I wrote this code to try and understand a bigger problem in some other larger code, but in the second iteration over the list in main prints garbage and I don't quite get what is going on here. Admittedly I tend to break pointers, but here it looks straight forward to me, anyone have any insight?
std::list<int *> myobjects;
const std::list<int *>& getMYObjects( void );
const std::list<int *>&
getMYObjects( void )
{
return( myobjects );
}
void fillMYObjects()
{
int myints[]={15,36,7,17,20,39,4,1};
for(int i=0;i<8;i++)
{
int *temp = &myints[i];
myobjects.push_back(temp);
std::cout << "list: " << *temp << std::endl;
std::cout << "list: " << temp << std::endl;
}
std::cout << "listBack: " << *myobjects.back() << std::endl;
for(std::list<int*>::iterator it=myobjects.begin(); it!=myobjects.end();++it)
{
std::cout << ' ' << **it << std::endl;
}
}
int main()
{
fillMYObjects();
std::list<int*> myobjects2 = getMYObjects();
for(std::list<int*>::iterator it=myobjects2.begin(); it!=myobjects2.end();++it)
{
std::cout << ' ' << **it << std::endl;
}
}
The variable myints is a local variable, and will go out of scope once the function returns. What happens to the pointers in the list then? The will point to memory now occupied and overwritten objects.
You are pointing to numbers that are local and thus are on the stack. They are failing out of scope when you leave the function. So you are still pointing to the right address but the data is gone and the addresses are being used for something else.
At the end of the fillMyObject function, the myInts array is destroyed since it was declated inside the function. at the end of this block, at the end of the function, the array do not live anymore, if you want to keep its values, you need to store the value and not the address. they won't be right anymore right after the end ;)
Related
This is most probably trivial and I'm confusing struct allocation and pointers somehow, I apologize for this. I have read answers to similar questions but it didn't help. The code is, as always, way more complicted, this is a reduction from 3000+ lines of code to the gist.
The output I expected was
prep 1
main 1
Instead, I get
prep 1
main 0
This is the code:
#include <iostream>
#include <vector>
using namespace std;
struct Entry
{
vector<int> list;
};
struct Registry
{
vector<Entry> entries;
void prep()
{
Entry* entry = new Entry();
entries.push_back(*entry);
entry->list.push_back(0);
cout << "prep " << entry->list.size() << "\n";
}
};
int main()
{
Registry registry;
registry.prep();
cout << "main " << registry.entries[0].list.size() << "\n";
return 1;
}
You don't store pointers in your vector<Entry> so you should not use new. Instead add a default constructed Entry using emplace_back.
A C++17 approach:
void prep()
{
Entry& entry = entries.emplace_back(); // returns a ref the added element
entry.list.push_back(0);
cout << "prep " << entry.list.size() << "\n";
}
Prior to C++17:
void prep()
{
entries.emplace_back(); // does NOT return a ref the added element
Entry& entry = entries.back(); // obtain a ref to the added element
entry.list.push_back(0);
cout << "prep " << entry.list.size() << "\n";
}
If you do want to create and maniplate your Entry before adding it to entries, you can do that too and then std::move it into entries.
void prep()
{
Entry entry;
entry.list.push_back(0);
cout << "prep " << entry.list.size() << "\n";
entries.push_back(std::move(entry)); // moving is a lot cheaper than copying
}
The problem is the order of the prep() function. If you change to push an element into the Element object, and then push it tho the entries vector, the behavior will be the expected.
void prep()
{
Entry* entry = new Entry();
entry->list.push_back(0);
entries.push_back(*entry);
cout << "prep " << entry->list.size() << "\n";
}
This is happening, because you uses a copy in the entries list.
It is also possible to store the pointer of the object therefore you can edit the actual instance after you pushed to the entries vector.
Edit:
As Ted mentioned, there is a memory leak, because the entry created with the new operator never deleted. Another approach could be to use smart pointers (however, in this small example it seems overkill, just use reference)
void prep()
{
std::unique_ptr<Entry> entry = std::make_unique<Entry>();
entry->list.push_back(0);
entries.push_back(*entry.get()); // get the pointer behind unique_ptr, then dereference it
cout << "prep " << entry->list.size() << "\n";
} // unique_ptr freed when gets out of scope
You need to change the implementation of prep():
void prep()
{
Entry entry;
entry.list.push_back(0);
entries.emplace_back(entry);
cout << "prep " << entries.back().list.size() << "\n";
}
There is no need to allocate a Entry on the heap just to make a copy of it.
I've tried to create a minimal example of my problem. I'm trying to check if the address of an void pointer is NULL or not. The address should be overgiven by constructing the class, and should be const. I wrote the class below.
