I am learning how classes/inheritances/pointers work in C++ and coding the following.
I have a class unit declared as such:
class unit{
public:
int locationX,locationY;
//genotype (does not change)
float agility, build,size,stamina, aggression;
//phenotype (changes based on genotype and stimuli)
float speed, strength, hunger;
};
When I create a new instance to pass into a void function (below, respectively), the memory is not allocated yet.
Instance
unit **units = 0;
Void Function Prototype
void initialize(grid *grids, unit **units /*, plant **plants, predator **predators */);
The memory is allocated within the void function using some parameters:
void initialize(grid *grids, unit **units,plant **plants,predator **predators)
{
units = new unit*[int(((grids->gridHeight)*(grids->gridWidth)*(grids->gridDivision))/20)];
for(register int i = 0; i<int(((grids->gridHeight)*(grids->gridWidth)*(grids->gridDivision))/20); i++)
{
units[i] = new unit;
units[i]->hunger = 5;
units[i]->locationX = (rand()%((grids->gridWidth)-0));
units[i]->locationY = (rand()%((grids->gridHeight)-0));
//etc, etc
}
}
However, once I exit the void function, the data I just stored gets deleted. Is there something wrong with the pointer declarations and passing into the function (which is as follows)?
initialize(&environment, units, plants, predators);
NOTE: I only have problems with the units variable declared under unit class. The environment variable is fine. The other two (plants and predators) are similar to units, so if this is fixed, I can fix the others.
Second Note: The main function is as follows (relevant parts):
int main()
{
unit **units = 0; //<--- Important one
plant **plants = 0;
predator **predators = 0;
grid environment(250,250,5); //Constructor for environment (don't mind this)
initialize(&environment, units, plants, predators); //<-- Void function
running = true;
return 0;
}
Thank you for any help/links/explanations you can provide.
You pass units into the function by value. This means that the units pointer within the function starts out as a copy of the one in the calling code. Within the function you assign a new value to the local units variable (i.e. the address of some newly created objects). Then when the function terminates, the local variable passes out of scope and the objects it pointed to are lost. The pointer in the calling code is never modified, and knows nothing about any of this.
Pass it by reference instead:
void initialize(grid *grids, unit ** &units)
You're passing "units" as a type "unit**" which is on the stack. When you assign a new value to the variable "units", you're affecting the variable local to the function. If you want the calling code to have it's variable "units" updated, you must pass it in by pointer or by reference.
Easiest way to fix this is to change your function signature to:
void initialize(grid *grids, unit **&units, etc.. );
this is passing the variable units as a reference to a pointer to a pointer to a unit.
Initializing a pointer (or array) A * a inside a method is a particular case of modifying your variable a (which is a pointer/array in this case). Modifying a variable inside a method can be done either by passing a pointer to it, or a reference to it. So in the case your variable is already a pointer, these two options are the following.
By passing a pointer to your pointer a:
void init(A * *a)
{
*a = new A(); // or "new A[n]" if "a" is an array
}
void main()
{
A * a;
init(&a);
}
Or passing a reference to your pointer:
void init(A * &a)
{
a = new A(); // or "new A[n]" if "a" is an array
}
void main()
{
A * a;
init(a);
}
So in your case, a is an array of unit* (which I think is a bad design by the way, it would be better to simply use an array of unit and save you from this loop to dynamically allocate separately all units), so basically a is of type unit ** so your A is actually the type unit*. Using the second method would lead to:
void init(unit* * &units)
But as I said, it is probably better to simply use an array of unit instead, and your whole code would look like:
void initialize(unit * &units /* etc */)
{
// note that I use unit instead of unit* here
units = new unit[n];
for(register int i = 0; i<n; i++)
{
// note that I remove the "new unit;" here
// note that I use "." instead of "->"
units[i].hunger = 5;
units[i].locationX = (rand()%((grids->gridWidth)-0));
units[i].locationY = (rand()%((grids->gridHeight)-0));
//etc, etc
}
}
void main()
{
unit * units;
initialize(units);
}
Related
I have a pretty standard class with some public member functions and private variables.
