CODEWITHSUNDEEP
c++vector
Aquarius_Girl
Aquarius_Girl

Reputation: 22916

error: void value not ignored as it ought to be

template <typename Z> Z myTemplate <Z> :: popFromVector ()
{
    if (myVector.empty () == false)
        return myVector.pop_back ();

    return 0;
}

int main ()
{
    myTemplate <int> obj;

    std :: cout << obj.popFromVector();

    return 0;
}

Error:

error: void value not ignored as it ought to be

AFAI can see, the return type of popFromVector is NOT void. What's the point that I am missing? The error disappears when I comment out this call in main().

Upvotes: 21

Views: 79011

Answers (3)

Jason
Jason

Reputation: 32510

That is because the definition of std::vector::pop_back has a void return type ... you are trying to return something from that method, which won't work since that method doesn't return anything.

Change your function to the following so you can return what's there, and remove the back of the vector as well:

template <typename Z> Z myTemplate <Z> :: popFromVector ()
{
    //create a default Z-type object ... this should be a value you can easily
    //recognize as a default null-type, such as 0, etc. depending on the type
    Z temp = Z(); 

    if (myVector.empty () == false)
    {
        temp = myVector.back();
        myVector.pop_back();
        return temp;
    }

    //don't return 0 since you can end-up with a template that 
    //has a non-integral type that won't work for the template return type
    return temp; 
}

Upvotes: 9

ltjax
ltjax

Reputation: 15997

It's the pop_back(). It has a void return type. You have to use back() to get the actual value. This is to avoid unnecessary copies.

Upvotes: 2

Puppy
Puppy

Reputation: 146930

std::vector<T>::pop_back() returns void. You attempt to return it as an int. This is not allowed.

Upvotes: 17

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