MyPointer.h:
public:
MyPointer(const void* activeApp) : m_activeApp(activeApp){
std::cout <<"adresse on construction: " << m_activeApp << std::endl;
};
MyPointer.cpp:
void printAdress();
private:
const void* m_activeApp;
};
The Methode "printAdress" should be able to print the correct address of the given pointer.
int main(void){
void* p_activeApp = nullptr;
std::cout << p_activeApp << std::endl;
MyPointer myPointer(p_activeApp);
std::cout << "Should be 0: " ;
myPointer.printAddress();
p_activeApp = new(bool);
std::cout << "Should be anything: ";
myPointer.printAddress();
}
void MyPointer::printAddress() {
std::cout << this->m_activeApp << std::endl;
};
Of course it doesn't work, because the m_activeApp still points to NULL, but how can I change this?
If you want to change what myPointer.m_activeApp points to, you have to set the pointer to a different value, simple as that. This pointer and p_activeApp are two distinct, independent pointers. Changing one does not change the other.
What you can do is to make one pointer a reference to the other pointer instead. Then, changing one would also change the other. This will work, though be warned, it won't be good programming style.
Pointers are integers (representing a memory location), so imagine void* to be a number. You set p_activeApp to 0, then you construct a MyPointer, and its internal pointer is set to 0.
Then you change the first number by using the new operator, p_activeApp now gets a value (to that new memory address), but there's no reason for m_activeApp to also change. It still points at 0, no matter what p_activeApp changes to.
I fixed it like this:
public:
MyPointer(void** activeApp) : m_pp_activeApp(activeApp) {
std::cout <<"address on construction: " << *m_pp_activeApp << std::endl;
};
void printAddress();
private:
void** m_pp_activeApp;
};
int main(void){
void* p_activeApp = nullptr;
void** pp_activeApp = &p_activeApp;
MyPointer myPointer(pp_activeApp);
std::cout << "Should be 0: " ;
myPointer.printAddress();
p_activeApp = new(bool);
std::cout << "Should be anything: ";
myPointer.printAddress();
}
void MyPointer::printAddress() {
std::cout << *m_pp_activeApp << std::endl;
};
(Disclaimer: Pointers in C++ is a VERY popular topic and so I'm compelled to believe that someone before me has already raised this point. However, I wasn't able to find another reference. Please correct me and feel free to close this thread if I'm wrong.)
I've come across lots of examples that distinguish between pointer to first element of array and pointer to the array itself. Here's one program and its output:
//pointers to arrays
#include <iostream>
using namespace std;
int main() {
int arr[10] = {};
int *p_start = arr;
int (*p_whole)[10] = &arr;
cout << "p_start is " << p_start <<endl;
cout << "P_whole is " << p_whole <<endl;
cout << "Adding 1 to both . . . " <<endl;
p_start += 1;
p_whole += 1;
cout << "p_start is " << p_start <<endl;
cout << "P_whole is " << p_whole <<endl;
return 0;
}
Output:
p_start is 0x7ffc5b5c5470
P_whole is 0x7ffc5b5c5470
Adding 1 to both . . .
p_start is 0x7ffc5b5c5474
P_whole is 0x7ffc5b5c5498
So, as expected, adding 1 to both gives different results. But I'm at a loss to see a practical use for something like p_whole. Once I have the address of the entire array-block, which can be obtained using arr as well, what can I do with such a pointer?
For single arrays, I don't think there's much point to it. Where it becomes useful is with multi-dimensional arrays, which are arrays of arrays. A pointer to one of the sub-arrays is a pointer to the row, and incrementing it gets you a pointer to the next row. In contrast, a pointer to the first element of the inner array is a pointer to a single element, and incrementing it gets you the next element.
int (*)[10] is a "stronger" type than int* as it keeps size of the array,
so you may pass it to function without passing additional size parameter:
void display(const int(*a)[10]) // const int (&a)[10] seems better here
{
for (int e : *a) {
std::cout << " " << e;
}
}
versus
void display(const int* a, std::size_t size) // or const int* end/last
{
for (std::size_t i = 0; i != size; ++i) {
std::cout << " " << a[i];
}
}
This may a stupid question, but I'm gonna ask it anyway:
Suppose you have a pointer: Object* pointer which points at a dynamically allocated object.
class PointClass
{
Array<Object*> m_array1;
Array<Object*> m_array2;
void Delete1()
{
for (int i = 0; i < m_array1.Length; i++)
{
delete m_array1[i];
}
}
void Delete2()
{
for (int i = 0; i < m_array2.Length; i++)
{
delete m_array2[i];
}
}
}
Now, you put your pointer both in m_array1 and in m_array2.
When you try to delete the arrays, in one of them you will have a pointer which points to a deallocated space, so you can't delete it again!
I can't just assign the pointers NULL after the deletion because it wouldn't affect the pointer in the other array.
How would you solve it?
Well the simplest way would be to use a reference-counting pointer, like those available in boost::smart_ptrs.
