Originally Posted by Ancient Dragon

lines 8 and 9. Files can not be opened outside a function like you have it. Move both those lines down to within main() (line 14).

line 16: eof() will not give you the file size, it just tells you that the end-of-file has been reached and nothing more. To get the file size

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// set file pointer to end-of-file
infile.seekp(0, ios::end);
// get file location
N = infile.tellg();

// set file pointer to end-of-file infile.seekp(0, ios::end); // get file location N = infile.tellg();

line 22: Not only doesn't that work with most compilers (I think it will when the newest c++ standards are implemented), but its also wrong. The value of N (see above) will not tell you the number of integers in the file, but how many characters are in the file. For example the number 100 will be counted as 4 or 5 depending on the operating system (3 digits plus '\n').

To get the actual number of integers in the file you will have to read each line and count them as you go along.

line 25: eof() doesn't work like that either because it will cause your program to process the last line twice. A better solution is like this

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int i = 0;
while( infile>>P[i]>>V[i] )
{
   // blabla
 
   ++i;
}

int i = 0; while( infile>>P[i]>>V[i] ) { // blabla ++i; }

Originally Posted by Narue

>How do we determine the value of n?
If you can't use dynamic memory (and I'm assuming any alternative that uses dynamic memory, like std::vector), your only option is to make the array large enough to handle any reasonable number of records:

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int P[1000], T[1000], V[1000];

int P[1000], T[1000], V[1000];

Of course, that size would pretty much be an arbitrary choice, and there are two immediate problems that arise from it:

1. You potentially waste a *lot* of space.
2. You have little choice but to ignore any records beyond N unless you're able to get clever and manage chunks of the file instead of the whole thing all at once.