArray() {
delete [] array;
}
Then realize that all of that is an ugly mess that will never really work well. In particular, the basic methodology is restricted to an array that's basically fixed in size. As long as you only store ints, it's fairly easy to overlook the problems, and make a dynamic array that (sort of) works. When/if you want to store some other type, however, you just about need to separate allocating memory from initializing objects in that memory, which means throwing away essentially all the code above, and replacing it with something that:
keeps track of the array size and allocation size separately
allocates memory with ::operator new, an Allocator object, or something else similar
uses placement new to initialize objects in the memory when needed.
uses explicit destructor calls to destroy the objects
uses ::operator delete to release memory
and so on. To summarize, std::vector is not a trivial piece of work.
The error message refers to these two lines:
int& operator[] (int y);
const int operator [] (const int y); // you need this one too.
Your error message says that (int y) and (const int y) are too similar to be two different overloads of the [] operator. You cannot overload on (int y) and (const int y) because the calls would all be ambiguous.
You probably meant to return a const int if your SafeArray is const, but return an int& if your SafeArray is not const. In that case, you declare the second function to apply to const SafeArray, by putting the word const after the parameter list. This is what you should write in SafeArray.h:
int& operator[] (int y);
const int operator [] (int y) const; // you need this one too.
You would then have to write both of these functions in SafeArray.cpp:
int& SafeArray::operator[] (int y) {
return array[y];
}
const int SafeArray::operator[] (int y) const { // you need this one too.
return array[y];
}
Related
I have written a DynamicArray class in the past analogous to vector which worked.
I have also written as a demo, one where the performance is bad because it has only length and pointer, and has to grow every time. Adding n elements is therefore O(n^2).
The purpose of this code was just to demonstrate placement new. The code works for types that do not use dynamic memory, but with string it crashes and -fsanitize=address shows that the memory allocated in the addEnd() method is being used in printing. I commented out removeEnd, the code is only adding elements, then printing them. I'm just not seeing the bug. can anyone identify what is wrong?
#include <iostream>
#include <string>
#include <memory.h>
using namespace std;
template<typename T>
class BadGrowArray {
private:
uint32_t size;
T* data;
public:
BadGrowArray() : size(0), data(nullptr) {}
~BadGrowArray() {
for (uint32_t i = 0; i < size; i++)
data[i].~T();
delete [] (char*)data;
}
BadGrowArray(const BadGrowArray& orig) : size(orig.size), data((T*)new char[orig.size*sizeof(T)]) {
for (int i = 0; i < size; i++)
new (data + i) T(orig.data[i]);
}
BadGrowArray& operator =(BadGrowArray copy) {
size = copy.size;
swap(data, copy.data);
return *this;
}
void* operator new(size_t sz, void* p) {
return p;
}
void addEnd(const T& v) {
char* old = (char*)data;
data = (T*)new char[(size+1)*sizeof(T)];
memcpy(data, old, size*sizeof(T));
new (data+size) T(v); // call copy constructor placing object at data[size]
size++;
delete [] (char*)old;
}
void removeEnd() {
const char* old = (char*)data;
size--;
data[size].~T();
data = (T*)new char[size*sizeof(T)];
memcpy(data, old, size*sizeof(T));
delete [] (char*)old;
}
friend ostream& operator <<(ostream& s, const BadGrowArray& list) {
for (int i = 0; i < list.size; i++)
s << list.data[i] << ' ';
return s;
}
};
class Elephant {
private:
string name;
public:
Elephant() : name("Fred") {}
Elephant(const string& name) {}
};
int main() {
BadGrowArray<int> a;
for (int i = 0; i < 10; i++)
a.addEnd(i);
for (int i = 0; i < 9; i++)
a.removeEnd();
// should have 0
cout << a << '\n';
BadGrowArray<string> b;
b.addEnd("hello");
string s[] = { "test", "this", "now" };
for (int i = 0; i < sizeof(s)/sizeof(string); i++)
b.addEnd(s[i]);
// b.removeEnd();
cout << b << '\n';
BadGrowArray<string> c = b; // test copy constructor
c.removeEnd();
c = b; // test operator =
}
The use of memcpy is valid only for trivially copyable types.
