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ask content of the file into array

Posted on 2004-10-14
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Last Modified: 2010-04-01
hi

this is wat my file looks like

6.34 2.0 1.0 10 20 10 10 -20 10 10 10 20 0 20 10 -10 50 -40 50 10

i wanna read that line into my prog
and put each number into array, so that i can call each of them instantly.

this is wat i trying(bleow), but i kina sruck,

any1 can suggest me a code of how to do this.
i'd be really appreciate. (i using : VC++ 6.0)

Thx
Tom

void load(char *filename)
{
      
      ifstream in;
      ofstream out;

      in.open(filename);

      if(in.fail())
      {
            cout << "fail2open\n";
            exit(1);
      }
      else
      {
            cout << "opened";
            cout << "\n\n\n";
      }


      char next;

      while(!in.eof())
      {
            cout << next;
            //cout << "\n";
            
            if(next == 'a')
            {
                  p_a();
            }
      
            if(next == 'b')
            {
                  p_b();
            }
      
            in.get(next);
      
            //in.getline(next,3); //where 100 is the size of the array
            //cout << next << endl;

      }



}
0
Comment
Question by:tom_mk
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7 Comments
 
LVL 13

Accepted Solution

by:
SteH earned 2000 total points
ID: 12306314
Code to fill an array would be.

int i=0;
while (in) { // should be equal to while (!in.eof())
   in >> ar[i];
   ++i;
}

I am not sure what p_a() and p_b () are, os I can't really comment on a solution within your code sample.
0
 
LVL 17

Expert Comment

by:rstaveley
ID: 12306353
You can also use the fact that your ofstream bool operator() returns false when you hit the end of file.

e.g.
--------8<--------
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;

int main(int argc,const char* argv[])
{
      if (argc != 2) return (cerr << "Usage: " << *argv << " {filename}\n"),1;
      ifstream fin(*++argv);if (!argv) return (cerr << "Error: Unable to open " << *argv << '\n'),1;
      vector<double> dList;
      double tmp;
      while (fin >> tmp) /* fin is false at eof */
            dList.push_back(tmp);
      for (int i = 0;i < dList.size();++i)
            cout << "Value[" << i << "] is " << dList[i] << '\n';
}
--------8<--------
0
 
LVL 17

Expert Comment

by:rstaveley
ID: 12306366
Actually SteH's does that too. I should put my reading glasses on.
0
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Author Comment

by:tom_mk
ID: 12306372
according to u guys,

will the space ' ' be filled into the array?
cos i only need the number in the array
0
 
LVL 17

Expert Comment

by:rstaveley
ID: 12306390
By using operator>> you skip the white space.
0
 

Author Comment

by:tom_mk
ID: 12306397
thx man..

i been away from c++ for too long...hahah
0
 
LVL 13

Expert Comment

by:SteH
ID: 12306421
I assumed that "ar" is a array of numbers (doubles in this case). So it can't contain characters any more.

Whitespaces will be removed by all ways of reading numbers from a file or string I know. But as rstaveley pointed out >> is removing them in any case. So if you need strings from that line you need to define

string ar[100];

in my sample or

vector<string> dList; // the name should perhaps become sList in that case
string tmp;

in rstaveleys case.
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