Getting characters past a certain point in a file in C - c

Question

I want to take all characters past location 900 from a file called WWW, and put all of these in an array:
//Keep track of all characters past position 900 in WWW.
int Seek900InWWW = lseek(WWW, 900, 0); //goes to position 900 in WWW
printf("%d \n", Seek900InWWW);
if(Seek900InWWW < 0)
printf("Error seeking to position 900 in WWW.txt");
char EverythingPast900[appropriatesize];
int NextRead;
char NextChar[1];
int i = 0;
while((NextRead = read(WWW, NextChar, sizeof(NextChar))) > 0) {
EverythingPast900[i] = NextChar[0];
printf("%c \n", NextChar[0]);
i++;
}
I try to create a char array of length 1, since the read system call requires a pointer, I cannot use a regular char. The above code does not work. In fact, it does not print any characters to the terminal as expected by the loop. I think my logic is correct, but perhaps a misunderstanding of whats going on behind the scenes is what is making this hard for me. Or maybe i missed something simple (hope not).

Answer 1

If you already know how many bytes to read (e.g. in appropriatesize) then just read in that many bytes at once, rather than reading in bytes one at a time.
char everythingPast900[appropriatesize];
ssize_t bytesRead = read(WWW, everythingPast900, sizeof everythingPast900);
if (bytesRead > 0 && bytesRead != appropriatesize)
{
// only everythingPast900[0] to everythingPast900[bytesRead - 1] is valid
}

Answer 2

I made a test version of your code and added bits you left out. Why did you leave them out?
I also made a file named www.txt that has a hundred lines of "This is a test line." in it.
And I found a potential problem, depending on how big your appropriatesize value is and how big the file is. If you write past the end of EverythingPast900 it is possible for you to kill your program and crash it before you ever produce any output to display. That might happen on Windows where stdout may not be line buffered depending on which libraries you used.
See the MSDN setvbuf page, in particular "For some systems, this provides line buffering. However, for Win32, the behavior is the same as _IOFBF - Full Buffering."
This seems to work:
#include <fcntl.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <stdio.h>
int main()
{
int WWW = open("www.txt", O_RDONLY);
if(WWW < 0)
printf("Error opening www.txt\n");
//Keep track of all characters past position 900 in WWW.
int Seek900InWWW = lseek(WWW, 900, 0); //goes to position 900 in WWW
printf("%d \n", Seek900InWWW);
if(Seek900InWWW < 0)
printf("Error seeking to position 900 in WWW.txt");
int appropriatesize = 1000;
char EverythingPast900[appropriatesize];
int NextRead;
char NextChar[1];
int i = 0;
while(i < appropriatesize && (NextRead = read(WWW, NextChar, sizeof(NextChar))) > 0) {
EverythingPast900[i] = NextChar[0];
printf("%c \n", NextChar[0]);
i++;
}
return 0;
}

Answer 3

As stated in another answer, read more than one byte. The theory behind "buffers" is to reduce the amount of read/write operations due to how slow disk I/O (or network I/O) is compared to memory speed and CPU speed. Look at it as if it is code and consider which is faster: adding 1 to the file size N times and writing N bytes individually, or adding N to the file size once and writing N bytes at once?
Another thing worth mentioning is the fact that read may read fewer than the number of bytes you requested, even if there is more to read. The answer written by #dreamlax illustrates this fact. If you want, you can use a loop to read as many bytes as possible, filling the buffer. Note that I used a function, but you can do the same thing in your main code:
#include <sys/types.h>
/* Read from a file descriptor, filling the buffer with the requested
* number of bytes. If the end-of-file is encountered, the number of
* bytes returned may be less than the requested number of bytes.
* On error, -1 is returned. See read(2) or read(3) for possible
* values of errno.
* Otherwise, the number of bytes read is returned.
*/
ssize_t
read_fill (int fd, char *readbuf, ssize_t nrequested)
{
ssize_t nread, nsum = 0;
while (nrequested > 0
&& (nread = read (fd, readbuf, nrequested)) > 0)
{
nsum += nread;
nrequested -= nread;
readbuf += nread;
}
return nsum;
}
Note that the buffer is not null-terminated as not all data is necessarily text. You can pass buffer_size - 1 as the requested number of bytes and use the return value to add a null terminator where necessary. This is useful primarily when interacting with functions that will expect a null-terminated string:
char readbuf[4096];
ssize_t n;
int fd;
fd = open ("WWW", O_RDONLY);
if (fd == -1)
{
perror ("unable to open WWW");
exit (1);
}
n = lseek (fd, 900, SEEK_SET);
if (n == -1)
{
fprintf (stderr,
"warning: seek operation failed: %s\n"
" reading 900 bytes instead\n",
strerror (errno));
n = read_fill (fd, readbuf, 900);
if (n < 900)
{
fprintf (stderr, "error: fewer than 900 bytes in file\n");
close (fd);
exit (1);
}
}
/* Read a file, printing its contents to the screen.
*
* Caveat:
* Not safe for UTF-8 or other variable-width/multibyte
* encodings since required bytes may get cut off.
*/
while ((n = read_fill (fd, readbuf, (ssize_t) sizeof readbuf - 1)) > 0)
{
readbuf[n] = 0;
printf ("Read\n****\n%s\n****\n", readbuf);
}
if (n == -1)
{
close (fd);
perror ("error reading from WWW");
exit (1);
}
close (fd);
I could also have avoided the null termination operation and filled all 4096 bytes of the buffer, electing to use the precision part of the format specifiers of printf in this case, changing the format specification from %s to %.4096s. However, this may not be feasible with unusually large buffers (perhaps allocated by malloc to avoid stack overflow) because the buffer size may not be representable with the int type.
Also, you can use a regular char just fine:
char c;
nread = read (fd, &c, 1);
Apparently you didn't know that the unary & operator gets the address of whatever variable is its operand, creating a value of type pointer-to-{typeof var}? Either way, it takes up the same amount of memory, but reading 1 byte at a time is something that normally isn't done as I've explained.

Answer 4

Mixing declarations and code is a no no. Also, no, that is not a valid declaration. C should complain about it along the lines of it being variably defined.
What you want is dynamically allocating the memory for your char buffer[]. You'll have to use pointers.
http://www.ontko.com/pub/rayo/cs35/pointers.html
Then read this one.
http://www.cprogramming.com/tutorial/c/lesson6.html
Then research a function called memcpy().
Enjoy.
Read through that guide, then you should be able to solve your problem in an entirely different way.
Psuedo code.
declare a buffer of char(pointer related)
allocate memory for said buffer(dynamic memory related)
Find location of where you want to start at
point to it(pointer related)
Figure out how much you want to store(technically a part of allocating memory^^^)
Use memcpy() to store what you want in the buffer