The compiler may even warn you on that, with something like:
warning: memcpy(data, old, size * sizeof(T));
writing to an object of non-trivially copyable type 'class string'
use copy-assignment or copy-initialization instead [-Wclass-memaccess]
Note that your code do not move the objects, but rather memcpy them, which means that if they have for example internal pointers that point to a position inside the object, then your mem-copied object will still point to the old location.
Trivially Copyable types wouldn't have internal pointers that point to a position in the object itself (or similar issues that may prevent mem-copying), otherwise the type must take care of them in copying and implement proper copy and assignemnt operations, which would make it non-trivially copyable.
To fix your addEnd method to do proper copying, for non-trivially copyable types, if you use C++17 you may add to your code an if-constexpr like this:
if constexpr(std::is_trivially_copyable_v<T>) {
memcpy(data, old, size * sizeof(T));
}
else {
for(std::size_t i = 0; i < size; ++i) {
new (data + i) T(std::move_if_noexcept(old[i]));
}
}
In case you are with C++14 or before, two versions of copying with SFINAE would be needed.
Note that other parts of the code may also require some fixes.
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;
Background: I'm stuck to arm-arago-linux-gnueabi-g++ (GCC) 4.3.3. Although answers that requires C++11 or later is also appreciated, please explicitly express any language requirement later than C++03.
The object's constructor fills values into tables to be used by the algorithm.
As those table does not change and are not supposed to be changed, I want the them to be const, how do I do that?
Difficulty #1, the values are computationally generated, and I don't want to hard code them in a source file.
Difficulty #2, the computing sometimes depends on inputs that are only available at runtime.
Difficulty #3, I don't know why but I don't want the array to be static, even though the values might be the same for all objects(cases where the values does not depend on runtime input).
Difficulty #4, it's an array, so initializer list in C++03 won't work.
Edit1:
A few weeks after this post, I found both std::array and std::vector are very good alternative to C-style array when std::array is not available.
You can encapsulate the tables in a private type, with a single const instance of that type in your object, then forward the relevant constructor parameters to the private object; this works because even a const object is non-const during its construction.
For example:
class MyClass {
const struct Tables {
double x[1000];
double y[200];
Tables(int i, double d) {
x[i] = d;
y[200 - i] = -d;
}
} tables;
public:
MyClass(int i, double d) : tables(i, d) {}
};
MyClass c(20, 5.5);
Another technique is to build the tables in an ephemeral mutable array whose lifetime is bounded by the lifetime of the constructor, then initialize the const array from those mutable arrays.
Using C++11 std::array (since array types can't be copy-initialized):
class MyClass {
static std::array<double, 1000> buildArray(...) {
std::array<double, 1000> array;
... // fill array
return array;
}
const std::array<double, 1000> mArray;
public:
MyClass(...) : mArray(buildArray(...)) {}
};
Note that std::array is easy to express in C++03; it doesn't depend on any C++11 language features.
If you're worried about the overhead of returning a large array, instrument it - even C++03 compilers are capable of optimising large array returns.
I think you could implement a class containing the actual non const array. That way you can easily compute the values in a constructor.
Then this class would only have to implement the operator[] to be usable as an array. Or it could also simply return a const reference to the array.
