CODEWITHSUNDEEP
c++
Alberto Tiraboschi
Alberto Tiraboschi

Reputation: 303

Add element in list inside function body, then return

Given the following:

#include<iostream>
#include<list>

class test{
    public:
    int t = 34;
};

std::list<test> list;

void func(){
    test t;
    list.push_back(t);
}

int main(){
    func();
    std::cout<<list.front().t<<std::endl;
}

It prints 34. Given that list.push_back(test &ref) is called (so the parameter is not copied), why does the object remains even after the exit of the function?

Upvotes: 0

Views: 180

Answers (1)

prehistoricpenguin
prehistoricpenguin

Reputation: 6326

list.push_back(test &ref) doesn't mean that the reference is stored. For containers, all elements are copied/moved, or construct in place.

Why does the object remain even after the exit of the function? The remained value is a copy.

See the reference:

void push_back( const T& value ); (1)  
void push_back( T&& value );(2)    (since C++11)

Appends the given element value to the end of the container.

  1. The new element is initialized as a copy of value.
  2. value is moved into the new element. No iterators or references are invalidated.

Upvotes: 1

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