What is the lifetime of local struct in c++

Question

I have following code

#include <iostream>
#include <vector>


struct M {
    int a = 0;
    int b = 0;

    ~M(){}
};

std::vector<M> ms;

void met(int i)
{
    // This struct 'm' does not get deleted after lifetime of function "met" is over  
    M m;
    m.a = i;
    m.b = i * i;
    ms.push_back(m);
}


int main()
{
 for(int i = 0; i < 10; i++){
   met(i);
 }
 for(int i = 0; i < 10; i++){
  std::cout << ms[i].a << " : " << ms[i].b << std::endl;
 }
 return 0;
}

// OUTPUT
0 : 0
1 : 1
2 : 4
3 : 9
4 : 16
5 : 25
6 : 36
7 : 49
8 : 64
9 : 81

As we know scope of local variable is lifetime of immediate code block. In above code, scope of struct M is lifetime function met. But in above code works fine as we can see output given below code. I was expecting undefined behavior Cause M m has to be deleted once function met return to main. Is is anything special about struct in this case?

possible duplicate of [Can a local variable's memory be accessed outside its scope?](http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope) — πάντα ῥεῖ, Jul 28 '14 at 20:06
You're storing instances of `M` by value not referencing by pointer. — Captain Obvlious, Jul 28 '14 at 20:08
Sorry wrong close reason! The answer simply is, that `std::vector::push_back()` takes a copy of `T`, not a reference. — πάντα ῥεῖ, Jul 28 '14 at 20:08
i was expecting undefined behavior but it clearly show working fine. i am calling it for 10 time and all call stack cant get allocated sequentially... — twid, Jul 28 '14 at 20:09
You seem to be confused about the difference between *types*, *variables* and *objects*. They're all different things. — Kerrek SB, Jul 28 '14 at 20:10
@twid: "I was expecting undefined behavior" - then you don't know how `vector` and copy-construction work in general. — Remy Lebeau, Jul 28 '14 at 20:12
@Remy: I know that just failed to notice `std::vector::push_back()` works on copy and not reference — twid, Jul 28 '14 at 20:15

score 4 · Answer 1 · answered Jul 28 '14 at 20:09

4

The structs in met() are, in fact, deleted. The ones in the vector are copies, made when push_back() is called.

answered Jul 28 '14 at 20:09

Ernest Friedman-Hill

80,601
10
150
186

score 2 · Answer 2 · answered Jul 28 '14 at 20:09

// This struct 'm' does not get deleted after lifetime of function "met" is over

Yes, it does. What you are not taking into account is that you are putting a copy of m inside of the vector, and your struct has a compiler-generated copy constructor to make a copy of the member values. m is gone when met() exits, but the vector is still alive since it has global scope, and you are outputting what is stored in the vector before it goes out of scope.

score 0 · Answer 3 · answered Jul 28 '14 at 20:15

When you store the instance variable m into the vector, that actual instance is not stored. What happens is this:

Create the struct m and change its properties.
Call the copy constructor of m. This is a special constructor method which is used whenever an implicit copy is requested.
Put this copy into the vector.
Delete the original value m as the function ends.

By default, the compiler creates a copy constructor for every struct. The default implementation is just to (shallow-)copy each member variable one-by-one. Because your member variables are just ints, they are copied over to the newly copied variable inside the vector.

This needs to be kept in mind when you have bigger structures, as the entire structure can be copied around multiple times implicitly.

What is the lifetime of local struct in c++

3 Answers3