how to read integers from a file( novice )

hey all,

I would like to know how one can read integers from a file,
I've have instanced an ifstream, but with fileName.get()
it seems that you can only read chars?

cheers

Paul

quick example, text.txt contains 3 4 5 on the first line.

Code:

---

#include <fstream.h>
 
void main( )
{
  ifstream fileIn( "test.txt" );
  int array[ 3 ];
  int i;

  for( i = 0; i < 3; i++ )
  {
      fileIn >>array[ i ];
      cout << array [ i ] << endl;
  }

  fileIn.close;
}

---

hey mate thx for the reply,
but how can you excactly read an integer when there are a lot of other chars in the same file:


char oldKar, kar, getal;
char in[20];
char uit[20];
int aantal;
while( !invoer.eof() )
{
if( kar == 92 )
{
oldKar = kar;
invoer.get( kar );
if( oldKar == kar ) {
if( invoer >> aantal )
{
for( int i = 0; i < getal ; i++ )
{
uitvoer.put( kar );
}
}
else
{
uitvoer.put( kar );
}
}

ok that was a very simple example, not very helpful it seems :(

Do you know the format of the file? Are the number separated from the characters by anything? If its is just contained as a single long string you will have to parse the whole string to find the integers :)

yeah I know, but basically it's an assignment for school, where you have to decode and encode a text file encoding works fine,
but it's the decoding that give me a headache, say you have this
string in the file:
e4

now this needs to be decoded to:

eeee

but I may only use the "get(.....)" method :(

so I guess I have to find something else?

Assuming that each item is stored in the file like so:

c4a16g12

You can use something like this to decode it:

Code:

---

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cctype>

void print_n ( char item, int n )
{
  for ( int i = 0; i < n; ++i )
    std::cout<< item <<std::flush;
}

int main()
{
  std::ifstream input ( "test.txt" );

  if ( !input.is_open() ) {
    std::cerr<<"Error opening file for reading\n";
    exit ( EXIT_FAILURE );
  }

  int count = 0;
  char i_val, save;

  while ( input.get ( i_val ) ) {
    if ( !isdigit ( i_val ) ) {
      print_n ( save, count );

      count = 0;
      save = i_val;
    }
    else
      count = 10 * count + ( i_val - '0' );

    if ( input.peek() == EOF )
      print_n ( save, count );
  }

  input.close();

  return 0;
}

---

If the strings are more complicated or surrounded by whitespace then you will have to handle several special cases. This job is dramatically harder since you are restricted to get(), but it can be done.

>void main( )
I officially demote you from a C++ black belt to a C++ white belt for using void main endo. ;)

Quote:

---

Originally posted by Prelude

>void main( )
I officially demote you from a C++ black belt to a C++ white belt for using void main endo. ;)

---

I get lazy when writing examples, I shouldn't really but if they want my help they get what they're given :)

>> get lazy when writing examples, I shouldn't really but if they want my help they get what they're given

Code:

---

int = (keystroke * 3);
void = (keystroke * 4);

---

I think you're a bit hazy on how to be lazy. At the moment you are listening to the master on laziness -- you're on my turf!