My problem originally stems from not being able to dynamically name object instances of my class so I created an array of pointers of the class type:
static CShape* shapeDB[dbSize];
I have some prompts to get info for the fields to be passed to the constructor (this seems to work):
shapeDB[CShape::openSlot] = new CShape(iParam1,sParam1,sParam2);
openSlot increments properly so if I were to create another CShape object, it would have the next pointer pointing to it. This next bit of code doesn't work and crashes consistently:
cout << shapeDB[2]->getName() << " has a surface area of: " << shapeDB[2]->getSA() << shapeDB[2]->getUnits() << endl;
The array of pointers is declared globally outside of main and the get() functions are public within the class returning strings or integers. I'm not sure what I'm doing wrong but something relating to the pointer set up I'm sure. I'm writing this code to try and learn more about classes/pointers and have gotten seriously stumped as I can't find anyone else trying to do this.
I'm also curious as to what the CShape new instances get named..? if there is any other way to dynamically create object instances and track the names so as to be able to access them for member functions, I'm all ears.
I've tried all sorts of permutations of pointer referencing/de-referencing but most are unable to compile. I can post larger chunks or all of the code if anyone thinks that will help.
class CShape {
int dim[maxFaces];
int faces;
string units;
string type;
string name;
bool initialized;
int slot;
public:
static int openSlot;
CShape();
CShape(int, string, string); // faces, units, name
~CShape();
void initialize(void);
// external assist functions
int getA(void) {
return 0;
}
int getSA(void) {
int tempSA = 0;
// initialize if not
if(initialized == false) {
initialize();
}
// if initialized, calculate SA
if(initialized == true) {
for(int i = 0; i < faces; i++)
{
tempSA += dim[i];
}
return(tempSA);
}
return 0;
}
string getUnits(void) {
return(units);
}
string getName(void) {
return(name);
}
// friend functions
friend int printDetails(string);
};
// constructor with values
CShape::CShape(int f, string u, string n) {
initialized = false;
faces = f;
units = u;
name = n;
slot = openSlot;
openSlot++;
}
My guess is you use the CShape constructor to increment CShape::openSlot?
You're probably changing the value before it's read, thus the pointer is stored in a different location.
Try replacing openSlot with a fixed value to rule out this CShape::option.
-- code was added --
I'm pretty sure this is the problem, the constructor is executed before the asignment, which means the lhs. will be evaluated after CShape::openSlot is incremented.
I have a shape class that I initialize from my main program and give the parameters in the constructor.
Shape *cusomShape = new CustomShape(float radius, float origin)
The shape class has some functions such as rollover and more.
When the rollover function inside the shape class is fired, I want to change a certain int value in the main program. This might similar to firing of an event that changes the value when the rollover function is fired, but I am not sure how to do that in C++. If at all, events is the ideal approach here, it would great to see a short example coming.
If using the event is not the correct, what would the ideal way to go about this?
I think what you need is to pass a value by pointer or reference to the function in Shape and then modify it. If the function is called not from main but from somewhere else passing the pointer is the better option you have. First pass the pointer to the class and store it using another method and then each time rollover is called make use of it.
EDIT: example:
class CustomShape {
void storePointer(int* _value) {
value = _value;
}
void rollover() {
.. do stuff
*value++; // for instance
... do stuff
}
int * value;
}
int main() {
int a;
CustomShape cs;
cs.storePointer(&a);
....
cs.rollover();
....
return 0;
}
Pass a reference to the variable in the constructor and save that reference. Change the value when needed.
I would suggest passing a reference to the variable to the member function that needs to change its value. Storing a reference in a class couples the Shape class to the reference. This means that each time you want to use the Shape, without updating the integer, you cannot, since the Shape constructor will expect a reference/pointer to the int as an argument (the Shape class will store the pointer/reference as an attribute). Passing a reference/pointer to the member function promotes Low Coupling.
#include <iostream>
class Shape
{
double shapeValue_;
public:
Shape (double value)
:
shapeValue_(value)
{}
void fireFunction(int& updateMe)
{
updateMe = 123;
}
};
using namespace std;
int main()
{
int update;
cout << update << endl;
Shape s(4.5);
s.fireFunction(update);
cout << update << endl;
return 0;
};
And in this case, you have an option for a main program that doesn't involve shape object calling on fireFunction:
int main()
{
Shape s(4.5);
// Main program that doesn't use fireFunction.
return 0;
};
In this case, if you have member functions changing input arguments, you should take on a style for defining such functions: e.g. make sure that the variable that gets changed by the member function is always the first input argument in its declaration.
