[Vermiculus]

void DisplayVector( vector<string> args )
{
	cout << args[0] << "\t" << args[1] << endl;
}

void DisplayAllVectors( vector<vector<string>> args )
{
	for each (vector<string> vect in args)
	{
		DisplayVector(vect);
	}
}

That code seems to work well, but there is a semantic error in it somewhere. I don't know why, but when I run the code it seems to simply cycle through the master vector... It's easier to see what I mean when it runs.. I have attached the source code to the entire program, but the problem definitely lies somewhere in those two functions.

I have attempted to erase vect (args) after I am done with it in DisplayVect, but that ends up with a nasty error. Logically, I shouldn't have to erase it at all since it should move on to the next one in DisplayAllVectors, but whoever said computers are logical...harhar

Vermiculus, on 25 Nov, 2009 - 01:23 PM, said:

I have attached the source code to the entire program, but the problem definitely lies somewhere in those two functions.

For some reason it did not attached, so I am trying again... hopefully it'll work...


By the way, it is possible to jerryrig the error and make it work by replacing it with the following

int derp = 0;
void DisplayVector( vector<string> args )
{
	cout << args[derp] << "\t" << args[derp+1] << endl;
	derp++;
}

void DisplayAllVectors( vector<vector<string>> args )
{
	for each (vector<string> vect in args)
	{
		DisplayVector(vect);
	}
}

But it's still confusing me why this error would occur in the first place.

Attached File(s)

•  Program.txt (3.55K)
  Number of downloads: 92

This post has been edited by Vermiculus: 25 November 2009 - 02:32 PM

[GWatt]

Your question doesn't seem to have anything to do with popping elements. When you pop a data structure, most likely a stack, you remove and return the last element you 'pushed' on to it.

vector vec;
for (int i = 0; i < 9; i++)
	vec.push_back(i);

int a = vec.pop_back();
// will get 9 as that was the last item 'pushed' on

Your problem is that your multidimensional vector is not printing properly. I would suggest using nested for loops in a single function instead of using different functions to print out the different layers of the vector.

This post has been edited by GWatt: 25 November 2009 - 03:07 PM

[Silhouette]

If you want to pop from the front, you need a queue or deque data structure.
Plus, if all you're trying to do is display the elements in a vector, you could just cycle through the elements in a for loop, i.e.

cout << myVector[i] << endl;

.
It also has the added benefit of being non-destructive, so you can display your vector elements AND keep using it.

[Vermiculus]

Ah, I probably should not have used the word pop. I realize that could be confusing ..

And yes, I am definitely over complicating this (which technically is the point of our project ), but at this point I am simply curious: how can you destroy the first element of a vector so that the latter vector elements are bumped up:

//semipsuedo
vector<int> vect;
vect.push_back(int1);
vect.push_back(int2);
vect.push_back(int3);
//vect now contains [int1, int2, int3]
vect.magicalFunctionKill(int1); // or an index or some kind of demonstrative
//vect now contains [int2, int3]

Is it possible to reverse the order of the vector, like a string.reverse()? Then I could use pop_back and then reverse it again...

[Bench]

the easiest way is to use the vector::erase function. it takes an iterator which points at the element you want to erase. the first element will be the same as vector::begin, so you can do

vect.erase( vect.begin() ); 

Though, as others have remarked, its easier and (usually) more efficient to use a std::deque ("deque" means Double-Ended Queue) instead of a std::vector if you want to push/pop at both ends. The only thing you lose in a deque is the guarantee that elements are stored contiguously like an array- behind the scenes, a deque is normally "paged" instead - but semantically you retain all the benefits of a vector (notably random access).


as for reversing the order of a vector, you can use an STL algorithm

#include <algorithm>
    // etc..
    std::reverse( vect.begin(), vect.end() ); 

This post has been edited by Bench: 26 November 2009 - 05:58 AM