Pointer to Integer Array in Class - c++

I have a class named Goal. This class has an array which stores the goal coordinates (x,y). It also has a function which generates a random goal, sets it and returns a pointer to the goal array.
While creating the pointer p, I get an error saying:
no suitable conversion function from "std::array<int, 2UL>" to "int *" exists
How can I fix this?
I'm also unsure if this pointer will be a shared or unique pointer if I were to use a smart pointer. Though I think it should be unique since p will be assigned to some pointer which will call the function.
class Goal{
protected:
std::array<int,2> goal;
public:
int* set_goal(){
srand((unsigned)time(NULL));
int flag1 = rand() % 2;
int flag2 = rand() % 2;
std::unique_ptr<int> p = std::make_unique<int>(Goal::goal);
if(flag1){
Goal::goal[0] = rand() % 16;
if(flag2){
Goal::goal[1] = 0;
std::cerr << "y - 0";
}
else{
Goal::goal[1] = 15;
std::cerr << "y - 15";
}
}
else{
goal[1] = rand() % 16;
if(flag2){
Goal::goal[0] = 0;
std::cerr << "x - 0";
}
else{
Goal::goal[0] = 15;
std::cerr << "x - 15";
}
}
return p;
}
};

This line of code fails:
std::unique_ptr<int> p = std::make_unique<int>(Goal::goal);
Because std::make_unique<int>() expects a pointer to the heap, that is, a pointer that was allocated with new, that you want std::unique_ptr<> to manage. Goal::goal is a member-variable array... NO pointer that would point to the contents of Goal::goal is a suitable input parameter for std::make_unique<int>(), nor would it be a suitable candidate for ANY type of managed or smart pointer.
Why do you make your life miserable with overly complex code?
You have a single coordinate (x, y) to store. Why not store that in a proper struct?
As in, for example:
// A simple 2d integer point.
struct point_2di
{
int x;
int y;
};
class Goal
{
protected:
point_2di goal;
public:
// returns a const reference to the goal.
// this will prevent the caller from changing our (protected) goal
// behind our back.
// returning a reference is always preferable to returning a pinter,
// it also tells the caller that the return value is always valid.
// By contrast, a returned pointer could be null, and the
// caller should then check if it's valid. As with anything
// that needs checking before use, it's a potential source of
// bugs. So, its best to avoid returning a pointer as
// much as possible.
// since point_2di is small, you could return a copy by value, if you prefer.
const point_2di& set_goal()
{
srand((unsigned)time(NULL));
int flag1 = rand() % 2;
int flag2 = rand() % 2;
if (flag1) {
goal.x = rand() % 16; // using named members make your code
if(flag2){ // easier to read and avoids silly bugs.
goal.y = 0;
std::cerr << "y <- 0";
}
}
// etc...
return goal;
}
};

Related

dereferencing a pointer to set a value vs. assigning an address to the pointer?