If you want complex objects to communicate updates between each other, you can make use of the Observer Pattern.
I am getting access violation error in the below code..i have pointed it out in the program.
void *pBuff = 0;
void set_data(void *pBuff)
{
int value = 70, i;
int *phy_bn = new int[8];
for(i=0; i<8; i++)phy_bn[i] = value;
pBuff =phy_bn;
cout<<((int*)pBuff)[0];//..accessing 0th element value..no error here..gives 70 as result..
}
int main()
{
set_data(pBuff);
cout<<((int*)pBuff)[0];//acces violation error
return 0;
}
Why that access violation even when i am not assigning it the address of a local variable...
Yes i can use vector or pass by reference.
But i want to know why pBuff is not getting assigned
Because it is a copy of the pointer being modified within set_data(). Pass the pointer by reference so the change is visible to the caller:
void set_data(void*& pBuff)
Note that the function variable pBuff hides the global variable pBuff in the function set_data().
That said, I am unsure of the reason for void* and why vector<int> is not being used which handles all dynamic memory allocation for you.
When you say
pBuff = phy_bn;
You're just changing the local value of pBuff, not the global value of pBuff. Either pass pBuff as a double pointer, or simply remove the argument to the function, as pBuff is global already.
void *pBuff = 0; /* This is the global pBuff, which isn't being changed */
void set_data(void *pBuff /* This is the local pBuff, which is being changed */)
{
...
pBuff = phy_bn;
...
}
'plz i want to avoid double pointers..its not required i guess...'
Guessed wrong, it is required! You'll need a pointer reference for the pBuff parameter then:
void set_data(void*& pBuff)
{
// ...
}
This is effectively the same as using a double pointer.
The only thing you're doing with
pBuff =phy_bn;
is manipulating the function parameter representation on the local stack.
The pBuff inside set_data is not the global pBuff. The value of the global pBuff never gets changed from 0. Since this is C++ code, set_data can take its pointer argument by reference, and assigning to it will change the value at the point of the function call.
In C++, pointers are passed by value, the same as other value types. It may be instructional to think of a pointer as literally an integer type; then it’s easy to see why pBuff = phy_bn; doesn’t accomplish anything, for the same reason that this code doesn’t:
#include <iostream>
void set(int x) {
x = 5;
}
int main(int argc, char** argv) {
int y = 0;
set(y);
std::cout << y << '\n';
return 0;
}
Here, x is a local variable. It is a copy of y, not y itself. You can change its value by assigning to it, sure, but you’re merely changing the value of a variable which will not exist outside the scope of set(). If you change the definition of set() to use a reference:
void set(int& x) {
x = 5;
}
Then y will indeed be updated, because you have explicitly requested that x be an alias for the name you pass to set(), instead of a copy. You were misled by the names: the pBuf in set_data() is not the same variable pBuf in main(), even though they happen to have the same value; they’re like two different people with the same name and the same amount of money.
I have a thread-class Buffer (own made class), and many derived classes such as BufferTypeA, BufferTypeB...
Since I have to synchronize them in a certain order, I'm giving any of them an integer which represents the order to run certain task. I also have to know inside each thread Buffer which one is next to run the task, so I'm passing every BufferType a reference to an integer which all of them must share and I didn't want to make it Global.
I got lost at any point and I don't see where.
First I create all the BufferTypes from a class where I also define that shared integer as:
int currentThreadOrder;
And when creating the BufferTypes:
int position = 0;
if (NULL == bufferA) {
bufferA = new BufferTypeA(¤tThreadOrder, ++position,
waitCondition);
}
if (NULL == bufferB) {
bufferB = new BufferPos(¤tThreadOrder, ++position,
waitCondition);
}
if (NULL == bufferC) {
bufferC = new BufferRtk(¤tThreadOrder, ++position,
waitCondition);
}
Then, in BufferTypeA header:
class BufferTypeA: public Buffer {
public:
BufferTypeA(int currentThreadOrder,
int threadConnectionOrder = 0,
QWaitCondition *waitCondition = NULL);
//..