Implementation example :
#include <iostream>
using namespace std;
class const_array {
int *arr;
size_t size;
public:
const_array(size_t size, int typ): size(size) {
arr = new int[size];
size_t i;
int val = 0;
for (i=0; i<size; i++) {
val += typ;
arr[i] = val;
}
}
const_array(const const_array & src): size(src.size) {
arr = new int[size];
size_t i;
for (i=0; i<size; i++) {
arr[i] = src.arr[i];
}
}
~const_array() {
delete[] arr;
}
const int * const getArray() const {
return arr;
}
int getSize() const {
return size;
}
const int& operator[](int i) {
return arr[i];
}
};
int main() {
const_array a(16, 4);
// int *arr = a.getArray(); error
const int *arr = a.getArray();
int j = a[2];
int k = arr[2];
// int * pj = &(a[2]); error
const int * pj = &(a[2]);
const int * pk = &(arr[2]);
cout << "a[2]=" << j << " (" << pj << ") - a.getArray[2]="
<< j << " (" << pj << ")" << endl;
return 0;
}
I have a class with a multidimensional array:
it is possible to create a one, two, ..., n dimensional array with this class
if the array has n dimensions, i want to use n operator[] to get an object:
example:
A a({2,2,2,2}];
a[0][1][1][0] = 5;
but array is not a vector of pointer which lead to other vectors etc...
so i want the operator[] to return a class object until the last dimension, then return a integer
This is a strongly simplified code, but it shows my problem:
The error i receive: "[Error] cannot convert 'A::B' to 'int' in initialization"
#include <cstddef> // nullptr_t, ptrdiff_t, size_t
#include <iostream> // cin, cout...
class A {
private:
static int* a;
public:
static int dimensions;
A(int i=0) {
dimensions = i;
a = new int[5];
for(int j=0; j<5; j++) a[j]=j;
};
class B{
public:
B operator[](std::ptrdiff_t);
};
class C: public B{
public:
int& operator[](std::ptrdiff_t);
};
B operator[](std::ptrdiff_t);
};
//int A::count = 0;
A::B A::operator[] (std::ptrdiff_t i) {
B res;
if (dimensions <= 1){
res = C();
}
else{
res = B();
}
dimensions--;
return res;
}
A::B A::B::operator[] (std::ptrdiff_t i){
B res;
if (dimensions <=1){
res = B();
}
else{
res = C();
}
dimensions--;
return res;
}
int& A::C::operator[](std::ptrdiff_t i){
return *(a+i);
}
int main(){
A* obj = new A(5);
int res = obj[1][1][1][1][1];
std::cout<< res << std::endl;
}
The operator[] is evaluated from left to right in obj[1][1]...[1], so obj[1] returns a B object. Suppose now you just have int res = obj[1], then you'll assign to a B object (or C object in the case of multiple invocations of []) an int, but there is no conversion from B or C to int. You probably need to write a conversion operator, like
operator int()
{
// convert to int here
}
for A, B and C, as overloaded operators are not inherited.
I got rid of your compiling error just by writing such operators for A and B (of course I have linking errors since there are un-defined functions).
Also, note that if you want to write something like obj[1][1]...[1] = 10, you need to overload operator=, as again there is no implicit conversion from int to A or your proxy objects.
Hope this makes sense.
PS: see also #Oncaphillis' comment!
vsoftco is totally right, you need to implement an overload operator if you want to actually access your elements. This is necessary if you want it to be dynamic, which is how you describe it. I actually thought this was an interesting problem, so I implemented what you described as a template. I think it works, but a few things might be slightly off. Here's the code:
template<typename T>
class nDimArray {
using thisT = nDimArray<T>;
T m_value;
std::vector<thisT*> m_children;
public:
nDimArray(std::vector<T> sizes) {
assert(sizes.size() != 0);
int thisSize = sizes[sizes.size() - 1];
sizes.pop_back();
m_children.resize(thisSize);
if(sizes.size() == 0) {
//initialize elements
for(auto &c : m_children) {
c = new nDimArray(T(0));
}
} else {
//initialize children
for(auto &c : m_children) {
c = new nDimArray(sizes);
}
}
}
~nDimArray() {
for(auto &c : m_children) {
delete c;
}
}
nDimArray<T> &operator[](const unsigned int index) {
assert(!isElement());
assert(index < m_children.size());
return *m_children[index];
}
//icky dynamic cast operators
operator T() {
assert(isElement());
return m_value;
}
T &operator=(T value) {
assert(isElement());
m_value = value;
return m_value;
}
private:
nDimArray(T value) {
m_value = value;
}
bool isElement() const {
return m_children.size() == 0;
}
//no implementation yet
nDimArray(const nDimArray&);
nDimArray&operator=(const nDimArray&);
};
The basic idea is that this class can either act as an array of arrays, or an element. That means that in fact an array of arrays COULD be an array of elements! When you want to get a value, it tries to cast it to an element, and if that doesn't work, it just throws an assertion error.