Otherwise, you need to assign owners to the pointers - you need to decide which class will be responsible for allocating/deleting that particular pointer. If for some reason you decide that should be this class, then you could remove the duplicates from the arrays by adding all the pointers to a set before enumerating them.
If you have to share pointers in this way, something like a ref counted pointer may work well.
See this site which gives an exposé of various 'smart-pointer' techniques.
Smart Pointers
My initial answer is: Don't do that.
If you absolutely have to for some reason, you could wrap it in a smart pointer
Best solved is not passing the same pointer to both arrays. :P If you really need to, and you also need to reflect that change to all other "same" pointers, a pointer-to-pointer will do.
#include <iostream>
struct Object{};
int main(){
Object* ptr = new Object;
Object** ptrptr = &ptr;
delete *ptrptr;
*ptrptr = 0;
// both print 0
std::cout << *ptrptr << std::endl;
std::cout << ptr << std::endl;
}
On Ideone.
Another way is with a reference-to-pointer.
int main(){
Object* ptr = new Object;
Object*& refptr = ptr;
delete refptr;
refptr = 0;
// both print 0
std::cout << refptr << std::endl;
std::cout << ptr << std::endl;
}
But the second best way is probably a ref-counted smart pointer.
How would you solve it?
By not storing the same pointer in two different places. Doing this creates a duplication of data, and confuses ownership semantics. Who owns the memory pointed to by pointer? Ownership is not clear.
Under normal circumstances, dynamically allocated objects should be owned by the same module that allocated it, and only that module will have direct access to the objects or delete the memory. That's not to say other modules can't get at the data.
As others have suggested use smart pointers to solve your problem. If you have to solve it by writing your own code, I would make each of the delete function also search the "other" array to delete all pointers in the first array that can be found in the other array. And it is a last option option as this would not be my first solution to implement anything as your approach
void Delete2()
{
for (int i = 0; i < m_array2.Length; i++)
{
for (int j = 0; j < m_array1.Length; j++)
{
if (m_array2[i] == m_array1[j])
{
delete m_array1[j]
m_array1[j] = NULL;
}
delete m_array2[i];
m_array2[i] = NULL;
}
}
Then look for ways to optimise it
If I understood your question, you have the same (valid) pointer stored in 2 different arrays.
The problem is that after you delete it on array1, you can't do it again in the second array.
One way to do this is change your array definition to store the memory address of the pointer itself, instead of storing the address of the allocated memory:
const int array_size = 3;
int** m_array1[array_size];
int** m_array2[array_size];
and the rest of the code could be implemented as:
void Delete1()
{
for (int i = 0; i < array_size - 1; i++) // delete all memory but leave the last intact
{
if (*(int*)m_array1[i])
{
cout << "Delete1: erasing #" << i << " with mem addr " << std::hex << *m_array1[i] << std::dec << endl;
delete *m_array1[i];
*m_array1[i] = NULL;
}
}
}
void Delete2()
{
for (int i = 0; i < array_size; i++)
{
if (*m_array2[i])
{
cout << "Delete2: erasing #" << i << " with mem addr " << std::hex << *m_array2[i] << std::dec << endl;
delete *m_array2[i];
*m_array2[i] = NULL;
}
else
{
cout << "Delete2: !!! memory at #" << i << " was already deleted." << endl;
}
}
}
int main()
{
int* num1 = new int(10);
int* num2 = new int(20);
int* num3 = new int(30);
cout << "main: storing " << std::hex << &num1 << " which points to " << num1 << std::dec << endl;
cout << "main: storing " << std::hex << &num2 << " which points to " << num2 << std::dec << endl;
cout << "main: storing " << std::hex << &num3 << " which points to " << num3 << std::dec << endl;
m_array1[0] = &num1;
m_array1[1] = &num2;
m_array1[2] = &num3;
m_array2[0] = &num1;
m_array2[1] = &num2;
m_array2[2] = &num3;
Delete1();
Delete2();
}
Outputs:
main: storing 0xbfc3818c which points to 0x87b6008
main: storing 0xbfc38188 which points to 0x87b6018
main: storing 0xbfc38184 which points to 0x87b6028
Delete1: erasing #0 with mem addr 0x87b6008
Delete1: erasing #1 with mem addr 0x87b6018
Delete2: !!! memory at #0 was already deleted.
Delete2: !!! memory at #1 was already deleted.
Delete2: erasing #2 with mem addr 0x87b6028
I'm working my way through Accelerated C++ and have decided to mess around with the one of structs that were defined in there. While doing so, I've come across a problem: creating a vector of these structs and modifying the elements in each one seems to modify the elements in all of them.
I realize that this probably means I've initialized all the structs in the vector to a struct at a single memory address, but I used the .push_back() method to insert "dummy" structs in to the vector. I was under the impression that .push_back() pushes a copy of its argument, effectively creating a new struct.