}
And in cpp file:
BufferTypeA::BufferTypeA(int currentThreadOrder, int threadConnectionOrder, QWaitCondition *waitCondition):
Buffer(currentThreadOrder, threadConnectionOrder, waitCondition) { }
Now I'll show Buffer header:
class Buffer: public QThread {
public:
Buffer(int ¤tThreadOrder,
int threadConnectionOrder = 0,
QWaitCondition *waitCondition = NULL);
//...
protected:
QWaitCondition *waitCondition;
int threadConnectionOrder;
int ¤tThreadOrder; // Shared address
}
And finally the cpp:
Buffer::Buffer(int ¤tThreadOrder, int threadConnectionOrder, QWaitCondition *waitCondition) {
this->threadConnectionOrder = threadConnectionOrder;
this->waitCondition = waitCondition;
this->currentThreadOrder = currentThreadOrder;
}
And the error I'm getting is error: uninitialized reference member Buffer::currentThreadOrder.
I'm embarrased to ask, because it's going to be a simple problem with pointers and addresses, but I can't see where the problem is, so please help.
When you create a class with a data-member that is a reference, the reference needs to be assigned a value in the constructor initializer list.
References have to be given a value when they are created, they are not pointers. They have to start with a value and that value cannot be changed (while the contents that is pointed to by that value can be changed).
Essentially you can think of a reference as an alias for an existing variable. You can't give a friend a nickname if you don't have a friend :)
RESPONSE TO COMMENT:
You don't "share a reference" between objects. Each object will have its own reference to the same variable. When you "pass by reference" you are telling the compiler that you want the variable in your function to actually be the variable in your outer scope, rather than creating a new variable by value. This means that you only have one variable at one memory location. The reference is just memory in some other place that forwards you to that same memory location.
Think of this as call forwarding... I can have 15 phone numbers in 15 different countries. I can set them all up to forward calls to my cell in the US. So, people are calling me no matter which number they call.
Each of your classes just has another reference to forward the "phone calls" or variable reads/writes to that same memory location. So, you're not sharing a reference between classes, you're making sure that each class HAS a reference to the same underlying memory location.
Back to the metaphore, each class won't have the same phone, but each class' phone will forward to the same number (variable) none-the-less which lets them all set/get the same value in the end.
RESPONSE II:
Here's a simple example to get your head going, it's pretty easy to apply to your classes. I didn't compile it but it should work minus a typo or two possibly.
class A
{
public:
A(int& shared) : m_shared(shared)
{
//No actions needed, initializer list initializes
//reference above. We'll just increment the variable
//so you can see it's shared in main.
m_shared += 7;
}
void DoSomethingWithIt()
{
//Will always reflect value in main no matter which object
//we are talking about.
std::cout << m_shared << std::endl;
}
private:
//Reference variable, must be initialized in
//initializer list of constructor or you'll get the same
//compiler error again.
int& m_shared;
};
int main()
{
int my_shared_integer = 0;
//Create two A instances that share my_shared_integer.
//Both A's will initialize their internal reference to
//my_shared_integer as they will take it into their
//constructors "by reference" (see & in constructor
//signature) and save it in their initializer list.
A myFirstA(my_shared_integer);
A mySecondA(my_shared_integer);
//Prints 14 as both A's incremented it by 7 in constructors.
std::cout << my_shared_integer << std::endl;
}
you pass a pointer int* as 1st argument to BufferTypeA, which expects and int, while you said in your question you meant to use a int&. To do this, the ctor of BufferTypeA should take a int& and initialise it in an initialisation list (i.e. not within the { } part of the ctor) like
class BufferType {
int &Ref;
public:
BufferTypeA(int& ref) : Ref(ref) { /* ... */ }
};
and in your construction of BufferA you must not pass an address, but the reference, i.e.
int counter;
Buffer = new BufferType(counter);
You want code like this:
Buffer::Buffer(
int ¤tThreadOrder0,
const int threadConnectionOrder0,
QWaitCondition *const waitCondition0
) :
threadConnectionOrder(threadConnectionOrder0),
waitCondition(waitCondition0),
currentThreadOrder(currentThreadOrder0)
{}
The reason is related to the reason you cannot write
const double pi;
pi = 3.14;
but can write
const double pi = 3.14;
A reference is typically implemented as a constant pointer, to which one cannot assign an address after one has initialized the pointer. Your version of the code assigns, as in the first pi example. My version of the code initializes, as in the second pi example.
