How is memory allocated to instances of class with array member? - c++

Question

In my understanding, under usual conditions, array is being allocated memory at compile time, but what happens when array is a member variable, and there is pretty much nothing to allocate memory to during compilation. Does it get implicitly dynamically-allocated, when instance of that class is created?
class Arr{
public:
int arr[10];
};
Arr doSomething(Arr &arg){
return arg; //copy of arg created; 'copy=new int[10]' at this point?
}
int main(){
Arr temp;
doSomething(temp);
//if returned object's array was dynamically initialized
//will it result the array in temp being released twice?
}
UPD. In my case, situation rises when I try to subract A-B where B>A. In these cases I can access array to read from it, but modifying its values leads to garbage. Moreover, final output in main is fine. Here's full code:
#include <iostream>
using namespace std;
const int MAX_ARRAY_SIZE=256;
char* getOperatorIndex(char* equationString);
bool isOperator(char characterDec);
int* toIntArray(char* firstInA, char* lastInA, int* firstInB);
class Number{
int* firstInNumber;
int* lastInNumber;
int number[MAX_ARRAY_SIZE];
public:
Number();
~Number();
bool operator<(Number& b);
int& operator[](int index);
Number operator-(Number& b);
void print();
int length(){return lastInNumber-firstInNumber;}
int*& lastPtr(){return lastInNumber;}
int*& firstPtr(){return firstInNumber;}
};
Number::Number(){
firstInNumber=number;
lastInNumber=number;
}
Number::~Number(){
cout<<"number destroyed"<<endl;
}
int& Number::operator[](int index){
return number[index];
}
bool Number::operator<(Number& b){
if(length()>b.length())return false;
if(length()<b.length())return true;
for(int a=0;a<length();++a){
if(number[a]>b[a])return false;
if(number[a]<b[a])return true;
}
return false;
}
void Number::print(){
for(int a=0; a<=length(); ++a){
cout<<number[a];
}
cout<<endl;
}
Number Number::operator-(Number& b){
Number result;
if(*this < b)
{
result=b-*this;
cout<<*result.lastPtr()<<endl;
result.print();
*result.lastPtr()*=-1; // GARBAGE HERE
cout<<*result.lastPtr()<<endl;
return result;
}
result[0]=0;
for(int q=0; q<=length(); ++q)
{
if(b.length()-q >= 0)
{
result[q]+=(*this)[length()-q]-b[b.length()-q];
if(result[q] < 0)
{
result[q]+=10;
result[q+1]=-1;
}
else
{
result[q+1]=0;
}
}
else
{
result[q]+=(*this)[length()-q];
}
++result.lastPtr();
}
do{
--result.lastPtr();
}while(!*result.lastPtr());
return result;
}
int main(){
char equationArray[MAX_ARRAY_SIZE*2+1]; // operandA(<=256) operator(1) operandB(<=256)
Number a,b;
cin>>equationArray;
char* operatorPtr=getOperatorIndex(equationArray);
a.lastPtr()=toIntArray(equationArray, operatorPtr, a.firstPtr());
b.lastPtr()=toIntArray(operatorPtr+1, operatorPtr+strlen(operatorPtr), b.firstPtr());
a.print();
b.print();
Number c;
switch(*operatorPtr){
case '-':
c=a-b;
break;
}
c.print();
}
char* getOperatorIndex(char* equationString){
while(!isOperator(*++equationString));
return equationString;
}
bool isOperator(char characterDec){
if(characterDec>='*' & characterDec<='/')return true;
return false;
}
int* toIntArray(char* firstInA, char* lastInA, int* firstInB){
while(lastInA-firstInA>0){
*firstInB=*firstInA-'0';
++firstInB;
++firstInA;
}
return --firstInB;
}

Answer 1

All data members of a class have the same storage-duration as the class' object created. Hence in:
int main(){
Arr temp;
doSomething(temp);
}
The data member temp.arr which is of type int[10]; also has automatic storage-duration.. (or what is inappropriately known as "stack object")
Arr doSomething(Arr &arg){
return arg; //copy of arg created; 'copy=new int[10]' at this point?
}
No, there is no heap allocation here. rather a member-wise copy by the implicit copy constructor. See How are C++ array members handled in copy control functions?