Moving through text file c++ - c++

Question

I'm trying to save numbers from first txt file to second one in reversed order.
To be clear, inside 1st txt I have typed numbers from 1 to 10 (decimal notation). When I try to count them, I get 5 or 7, depending on what's between them (space or enter).
Then, another error is that inside 2nd txt program saves as much "0s" as dl's variable value is equal to instead of loaded numbers in reversed order.
I paste the whole code, because I don't know file operation rules good enough to determine which exact part could be the source of problem. Thank You in advance.
#include <fstream>
#include <iostream>
using namespace std;
int main() {
fstream plik1;
plik1.open("L8_F3_Z2a.txt", ios::in | ios::binary);
fstream plik2;
plik2.open("L8_F3_Z2b.txt", ios::out);
if(!plik1.good() || !plik2.good()) {
cout << "file(s) invalid" << endl;
return 1;
}
plik1.seekg(0, ios::end);
int dl = plik1.tellg() / sizeof(int);
cout << "length = " << dl << endl;
int a;
for(int i = 0; i < dl; i++) {
plik1.seekg((i + 1) * sizeof(int), ios::end);
plik1 >> a;
plik2 << a;
cout << i + 1 << ". a = " << a << endl;
}
plik1.close();
plik2.close();
return 0;
}
edit the output is:
length = 7
1. a = 0
2. a = 0
3. a = 0
4. a = 0
5. a = 0
6. a = 0
7. a = 0
--------------------------------
Process exited after 0.03841 seconds with return value 0
Press any key to continue . . .

Answer 1

Problem
When a file is encoded as text the binary size of the data is irrelevant.
int dl = plik1.tellg() / sizeof(int);
will get you the side of the file in integers, but the file isn't storing integers. It is storing a stream of characters. Say for example the file holds one number:
12345
which is five characters long. Assuming the file is using good ol ASCII, that's 5 bytes. When 12345 is converted to an int it will probably be 4 or 8 bytes and almost certainly not 5 bytes. Assuming the common 32 bit (4 byte) int
int dl = plik1.tellg() / sizeof(int);
int dl = 5 / 4;
int dl = 1;
Yay! It worked! But only by the grace of whatever deity or cosmic entity you worship. Or don't worship. I'm not going to judge. To show why you can't count on this, lets look at
123
this is three characters and 3 bytes, so
int dl = plik1.tellg() / sizeof(int);
int dl = 3 / 4;
int dl = 0;
Whoops.
Similarly
1 2 3 4 5
is five numbers. The file length will probably be the sum of one byte per digit and one byte per space, 9 bytes.
Where this gets weird is some systems, looking at you Windows, use a two character end of line marker, carriage return and a line feed. This means
1
2
3
4
5
will sum up to 13 bytes.
This is why you see a different size depending on whether the numbers are separated with spaces or newlines.
Solution
The only way to find out how many numbers are in the file is to read the file, convert the contents to numbers, and count the numbers as you find them.
How to do that:
int num;
int count = 0;
while (plik1 >> num) // read numbers until we can't read any more
{
count++;
}
From this you can determine the size of the array you need. Then you rewind the file, seek back to the beginning, allocate the array and read the file AGAIN into the array. This is dumb. File IO is painfully slow. You don't want to do it twice. You want to read the file once and store as you go without caring how many numbers are in the file.
Fortunately there are a number of tools built into C++ that do exactly that. I like std::vector
std::vector<int> nums;
int num;
while (plik1 >> num)
{
nums.push_back(num);
}
vector even keeps count for you.
Next you could
std::reverse(nums.begin(), nums.end());
and write the result back out.
for (int num: nums)
{
plik2 << num << ' ';
}
Documentation for std::reverse
If your instructor has a no vector policy, and unfortunately many do, your best bet is to write your own simple version of vector. There are many examples of how to do this already on Stack Overflow.
Addendum
In binary 5 integers will likely be 20 or 40 bytes no matter how many digits are used and no separators are required.
It sounds like storing data as binary is the bees knees, right? Like it's going to be much easier.
But it's not. Different computers and different compilers use different sizes for integers. All you are guaranteed is an int is at least 2 bytes and no larger than a long. All of the integer types could be exactly the same size at 64 bits. Blah. Worse, not all computers store integers in the same order. Because it's easier to do some operations if the number is stored backwards, guess what? Often the number is stored backwards. You have to be very, very careful with binary data and establish a data protocol (search term for more on this topic: Serialization) that defines the how the data is to be interpreted by everyone.