So I'm still rather new to programming/C++, and still trying to wrap my head around pointers and passing by reference and everything. A program I'm trying to figure out now needs to pass an array of structs to another function. I've gotten it working by just passing the array directly there. It seems to work fine. However, what I'm concerned about is that I believe I'm passing it by value, and I understand that it's better to pass structs by reference, so you're not making a copy of the struct every time...
Anyway, here's a basic example of what I'm doing:
struct GoldenHelmet {
int foo;
string bar;
};
void pass (GoldenHelmet ofMambrino[], int size);
int main () {
GoldenHelmet ofMambrino[10];
int size = sizeof(ofMambrino) / sizeof(ofMambrino[0]);
ofMambrino[1].foo = 1;
pass(ofMambrino, size);
cout << ofMambrino[2].foo << endl;
return 0;
}
void pass (GoldenHelmet ofMambrino[], int size) {
ofMambrino[2].foo = 100;
ofMambrino[2].bar = "Blargh";
}
From what I understand, it works because arrays are already pointers, right? But the way I have that configured, am I still passing a copy of the struct and everything to the pass() function? I've tried to pass it by reference, but it doesn't seem to want to work any way I've tried.
The C++ way:
#include <array>
typedef std::array<GoldenHelmet, 10> Helmets;
void pass(Helmets &);
int main()
{
Helmets h;
h[1].foo = 1;
pass(h);
//...
}
void pass(Helmets & h)
{
h[2].foo = 100;
// ...
}
Indeed, we pass the array by reference.
This syntax:
void pass (GoldenHelmet ofMambrino[], int size)
is actually quite confusing. Because you are not passing an array, you are passing a pointer. They are not the same thing though, don't get confused. This oddity only applies to function parameters. The above is exactly identical to this:
void pass (GoldenHelmet * ofMambrino, int size)
It's actually impossible to pass an array by value, unless it is a sub-object of another object. You can pass them by reference, you need to include the size though, but you can do that using a template:
template<int N>
void pass (GoldenHelmet (&ofMambrino)[N])
These are all possible, but none of them are pass by value. Just think of ofMambrino as being the address of the beginning of the array, and that is what you are passing.
void pass (GoldenHelmet ofMambrino[], int size)
void pass (GoldenHelmet ofMambrino[10], int size)
void pass (GoldenHelmet *ofMambrino, int size)
void pass (GoldenHelmet (&ofMambrino)[10], int size)
Arrays are represented and passed as pointers, so you are not copying anything here. In contrast, if you were passing a single struct, it would be passed by value.
Below is a code snippet to illustrate this last point:
void passByVal (GoldenHelmet ofMambrino) {
ofMambrino.foo = 100;
ofMambrino.bar = "Blargh";
}
void passByRef (GoldenHelmet& ofMambrino) {
ofMambrino.foo = 100;
ofMambrino.bar = "Blargh";
}
int main() {
GoldenHelmet h;
passByVal(h); // h does not change
passByRef(h); // fields of h get assigned in the call
}
First of all array is not pointers. We refer this as a pointer in the argument list because when we use
int x[ ]
x is actually const pointer that points the beginning of the array. And when you pass this to a function you send the adress of the memory that is beginning of the array. Thats why when you make a change in your function, you make change in the adress of your variable in the caller section actually. This is actualy simulated call by reference not call by reference. But effect is same with call by reference because you are working on memory locations. For this reason when you send array of your struct you pass actually adress of your array of structs. Thats why when you change value on this, you actually change your structs.
To use call by reference, one thing you must to do is to define your function prototype like
void f(int ¶m)
and when calling function, it is same with the others.
To summarize:
int main()
{
int x;
// simulated call by reference that use adress of variable,
// lets say adress of x is 19ff
f(&x); // actually you send 19ff
f(x); // call by reference that use reference of variable
}
// simulated call by reference
void f(const int *y)
{
// when you use like *y=10, you are writing on memory area 19ff, you actually
// change memory area that is belong to x in the main
}
// call by reference
void f(const int &y)
{
}