inspired by:
https://mazzo.li/posts/fast-pipes.html
I was curious to see how things stacked up on plan 9.
The machines are apples to oranges (my 9front box
is a 2015-ish era Zbox with a Intel Core i5-7300HQ
processor, and my work machine is a Linux with a
12 core Ryzen 9 3900X).
Transferring 128 gigs instead of the 10 in the initial
benchmark, here are the results:
# 9front 595684fd8a2f08e12d5df48152d93fb8ab800fe3 amd64
% time rc -c '6.write | 6.read'
0.33u 23.28s 11.64r rc -c 6.write | 6.read
# Linux 5.13.0-40-generic #45~20.04.1-Ubuntu SMP
$ time sh -c './write | ./read'
real 0m32.039s
user 0m0.231s
sys 0m34.576s
here's the plan 9 version of the program:
write.c:
#include <u.h>
#include <libc.h>
void
main(void)
{
usize sz;
char* buf;
sz = 1 << 18;
buf = malloc(sz);
memset((void*)buf, 'X', sz);
while(1){
n = write(1, buf, sz) != sz)
break;
exits(nil);
}
read.c:
#include <u.h>
#include <libc.h>
enum {
KiB = 1024ULL,
MiB = 1024*KiB,
GiB = 1024*MiB,
};
void
main(void)
{
vlong sz, r, n;
char* buf;
r = 0;
sz = 1 << 18;
buf = malloc(sz);
while(r <= 128ULL*GiB){
n = read(0, buf, sz);
if(n <= 0)
break;
r += n;
}
exits(nil);
}
And the linux version:
write.c:
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
int
main(int argc, char **argv)
{
size_t sz;
char* buf;
sz = 1 << 18;
buf = malloc(sz);
memset(buf, 'X', sz);
while(1)
if(write(1, buf, sz) != sz)
break;
return 0;
}
read.c:
#include <stdlib.h>
#include <unistd.h>
enum {
KiB = 1024ULL,
MiB = 1024*KiB,
GiB = 1024*MiB,
};
int
main(int argc, char **argv)
{
ssize_t sz, r, n;
char* buf;
r = 0;
sz = 1 << 18;
buf = malloc(sz);
while(r <= 128ULL*GiB){
n = read(0, buf, sz);
if(n <= 0)
break;
r += n;
}
return 0;
}
------------------------------------------
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