how to memcpy iterator element in C++?

Question

I have the following code in C++:

typedef struct 
{
   int age;
   int roomNumber;
} Student;

Student student;
student.age = 20;
student.roomNumber = 10;

vector<Student> studentV;
student.push_back(student);
student.push_back(student);
student.push_back(student);

Student student1[3];

int i;

for(vector<Student>::const_iterator iterator = studentV.begin();
    iterator != studentV.end(); ++iterator,++i)
{
   memcpy(&student1[i], iterator, sizeof(Student));
}

It show the following message for the memcpy part:

error: cannot convert 'std::vector<Student>::const_iterator {aka __gnu_cxx::__normal_iterator<const Student*, std::vector<Student> >}' to 'const void*' for argument '2' to 'void* memcpy(void*, const void*, size_t)'

What is the problem and how to fix it? is iterator can not be copied like this?

an iterator is not a pointer.. you could get the value *pointed* by the iterator `(*iterator)` and then get its address `&(*iterator)`... haven't really tried it. — jsantander, Mar 27 '14 at 14:01

score 4 · Answer 1 · answered Mar 27 '14 at 14:02

4

The second argument of memcpy should be address of Student object, so you should use &*iterator instead of iterator. Because of operator overloading those two are not equivalent. But I would recommend not to use memcpy at all.

answered Mar 27 '14 at 14:02

Wojtek Surowka

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why not to use memcpy? then what to use? – ratzip Mar 27 '14 at 14:05
Using `memcpy` for structs copying is a low-level mechanism inherited from C. In C++ instead of your loop use `std::copy(studentV.begin(), studentV.end(), student1);` which is much simpler and makes many errors not possible - for instance in your code `i` is not initialised, which would cause undefined behaviour, while with `copy` you do not need any `i` at all. – Wojtek Surowka Mar 27 '14 at 14:11

score 4 · Answer 2 · answered Mar 27 '14 at 14:03

4

What you want to do is copy what the iterator refers to. Use *iterator to get the object referenced by the iterator and & to get its address. By the way DO NOT use memcpy but the copy constructor:

student1[i] = *iterator;

answered Mar 27 '14 at 14:03

neuro

14,948
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Matthieu Poullet · Answer 3 · 2014-03-27T16:17:35.933

You should take advantage of the actual C++11 facilities and avoid using memcpy at all:

#include <iostream>
#include <vector>
#include <array>

typedef struct
{
    int age;
    int roomNumber;
} Student;

int main() {
    Student student{ 20, 10 };
    std::vector<Student> studentV{ student, student, student };
    std::array<Student, 3> student1;

    if (studentV.size() <= student1.size())
        std::copy(studentV.begin(), studentV.end(), student1.begin());
    for (const auto& s : student1) {
        std::cout << s.age << " " << s.roomNumber << std::endl;
    }
}

score 0 · Answer 4 · answered Mar 27 '14 at 14:37

Use std::copy algorithm. It will use plain memory copy routines (like memcpy) if the object is POD or will use copy constructor otherwise. If will also protect you from copying non-copyable objects (by printing an error during compilation), something your code may do and which may cause hard to trace errors (like double free inside copied non-copyable objects).

how to memcpy iterator element in C++?

4 Answers4