while not '\n'

suppose you will get the average of N numbers.
the input would just be the N numbers

e.g.

1 2 3 4
2.5

1 2 3 4 5
3

my problem is how to terminate when the enter is pressed.

Code:

while ( \* right code *\ != '\n' )
{
   cin >> x;
   ave+=x;
   n++;
}
cout<< ave/n;

You should use std::getline for that:
getline - C++ Reference

First read a line, then parse it...
To be honest I consider this to be one great weakness in the C++ streams. I used to be a C coder, and I just never got the hang of those C++ streams, even though I know most of the stl....

theres 2 ways to do this that come to mind.

one of them involve stdio.h

(the main idea is that you know how many numbers (N) there are)

Code:

#include <stdio.h>

const int n = 5;

int main()
{

double avg;
for(int i=0;i<n;++i)
{
  int input;
  scanf("%d", &input);
  avg+=input;
}
avg=avg/n;
printf("%f",avg);

return 0;
}

and

Code:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

const int n = 5;

int main()
{

string input;
double avg;
getline(cin,input);
istringstream iss(input);
for(int i=0;i<n;++i)
{
  int token;
  iss >> token;
  avg+=token;
}
avg = avg / n;

cout << avg;

return 0;
}

...now I guess if you dont know N before hand..

try this, im not 100% this is correct though

Code:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;
int main()
{
int n = 0;
string input;
double avg;
getline(cin,input);
istringstream iss(input);

int token;
while(  iss >> token )
{
  avg+=token;
  ++n;
}
avg = avg / n;

cout << avg;

return 0;
}

EDIT: i tried it, it works

@rodrigorules:
If you wouldn't have forgotten to initialize "avg" to 0 it would even be legal C++ code :P (I think, that's the only bug I can spot by just looking at it for a few seconds).