How to store a value obtained from a vector `pop_back()` in C++?

Question

I'm not sure how to cast a value obtained from the pop_back() function of vector. The following is a simple code to illustrate the problem.

#include<vector>
#include<iostream>

using namespace std;

int main()
{
  vector<int> a,b;
  int val;

  a.push_back(1);
  a.push_back(2);
  a.push_back(3);
  a.push_back(4);

  for(int i=0; i<4; i++)
  {
    val = a.pop_back();
    b.push_back(val);
  }

  vector<int>::iterator v = b.begin();
  while( v != b.end())
  {
    cout << *v << " ";
    v++;
  }
  return 0;
}

This is the error I get.

pushback.cpp:18:9: error: assigning to 'int' from incompatible type 'void'
    val = a.pop_back();
        ^ ~~~~~~~~~~~~
1 error generated.

I tried casting as well; (int)a.pop_back(), but it throws an error stating C-style cast from 'void' to 'int' is not allowed.

May I know if there is a standard way to store the value from the pop_back() function?

Answer 1

It may sound as pop as in returning a value. But it actually doesn't. The standard says that vector::pop_back should erase the last value, with no return value.

You can do:

auto val = a.back();
a.pop_back();

Answer 2

As stated in documentation std::vector::pop_back() does not return any value, you just need to call std::vector::back() right before:

val = a.back();
a.pop_back();

Answer 3

In case you have a vector of objects and not just primitive types, moving out from the vector can be done with std::move:

auto val = std::move(a.back()); // allow call of move ctor if available
a.pop_back();

Note that wrapping the call to back() with std::move is done only because we know that we are about to erase this element from the vector on the next line.

Answer 4

According to http://www.cplusplus.com/reference/vector/vector/pop_back/

The pop_back is void function, it is nonvalue-returning.