Could someone explain the difference between the following two snippets of code?
In the function below, *frame_id maybe a nullptr, it's basically the output parameter in the function.
bool LRUReplacer::Victim(frame_id_t *frame_id) {
long oldest_ts = std::numeric_limits<long>::max();
for (auto & it : this->LRUCache){
if (it.ts < oldest_ts){
oldest_timestamp = it.ts;
*frame_id = it.frame_id;
}
}
vs.
bool LRUReplacer::Victim(frame_id_t *frame_id) {
long oldest_ts = std::numeric_limits<long>::max();
for (auto & it : this->LRUCache){
if (it.ts < oldest_ts){
oldest_timestamp = it.ts;
frame_id = &it.frame_id;
}
}
in the first case, we are assigning frame_id the value of it.frame_id and in the second case we are assigning the pointer the address of it.frame_id?
Let's consider a simplified example doing basically the same as reading/writing/dereferencing is concerned:
void f1(int* value)
{
*value = 1;
}
int g_value = 1;
void f2(int* value)
{
value = &g_value;
}
int main()
{
{
int value = 2;
f1(&value);
std::cout << value << '\n'; // prints 1
}
{
int value = 2;
f2(&value);
std::cout << value << '\n'; // prints 2
}
}
Note that you just reassign the parameter in the second alternative. The parameter was a pointer value that was passed by value and therefore there is absolutely no effect for the variable the pointer the caller passed a points to.
In C++ using a reference is preferrable:
void f3(int& value)
{
value = 1;
}
int main()
{
int value = 2;
f3(value);
std::cout << value << '\n'; // prints 1
}
In the first loop, if it.ts < oldest_ts you will assign the value of it.frame_id to the frame_id_t to which frame_id points. After the loop is done, the frame_id_t instance (*frame_id) will hold the last value that was assigned to it.
In the second loop, you will instead assign the address of it.frame_id to the frame_id_t* (frame_id). After the loop is done, frame_id will point at the last it.frame_id it was set to point at.
The first one is replacing a value pointed to by frame_id with a new value. The second is trying to replace the given pointer, to make it point to something else. But it will not work.
You are given a pointer on the stack and resetting it. That's not going to get passed back to the caller. He is going to be stuck with the value he passed you. Unless you pass it by reference
If you wanted the second one to work you would need to change the signature to be this:
bool LRUReplacer::Victim(frame_id_t*& frame_id) {

Size of an object without using sizeof in C++

This was an interview question:
Say there is a class having only an int member. You do not know how many bytes the int will occupy. And you cannot view the class implementation (say it's an API). But you can create an object of it. How would you find the size needed for int without using sizeof.
He wouldn't accept using bitset, either.
Can you please suggest the most efficient way to find this out?
The following program demonstrates a valid technique to compute the size of an object.
#include <iostream>
struct Foo
{
int f;
};
int main()
{
// Create an object of the class.
Foo foo;
// Create a pointer to it.
Foo* p1 = &foo;
// Create another pointer, offset by 1 object from p1
// It is legal to compute (p1+1) but it is not legal
// to dereference (p1+1)
Foo* p2 = p1+1;
// Cast both pointers to char*.
char* cp1 = reinterpret_cast<char*>(p1);
char* cp2 = reinterpret_cast<char*>(p2);
// Compute the size of the object.
size_t size = (cp2-cp1);
std::cout << "Size of Foo: " << size << std::endl;
}
Using pointer algebra:
#include <iostream>
class A
{
int a;
};
int main() {
A a1;
A * n1 = &a1;
A * n2 = n1+1;
std::cout << int((char *)n2 - (char *)n1) << std::endl;
return 0;
}
Yet another alternative without using pointers. You can use it if in the next interview they also forbid pointers. Your comment "The interviewer was leading me to think on lines of overflow and underflow" might also be pointing at this method or similar.
#include <iostream>
int main() {
unsigned int x = 0, numOfBits = 0;
for(x--; x; x /= 2) numOfBits++;
std::cout << "number of bits in an int is: " << numOfBits;
return 0;
}
It gets the maximum value of an unsigned int (decrementing zero in unsigned mode) then subsequently divides by 2 until it reaches zero. To get the number of bytes, divide by CHAR_BIT.
Pointer arithmetic can be used without actually creating any objects:
class c {
int member;
};
c *ptr = 0;
++ptr;
int size = reinterpret_cast<int>(ptr);
Alternatively:
int size = reinterpret_cast<int>( static_cast<c*>(0) + 1 );

Changing the value of a int variable through pointers passed as arguments?

