While Loop Error - c++

Question

I have an assignment to merge two sorted vectors into a third sorted vector. I'm sure the solution to this problem is pretty simple, but my brain is fried at the moment and could use your help.
Basically vectors A and B have a size of 3. They will hold integers such as 1, 2, 3 and 4, 5, 6 respectively. I can't seem to get the syntax of my while loop correctly. I've tried making it a do/while loop, putting parentheses around the cases, and a few other things. It just doesn't seem to be reading the part after the &&.
The first for loop just makes the R vector have the right size. And the second for loop just displays the values of the R vector. The result should print out from 1-6, but I'm only seeing 1-3.
Any help would be appreciated!
void::combine(vector<int> A, vector<int> B, vector<int> R) {
int ia = 0, ib = 0, ir = 0;
for (int i = 0; i < A.size() + B.size(); i++) {
R.push_back(0);
}
while (ia != A.size() && ib != B.size()) {
if (A[ia] < B[ib]) {
R[ir] = A[ia];
ia += 1;
}
else {
R[ir] = B[ib];
ib += 1;
}
ir += 1;
}
for (int i = 0; i < R.size(); i++) {
cout << "L3[" << i << "] = " << R[i] << endl;
}
}

Answer 1

Assuming A contains [1,2,3] and B contains [4,5,6] as you say, this will not add any of the element in the B vector to the R vector.
This is because on the 4th iteration, ia == 3, and so the conjunctive condition is no longer true..
Try changing it to while(ia != A.size() || ib != B.size())

Answer 2

Probably you should avoid the loop altogether:
void combine(vector<int> const& A, vector<int> const& B, vector<int> & R) {
R.resize( A.size() + B.size() );
std::copy( A.begin(), A.end(), R.begin() );
std::copy( B.begin(), B.end(), R.begin()+A.size() );
std::sort( R.begin(), R.end() );
for ( int i = 0; i < R.size(); ++i )
{
cout << "L3[" << i << "] = " << R[i] << endl;
}
}
This is suboptimal as you are first copying and then ordering, but for a small size it will have no impact.
On the actual issues with your code: try to avoid pass-by-value, use resize instead of multiple push_back() to fix the size (note that if the R argument to your function was a non-empty vector then the final size would be bigger than you want). Consider using a return value instead of a reference argument --easier to read. You looped until the first of the counters reached the end, but left the rest of the elements in the other container without copying.
A manual implementation, using iterators would also be simpler:
typedef std::vector<int> vector_t;
vector_t combine( vector_t const & a, vector_t const & b ) {
vector_t r( a.size() + b.size() ); // [*]
vector_t::const_iterator ita = a.begin(), enda = a.end();
vector_t::const_iterator itb = b.begin(), endb = b.end();
vector_t::iterator itr = r.begin();
while ( ita != enda && itb != endb ) {
if ( *ita < *itb )
*itr++ = *ita++;
else
*itr++ = *itb++;
}
if ( ita != enda )
std::copy( ita, enda, itr );
else
std::copy( itb, endb, itr );
return r;
}

Answer 3

I don't know what you're trying to do in the while loop. But if you're just populating the vector R with the elements in A and B, without giving any regard to the order how they're added, then you can use insert function, as:
void::combine(const vector<int> &A, const vector<int> &B, vector<int>& R)
{
R.insert(R.end(), A.begin(), A.end());
R.insert(R.end(), B.begin(), B.end());
}
And if you want to order the vector R, then you can add the following line to the above function:
sort( R.begin(), R.end()); //sorts in increasing order.
You've to #include<algorithm> if you do so. If you want to sort in decreasing order then do this:
bool compare( int a, int b ) { return a > b; }
sort( R.begin(), R.end(), compare); //sorts in decreasing order.

Answer 4

First of all, this smells like a classic merge sort or a piece of it:
http://en.wikipedia.org/wiki/Merge_sort
The objective is to examine elements of the first vector, A, to elements of vector B, and append the elements to a resulting vector R, in order.
Thus if A[i] < B[j], append A[i] to R. With std::vector, there is no need to load the result vector with zeros before starting. See std::vector::push_back().
The following is untested code:
void Sort_Vector(const std::vector& A, const std::vector& B, std::vector& result)
{
unsigned int index_a = 0; // Point to first element in first vector.
unsigned int index_b = 0; // Point to first element in second vector.
result.clear(); // Clear all elements, so size == 0.
while ((index_a < A.size()) && (index_b < B.size())
{
if (A[index_a] < B[index_b])
{
result.push_back(A[index_a]);
++index_a;
}
else // B[index_b] <= A[index_a]
{
result.push_back(B[index_b]);
++index;
}
}
// Append any remaining elements to the result vector.
// Only one of the two loops should be executed,
// the other will fail the comparison expression
// and not execute.
for (; index_a < A.size(); ++index_a)
{
result.push_back(A[index_a]);
}
for (; index_b < B.size(); ++index_b)
{
result.push_back(B[index_b]);
}
return;
}
Notice the difference between my function's signature (parameters) and yours. I'm passing the source vectors as constant data and the result vector is passed by reference which enables my function to modify the caller's result vector.
Also, I'm lazy and using the std::vector::push_back() method. I believe this is more readable than assigning to an element in the result vector. The push_back method implies that I am appending to the vector. The assignment route may catch readers off-guard if they forget that a vector will expand in order resolve the assignment.

Answer 5

that's because of your && condition. as soon as ia == A.size(), the first part of the condition (ia != A.size()) will always evaluate to false and the ib part of your loop won't execute.
to fix it, you can change the condition to : (iA < A.size()) || (iB < B.size())

Answer 6

e.g. with 123 and 456 what is being done right now.
Initialize R to {0,0,0,0,0,0}
until all values from one vector are inserted into R compare candidates and insert smaller one into R, get next candidate from winner list.
and that's all - I don't know that is the purpose of the function if you'll write it I/We can provide solution otherwise just review 2nd point in above form.

Answer 7

The syntax is fine (the code wouldn't compile otherwise), but your loop condition isn't. Given your example A = { 1, 2, 3 } and B = { 4, 5, 6 }, the loop will enter the A[ia] < B[ib] branch three times and increment ia to 3. Once this happens, the first part of the loop condition, ia != A.size() won't be true, and so the loop ends.
Since you already know how many items you need to place in the result vector, I'd recommend replacing your while loop with a simple for one with the same body.