Hopefully it makes sense, and of course if you have any questions ask away! In fact, I hope you do ask because the scope of the problem you describe is greater than you probably think it is.
It could be fun to use a Russian-doll style template class for this.
// general template where 'd' indicates the number of dimensions of the container
// and 'n' indicates the length of each dimension
// with a bit more template magic, we could probably support each
// dimension being able to have it's own size
template<size_t d, size_t n>
class foo
{
private:
foo<d-1, n> data[n];
public:
foo<d-1, n>& operator[](std::ptrdiff_t x)
{
return data[x];
}
};
// a specialization for one dimension. n can still specify the length
template<size_t n>
class foo<1, n>
{
private:
int data[n];
public:
int& operator[](std::ptrdiff_t x)
{
return data[x];
}
};
int main(int argc, char** argv)
{
foo<3, 10> myFoo;
for(int i=0; i<10; ++i)
for(int j=0; j<10; ++j)
for(int k=0; k<10; ++k)
myFoo[i][j][k] = i*10000 + j*100 + k;
return myFoo[9][9][9]; // would be 090909 in this case
}
Each dimension keeps an array of previous-dimension elements. Dimension 1 uses the base specialization that tracks a 1D int array. Dimension 2 would then keep an array of one-dimentional arrays, D3 would have an array of two-dimensional arrays, etc. Then access looks the same as native multi-dimensional arrays. I'm using arrays inside the class in my example. This makes all the memory contiguous for the n-dimensional arrays, and doesn't require dynamic allocations inside the class. However, you could provide the same functionality with dynamic allocation as well.
I have a simple class aClass:
class aClass
{
public:
aClass(int size)
{
condition = new bool[size];
}
~aClass()
{
delete condition;
}
bool getCondition(int i) const
{
return condition[i];
}
void setCondition(bool* condition, int i)
{
*(this->condition + i) = *condition;
}
private:
bool* condition;
};
In fact I defined a bool pointer and using constructor to allocate memory.
#include <iostream>
#include "aClass.h"
using namespace std;
int main()
{
aClass tempVar(10);
bool *pC;
for (int i = 0; i < 10; i++)
{
*pC = 0;
tempVar.setCondition(pC, i);
}
for (int i = 0; i < 10; i++)
{
cout << tempVar.getCondition(i);
}
return 0;
}
I do not know what is the problem in this code.
I used gcc version 4.6.3 to compile the code.
You are trying to dereference pC although you never initialized the pointer. Probably pC should have type bool, not bool*, at the same time *pC = 0 should likely be pC = 0 and setCondition should likely take bool not bool*.
You delete although it should be delete[], see here why: delete vs delete[]
Your code will try to delete allocated memory twice if an instance of aClass is copied somewhere. See Rule of three.
You should use std::vector instead of the manually allocated array.
There are two mutually exclusive problems here:
You haven't initialized pC. Do this:
int main() {
⋮
bool * pC = new bool;
⋮
delete pC;
}
Actually, raw pointers aren't such a good idea, so this is better:
std::unique_ptr<bool> pC(new bool);
// No delete required.
setCondition() doesn't need a pointer parameter at all:
class aClass {
⋮
void setCondition(bool cond, int i) {
condition[i] = cond;
}
⋮
};
int main() {
⋮
for (int i = 0; i < 10; ++i) {
tempVar.setCondition(false, i);
}
⋮
}