I want to modify values of some variables of a particular class by accessing address of these variables from another different class through a function. So, to access this address I try to pass pointers-to-variables as arguments to a function, where these pointers-to-variables will be set with the address of the variables. To learn how to do it, I'm trying to mimic in a simple program.
Here is my code:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int *a, int *b)
{
*a = numberA;
*b = numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA = 0;
int *testB = 0;
referenceSetter(testA, testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
As you could see I declare numberA and numberB as global variables and set their values. The I try to get the address of these two variables through the function referenceSetter function and then after that I try to modify the values in those variables using the references. Apparently, I'm doing something wrong which leads to to have Unhandled Exception error exactly when I try to modify the values and try to set them as 30 and 40 resepectively.
Alternatively I tried the following approach:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int *a, int *b)
{
a = &numberA;
b = &numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA;
int *testB;
referenceSetter(testA, testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
But this approach throws up the error uninitialized local variables testA and testB. Do I have to initialize pointers too?
Please help me find my mistake. Thanks.
The thing you're not understanding is that pointers are passed by value, just like any other variable. If you want the passed pointer to be changed, you need to pass a pointer to a pointer (or a reference to a pointer, but I'll leave that alone, as explaining references at this point will confuse you further).
Your main() is passing NULL pointers to referenceSetter(). The assignment *a = numberA copies the value of numberA (i.e. 100) into the memory pointed to by a. Since a is a NULL pointer, that has the effect of overwriting memory that doesn't exist as far as your program is concerned. The result of that is undefined behaviour which means - according to the standard - that anything is allowed to happen. With your implementation, that is triggering an unhandled exception, probably because your host operating system is detecting that your program is writing to memory that it is not permitted to write to.
If, after the call of referenceSetter() you want testA and testB to contain the addresses of numberA and numberB respectively, you need to change referenceSetter() to something like;
void referenceSetter(int **a, int **b)
{
*a = &numberA;
*b = &numberB;
}
This allows the values passed to be addresses of pointers. *a then becomes a reference to the pointer passed. &numberA compute the address of numberA, rather than accessing its value 100. Similarly for numberB.
The second change is to change main() so it calls the function correctly;
referenceSetter(&testA, &testB);
which passes the address of testA (and testB) to the function, so those pointers can be changed
You are trying to set the contents of address 0 to be equal to the other numbers, so when you're doing *a = numberA you're assigning a value of numberA to memory address 0.
Not sure, but I think what you're trying to achieve is this:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int **a, int **b)
{
*a = &numberA;
*b = &numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA = 0;
int *testB = 0;
referenceSetter(&testA, &testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
This way, using pointers to pointers as arguments for referenceSetter(), you are actually modifying the address that your passed pointers are pointing to.
You are close, but the key is you need to pass the address of the value you want to set. You declare the values as int in main and pass the address by using the & operator:
int *testA = 0;
int *testB = 0;
referenceSetter(&testA, &testB);
*testA = 30;
*testB = 40;
numberOutput();
If you declare testA and testB as pointers in main and pass the pointer, the function gets a copy of the pointer instead of the address of the value you want to set.

