推向量 <double> 到矢量 <vector<double> &gt;

Question

So, I have this situation:

vector<vector<double>> myVector; 
myVector.resize(somePreviousObject.size());
for(int i = 0; i < myVector.size(); i++)
{
    vector<double> tempVector;
    //Do some stuff that fills tempVector, in a loop
    //After loop:
    myVector[i].push_back(tempVector);
}

However, this produces a compile error, stating that:

no matching function for call to 'std:;vector<double, std::allocator<double > >::push_back(std::vector<double, std::allocator<double> >&)'
...stl_vector.h:733 note: candidates are: void std::vector<_Tp, _Alloc>::push_back(const _Tp&) [with _Tp = double, _Alloc = std::allocator<double>]

I realize that using a vector of vectors is not performance-friendly, but it still should compile and execute, correct? Why doesn't it?

Thanks for your help.

Answer 1

In the expression:

myVector[i].push_back(tempVector);

The subexpression myVector is a std::vector<std::vector<double>> , myVector[i] is a std::vector<double>& , and push_back on that vector requires a double , not a std::vector<double> .

It looks like you want to set the i -th element to be tempVector , if that is the case you should do (on the assumption that you don't need tempVector after this statement):

myVector[i].swap(tempVector);

(While myVector[i] = tempVector; will have the same net effect, it will be more expensive as it will allocate and copy all of the contents. The swap will move the contents of tempVector into myVector[i] at almost no cost (three pointer swaps))

I know someone will come along and suggest using myVector[i] = std::move(tempVector) , which is also fine in C++11, but not available in C++03. The difference between the two is that in the case of swap the original contents of myVector[i] are guaranteed to be inside tempVector after the operation (in this case myVector[i] was empty, so it won't matter). On the other hand, the state of tempVector after myVector[i] = std::move(tempVector); is undefined in the standard, it could leave the vector empty, or it could swap it or... the only guarantee is that the tempVector object can be destroyed after the move.

Answer 2

It's either

myVector[i] = tempVector;

or

myVector.push_back(tempVector);

Since you resized your vector beforehand, I'm assuming you want the first version.

In your code, remember that myVector[i] is itself a vector<double> , that's why it doesn't work.

Answer 3

vector<vector<double>> myVector;  

should be:

vector<vector<double> > myVector;  

and then

myVector[i].push_back(tempVector); 

should be

myVector.push_back(tempVector);