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C++  Pass multidimensional vector  to a fucntion

Posted on 2007-11-21
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Last Modified: 2010-04-21
In visual C++, I have created a multidimensional vector.  vector<vector<int> > vI(10, vector<int>(3));
I would like to know how to pass this vector to the function below.
=========================================================
void display (vector<vector<int> >& vy)       //vy is instance of int vector
{  for (int i = 0; i < vy.size(); i++)           // loops through each row of vy
   {  for (int j = 0; j < vy[i].size(); j++) // loops through each element of each row
          cout << vy[i][j] << " ";           // prints the jth element of the ith row
      cout << endl;
   }
}
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Question by:luling1972
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6 Comments
 
LVL 86

Expert Comment

by:jkr
ID: 20332527
Well, just use

void display (vector<vector<int> >& vy)       //vy is instance of int vector
{  for (int i = 0; i < vy.size(); i++)           // loops through each row of vy
   {  for (int j = 0; j < vy[i].size(); j++) // loops through each element of each row
          cout << vy[i][j] << " ";           // prints the jth element of the ith row
      cout << endl;
   }
}

voif foo()
{
vector<vector<int> > vy;

// fill vector

display(vy);

}
0
 
LVL 7

Accepted Solution

by:
lucky_james earned 125 total points
ID: 20332900
pass it like any other data structure you use.

By using reference :

sign:
void display (vector<vector<int> >& vy)
Call:
display(vy);

By using pointers:
void display (vector<vector<int> >* vy)
Call:
display(&vy);


Hope it helps.
0
 
LVL 11

Expert Comment

by:DeepuAbrahamK
ID: 20333210
Typedef will make it more clear too.
Have a look at this:
http://www.dcs.bbk.ac.uk/~roger/cpp/week7.htm
Best Regards,
DeepuAbrahamK
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LVL 11

Expert Comment

by:DeepuAbrahamK
ID: 20333225
Well, looks like your problem is also from this site :P
0
 
LVL 17

Expert Comment

by:rstaveley
ID: 20333716
> Well, looks like your problem is also from this site :P

Good old Google. 8-)
0
 

Author Closing Comment

by:luling1972
ID: 31410515
This solution solved my problem completely.
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