Class accept and return array

I try to create a class that accept and return an array but I got some problem. I'm not sure if it is legal to return an array from a class. Or it could be done by returning an pointer to the array. Thank for any solution to the problem.
#include <iostream>
using namespace std;
class myclass {
private:
int Array[10];
public:
myclass (int temp[10]) {
for (int i = 0; i < 10; i++) {
Array [i] = temp [i];
}
}
int returnArray () {
return Array; // error here, I'm not sure if it is legal to return an array.
}
int* returnArray2 () {
return this->Array; // hope it will return a pointer to the array
}
};
int main () {
int Array[10] = {1,2,3,4,5,6,7,8,9};
myclass A(Array);
cout << A.returnArray() << endl; // try to return an array and print it.
myclass* ptr = &A;
cout << *ptr->returnArray2 << endl; // error here
return 0;
}
First of all it is better to write the constructor either like
myclass ( const int ( &temp )[10] ) {
for (size_t i = 0; i < 10; i++) {
Array [i] = temp [i];
}
}
or like
myclass ( int temp[], size_t n ) : Array {} {
if ( n > 10 ) n = 10;
for (size_t i = 0; i < n; i++) {
Array [i] = temp [i];
}
}
Or even you may define the both constructors.
As for the returning value then you may not return an array. You may return either a reference to an array or a pointer to the entire array or a pointer to its first element
For example
int ( &returnArray () )[10] {
return Array;
}
In this case you can write in main
for ( int x : A.returnArray() ) std::cout << x << ' ';
std::cout << std::endl;
As for this statement
cout << *ptr->returnArray2 << endl; // error here
then you forgot to place parentheses after returnArray2. Write
cout << *ptr->returnArray2() << endl;
And the following member function is wrong because the expression in the return statement has type int * while the return type of the function is int
int returnArray () {
return Array; // error here, I'm not sure if it is legal to return an array.
}
So either the function will coincide with the the second member function if you specify its return type like int *. Or you could change the return expression to *Array
int returnArray () {
return Array; // error here, I'm not sure if it is legal to return an array.
}
This is illegal because Array is not of int type. Your returnArray2 is valid, however. As for this line:
cout << *ptr->returnArray2 << endl; // error here
This is illegal because returnArray2 is a function; you must call it to return the int*:
cout << *ptr->returnArray2() << endl; // prints the first value in the array
Other notes:
Your capitalization is backwards; you should call your class MyClass and your member array arr or arr_, or you will confuse a lot of people.
return this->Array; this is redundant, you can simply return Array;
If you haven't heard of std::vector and std::array you should research those, as they are generally superior to C-style arrays.
In general, I would suggest to read a c++ book to get your basics correct as there are lot of issues in the code you posted.
Regarding your main question about exposing C style arrays in class public API, this is not a very robust mechanism. Do it if it is absolutely essential because of existing code but if possible prefer to use std::vector. You will mostly always end up with better code.
Other answers have corrected your coding errors, so i won't repeat that.
One other thing, your code suggests that the array size is fixed. You can pass and return the array by reference as well. Refer to: General rules of passing/returning reference of array (not pointer) to/from a function?

Set pointer to element in vector to null, then check whether pointer is null (C++)

