with vector, how to make a function that returns the container number

[jasonSung]

I have a vector, say, DL.

And i have a function

int ChooseVertex(const vector<int>& DL){
	// given DL, return the vertex with minimum element
	vector<int>::iterator min_DL;
	min_DL = min_element(DL.begin(), DL.end());
	return min_DL; 
}

when i called the function like this

v = ChooseVertex(DL);

I got an error on line 4,

error C2679: binary '=' : no operator found which takes a right-hand operand of type 'std::_Vector_const_iterator<_Ty,_Alloc>' (or there is no acceptable conversion)

Does anybody know the reason? Thank you in advance. :)

[Disch]

because you have a const vector, you need to use a const_iterator:

vector<int>::const_iterator min_DL;   // <- 

Also note that you can't return an iterator as an int, so you'll probably have to change the return type from int to vector<int>::const_iterator

[jasonSung]

Thank you.

But I need to return 'min_DL' as integer because in the main() I will store this return value in a vector<int> container.

Is this possible?

[Disch]

not with iterators. No way to get an integer index from an iterator as far as I know.

You'll have to use min_element another way, and use some pointer math:

int ChooseVertex(const vector<int>& DL)
{
    const int* start = &DL[0];
    return min_element(start,start + DL.size()) - start;
}

[jasonSung]

wow!! it works!! you are so smart!

But i have another question.

what does &DL[0] mean? Is this indicate the address of the first element of vector DL?

I thought in vector container, the index begins from 1, unlike array whose index begins from 0.

[jsmith]

More generically

#include <iterator>  // std::distance

return std::distance( DL.begin(), std::min_element( DL.begin(), DL.end() );

This is container and iterator agnostic. Above works only for vector.

[Disch]

oh crap. Nice.

I forgot <iterator> existed XD

EDIT: but wait... wouldn't that step through the vector element at a time?

EDIT 2: of course! Unless it's specialized!

templates are awesome

[jasonSung]

thank you guys. You are awesome!