returning address or local variable or temporary C++ warning [duplicate]

Question

Possible Duplicate:
c++ warning: address of local variable

Hi, When i write this code:

//Returns the transpose matrix of this one
SparseMatrix& SparseMatrix::transpose()const{
    vector<Element> result;
    size_t i;
    for(i=0;i<_matrix.size();++i){
        result.push_back(Element(_matrix.at(i)._col, _matrix.at(i)._row, _matrix.at(i)._val));
    }

    return SparseMatrix(numCol,numRow,result);
}

I get the warning "returning address or local variable or temporary". The last line calls the SparseMatrix constructor. I don't understand what's wrong with this code, and how can i fix it so i can return a SparseMatrix object as i want.

Answer 1

You're returning a reference, not an actual object - note the & here:

SparseMatrix& SparseMatrix::transpose()const{

If you want to return the actual object, remove that &.

The last line does indeed call the constructor, but it doesn't return the resulting object. That object immediately gets destroyed, and an invalid reference to it gets returned.

Answer 2

In C++, local variables are 'automatically' destructed when going out of scope. Your return statement will create a nameless, temporary variable of type SparseMatrix that will immediately go out of scope. Hence, returning a reference to it doesn't make sense.

It may be easier to return by value: then a copy of the temporary will be returned. The compiler can optimize that away (copy elision).

If you really want to pass an object out of a function, you should create it on the heap, using new:

SparseMatrix* SparseMartix::transopse()const{


     //...
     return new SparseMatrix(...);

}

But then, you need to take care of the lifetime of the returned object yourself.