What does "file system outputs" mean with time -v?

Question

What is 'file system outputs' counting when using the Linux 'time' command with dd?

It doesn't equal dd 'count' (presumably the number of calls to fwrite?), nor the size of the output in 4096-byte pages (which should be 1024000 in this example).

An example:

> /usr/bin/time -v dd if=/dev/zero of=/tmp/dd.test bs=4M count=1000
1000+0 records in
1000+0 records out
4194304000 bytes (4.2 GB) copied, 4.94305 s, 849 MB/s
Command being timed: "dd if=/dev/zero of=/tmp/dd.test bs=4M count=1000"
User time (seconds): 0.00
System time (seconds): 4.72
Percent of CPU this job got: 95%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:04.94
Average shared text size (kbytes): 0
Average unshared data size (kbytes): 0
Average stack size (kbytes): 0
Average total size (kbytes): 0
Maximum resident set size (kbytes): 5040
Average resident set size (kbytes): 0
Major (requiring I/O) page faults: 0
Minor (reclaiming a frame) page faults: 1322
Voluntary context switches: 32
Involuntary context switches: 15
Swaps: 0
File system inputs: 240
File system outputs: 8192000
Socket messages sent: 0
Socket messages received: 0
Signals delivered: 0
Page size (bytes): 4096
Exit status: 0

Answer 1

The command time is printing out values from the rusage struct (see getrusage(2)).

And according to the source:

/*
 * We approximate number of blocks, because we account bytes only.
 * A 'block' is 512 bytes
 */
static inline unsigned long task_io_get_oublock(const struct task_struct *p)
{
    return p->ioac.write_bytes >> 9;
}

So (at least on Linux) "File system outputs" in time output is the total number of bytes written / 512.