I would like to set pointers to some elements in my vector array to NULL (based on a criteria), and then check whether an element pointer is NULL. If the pointer pointing that element is NULL, I remove the element from my vector array.
My compiler is giving me an error, saying that the address expression must be an lvalue or function designator and I do not understand why (line location commented in code). Since I am taking the address of the value using &, am I not seeing if the pointer pointing to that element is NULL?
I included the preceding code as the error may lie there,
Relevant code:
vector<particle> pl = c.particlelist;
vector<particle> noncollision = c.particlelist;
vector<vector<particle>> collisionlist = new vector<vector<particle>>();
for (int i = 0; i < c.numparticles-1; i++){
particle first = pl[i];
for (int j = i+1; j < c.numparticles; j++)
{
particle second = pl[j];
double d = distance(first, second);
if (d==0)
{
vector<particle> temp = {pl[i], pl[j]};
collisionlist.push_back(temp);
noncollision[i].setxposint(NULL);
noncollision[j].setxposint(NULL);
}
else
{
}
}
}
int j = 0;
for (int i = 0; i < noncollision.size(); i++)
{
if (&(noncollision[i].getxpos()) == NULL) ////// ERROR HERE
{
noncollision.erase(noncollision.begin()+i);
}
else
{
j++;
}
}
I am new to C++, and if you could suggest a more elegant way to do this, or a fix, it would be much appreciated. I also assume that my method of setting the pointer to an element, noncollision[i].setxposint(NULL); is correct? Can I return an integer using a function, and take the address?
Functions for getxpos and setxposint:
int particle::getxpos(){
return xpos;
}
void particle::setxposint(int b){
xpos = b;
}
You're using & to take a pointer to a temporary vale (the return from getxpos) which isn't allowed; since a temporary will be going away, the address won't be useful in any way so the language doesn't allow it. It certainly wouldn't ever be NULL even if you could get its address.
noncollision[i].setxposint(NULL);
All that line is doing is setting xpos to zero. Generally the term NULL is used with pointers, and 0 is used with things like integers. NULL is usually a macro for 0L anyway.
&(noncollision[i].getxpos()) == NULL
What this is doing, which is incorrect, is attempting to take the address of the return value from the member method getxpos() and compare it to NULL. Whereas what you really want to do is simply see if the function returns zero. So simply change this line to:
noncollision[i].getxpos() == 0
I'll explain why the compiler doesn't understand what you mean.
When you write
&(someFunction())
you are asking for the address of the thing that the function returns. But functions return values. A value doesn't have an address. Variables have addresses.
When something is a word of memory (which will contain a value), it can be used as an lvalue (left-value), because you can put things into that word of memory:
int b = 1; //make room for an `int` on the stack, then put a `1` there.
When something is just a value, it can only ever be used as an rvalue. The following would not compile, for the same reason that your code would not:
int b; //make room for an `int` on the stack.
42 = b; //ERROR, this makes no sense.
if (42 == NULL) { std::cout << "this is never true" << std::endl; }
&42; //ERROR, 42 isn't a piece of memory, it's a value.
(Caveat: you can use values to refer to words in memory: this usage is called a pointer, e.g.
int b = 1;
*((int *)(42)) = b;
meaning "put the value of b into the memory which has the address 42. This compiles fine (but crashes if you're not allowed to write to the memory at 42.)
It looks to me you're trying to keep track of 'visited' items, not sure exactly in which way.
Instead of "modifying" the items, you could use an "external" mark. A set looks to be fine here. You could use a set of iterators into the particle list, or in this case a set of indices (i,j) which will likely be more stable.
Here's a start:
#include <vector>
#include <set>
struct particle { };
double distance(particle const&, particle const&) { return 1.0; }
struct context
{
std::size_t numparticles;
std::vector<particle> particlelist;
context() : numparticles(100), particlelist(numparticles) {}
};
static context c;
int main()
{
using std::vector;
using std::size_t;
vector<particle> pl = c.particlelist;
vector<vector<particle>> collisionlist;
std::set<size_t> collision;
for(size_t i = 0; i < c.numparticles-1; i++)
{
particle first = pl[i];
for(size_t j = i+1; j < c.numparticles; j++)
{
particle second = pl[j];
double d = distance(first, second);
if(d < 0.0001)
{
collisionlist.push_back({pl[i], pl[j]});
collision.insert(i);
collision.insert(j);
}
else
{
}
}
}
for(size_t i = 0; i < pl.size(); i++)
{
if(collision.end() != collision.find(i))
{
// do something
}
}
// alternatively
for (int index : collision)
{
particle& p = pl[index];
// do something
}
}
NOTE Be very very wary of floating point comparison like
if (d==0.0) // uhoh
because it will likely not do what you expect
How dangerous is it to compare floating point values?
What is the most effective way for float and double comparison?
Is floating-point == ever OK?
It seems that you are trying to check pairs of points for collisions. You then record for each point whether it has any collision. This is best handled by a simple list of flags:
std::vector<bool> has_collision(c.numparticles, false); // init: no collisions found
Afterwards:
if (d==0)
{
has_collision[i] = true;
has_collision[j] = true;
}
At the end, iterate over the list of flags and get the points that have no collisions:
for (size_t i = 0; i < c.numparticles; ++i)
{
if (!has_collision[i])
{
// whatever
// possibly push_back pl[i] into some list
}
}
In addition: using a vector to hold a pair (i,j) of points is confusing. Standard library has the std::pair type for purposes such as this.
Also: you don't need explicit dynamic allocation (new); let Standard Library manage memory for you in a safe, non-confusing way. Instead of
vector<vector<particle>> collisionlist = *new vector<vector<particle>>();
Use
vector<vector<particle>> collisionlist;
(or vector<pair<particle, particle>>, as described above).