Scope of struct object in c++ created without new keyword - c++

Question

Coming from a Java background, following code is confusing me. Confusion here is C++ object creation without new keyword. I am creating this object Student_info info; and then adding it to a vector. Since I am not creating it using new keyword, will it be allocated on stack and be destroyed after the loop exits? If that is the case, how the last loop is able to extract the info from vector correctly? Shouldn't it throw exception as all the object added to the vector has been destroyed?
struct Student_info {
string name;
vector<double> marks;
};
int main()
{
int num = 2;
vector<Student_info> student_infos;
for (int i = 0; i < 3; i++) {
Student_info info; //will this object be destroyed once this loop exits?
cout << "Enter name:";
cin >> info.name;
double x;
cout << "Enter 2 marks";
for (int j = 0; j < num; j++) {
cin >> x;
info.marks.push_back(x);
}
student_infos.push_back(info);
}
for (int i = 0; i < 3; i++) {
cout << student_infos[i].name;
for (int j = 0; j < num; j++) {
cout << student_infos[i].marks[j];
}
cout << endl;
}
return 0;
}

Answer 1

You are correct in that the info object will be destroyed, but it is copied into the vector when using push_back. The rest of the code reads the copy from the vector and not the original object.
http://www.cplusplus.com/reference/vector/vector/push_back/
The copying is done using the object's copy constructor which in this case is compiler-provided. It will in turn call the copy constructors on the string and vector members of the struct.

Answer 2

The scope of variable info is the block scope of the for statement starting from the point of declaration of the variable.
for (int i = 0; i < 3; i++) {
Student_info info; //will this object be destroyed once this loop exits?
//...
}
Each time a new iteration of the loop is executed a new object with name info is created.
The scope of variable student_infos is the outer most block scope of function main where the variable is declared.
int main()
{
vector<Student_info> student_infos;
//...
}
When the loop is executed a new object is added to the vector that is a copy of variable info. The vector does not contain a reference to the original object with name info. It creates a copy of the object and stores it within itself internally.
Take into account that in general these loops
for (int i = 0; i < 3; i++) {
cout << student_infos[i].name;
for (int j = 0; j < num; j++) {
cout << student_infos[i].marks[j];
}
cout << endl;
}
are unsafe. It is much better to rely on the actual number of elements in the vectors. You could write instead
for ( vector<Student_info>::size_type i = 0; i < student_infos.size(); i++)
{
cout << student_infos[i].name;
for ( vector<double>::size_type j = 0; j < student_infos[i].marks.size(); j++)
{
cout << student_infos[i].marks[j];
}
cout << endl;
}

Answer 3

Yes, info will be destroyed at the end of the loop. push_back creates a new copy from info, so even if info is destroyed, its copy still exist in the vector.

Answer 4

It will copy the temp Student_info info to the student_infos when push_back is done. If you can write a custom copy constructor for Student_info and print something there you will see the copy construct will invoked.
Consider using,
http://en.cppreference.com/w/cpp/container/vector/emplace_back