+= on a vector without boost

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后悔当初
后悔当初 2021-01-22 11:34

Is there any way to use the += operator with a vector without using boost or using a derivated class?

Eg.

somevector += 1, 2, 3, 4, 5, 6, 7;
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  • 2021-01-22 12:03

    Not with syntax like that, no. But you could do something like this:

    int tmparray[] = {1, 2, 3, 4, 5, 6, 7};
    
    somevector.insert(somevector.end(), 
                      tmparray, 
                      tmparray + (sizeof(tmparray) / sizeof(tmparray[0])));
    
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  • 2021-01-22 12:26

    With a little ugly operator overloading, this isn't too difficult to accomplish. This solution could easily be made more generic, but it should serve as an adequate example.

    #include <vector>
    

    Your desired syntax uses two operators: the += operator and the , operator. First, we need to create a wrapper class that allows us to apply the , operator to push an element onto the back of a vector:

    template <typename T>
    struct push_back_wrapper
    {
        explicit push_back_wrapper(std::vector<T>& v) : v_(&v) { }
    
        push_back_wrapper& operator,(const T& x)
        {
            v_->push_back(x);
            return *this;
        }
    
        std::vector<T>* v_;
    };
    

    Then, in order to use this in conjunction with += on a vector, we overload the += operator for a vector. We return a push_back_wrapper instance so that we can chain push backs with the comma operator:

    template <typename T, typename U>
    push_back_wrapper<T> operator+=(std::vector<T>& v, const U& x)
    {
        v.push_back(x);
        return push_back_wrapper<T>(v);
    }
    

    Now we can write the code you have in your example:

    int main()
    {
        std::vector<int> v;
        v += 1, 2, 3, 4, 5, 6, 7;
    }
    

    The v += 1 will call our operator+= overload, which will return an instance of the push_back_wrapper. The comma operator is then applied for each of the subsequent elements in the "list."

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