Here is the header for the struct:
#ifndef _STUDENT_INFO__CHAPTER_9_H
#define _STUDENT_INFO__CHAPTER_9_H
#include <string>
#include <iostream>
#include <vector>
class Student_info9{
public:
Student_info9(){homework = new std::vector<double>;};
Student_info9(std::istream& is);
std::string getName() const {return name;};
double getMidterm() const {return midterm;};
double getFinal() const {return final;};
char getPassFail() const {return passFail;};
std::vector<double> *getHw(){return homework;};
void setName(std::string n) {name = n;};
void setMidterm(double m) {midterm = m;};
void setFinal(double f) {final = f;};
private:
std::string name;
double midterm;
double final;
char passFail;
std::vector<double> *homework;
};
#endif /* _STUDENT_INFO__CHAPTER_9_H */
And here is the code that i'm fooling around with (excuse the excessive print statements... the result of some time trying to debug :) ):
vector<Student_info9> did9, didnt9;
bool did_all_hw9(Student_info9& s)
{
vector<double>::const_iterator beginCpy = s.getHw()->begin();
vector<double>::const_iterator endCpy = s.getHw()->end();
return(find(beginCpy, endCpy, 0) == s.getHw()->end());
}
void fill_did_and_didnt9(vector<Student_info9> allRecords)
{
vector<Student_info9>::iterator firstDidnt = partition(allRecords.begin(), allRecords.end(), did_all_hw9);
vector<Student_info9> didcpy(allRecords.begin(), firstDidnt);
did9 = didcpy;
vector<Student_info9> didntcpy(firstDidnt, allRecords.end());
didnt9 = didntcpy;
}
int main(int argc, char** argv) {
vector<Student_info9> students;
Student_info9 record;
for(int i = 0; i < 5; i++)
{
students.push_back(record);
}
for(int i = 0; i < students.size(); i++)
{
students[i].setMidterm(85);
students[i].setFinal(90);
students[i].getHw()->push_back(90);
std::cout << "student[" << i << "]'s homework vector size is " << students[i].getHw()->size() << std::endl;
students[i].getHw()->push_back(80);
std::cout << "student[" << i << "]'s homework vector size is " << students[i].getHw()->size() << std::endl;
students[i].getHw()->push_back(70);
std::cout << "student[" << i << "]'s homework vector size is " << students[i].getHw()->size() << std::endl;
std::cout << "Just pushed back students[" << i << "]'s homework grades" << std::endl;
if(i == 3)
students[i].getHw()->push_back(0);
}
std::cout << "student[3]'s homework vector size is " << students[3].getHw()->size() << std::endl;
for(vector<double>::const_iterator it = students[3].getHw()->begin(); it != students[3].getHw()->end(); it++)
std::cout << *it << " ";
std::cout << std::endl;
std::cout << "students[3] has " << ( ( find(students[3].getHw()->begin(),students[3].getHw()->end(), 0) != students[3].getHw()->end()) ? "atleast one " : "no " )
<< "homework with a grade of 0" << std::endl;
fill_did_and_didnt9(students);
std::cout << "did9's size is: " << did9.size() << std::endl;
std::cout << "didnt9's size is: " << didnt9.size() << std::endl;
}
As you can see by the print statements, it seems that the homework grades are being added only to one Student_info9 object, copies of which seem to be populating the entire vector. I was under the impression that if you were to use consecutive copies of .push_back() on a single object, it would create copies of that object, each with different memory addresses.
I'm not sure if that's the source of the problem, but hopefully someone could point me in the right direction.
Thanks.
When you push a StudentInfo onto the vector, it is indeed copied, so that's not the problem. The problem is the vector containing the homework grades. Since you only store a pointer to that vector in StudentInfo, only the pointer, not the vector, is copied when you copy a StudentInfo. In other words you have many different StudentInfos that all have a pointer to the same homework vector.
To fix this you should define a copy constructor which takes care of copying the homework vector.
Have you learned about the copy constructor yet? If so, think about what is happening with vector<Student_info9> students on push_back().
Specifically, what happens with this pointer.
std::vector<double> *homework;
The line Student_info9 record; constructs a Student_info9 using the first constructor. This first constructor creates a vector and stores a pointer to it as a member variable. You then proceed to add a copy of this Student_info9 to a vector 5 times. Each copy has a pointer to the same vector.
Your StudentInfo9 class contanis a pointer to a std::vector<double>, which means in the default copy constructor (which will be called when you add a StudentInfo9 object to your vector), the pointer itself is copied. That means all of your StudentInfo9 objects have the same homework vector.
Does that make sense? Please refer to http://pages.cs.wisc.edu/~hasti/cs368/CppTutorial/NOTES/CLASSES-PTRS.html for a more in depth look at pointers and copy constructors.