Here's a simple memset bandwidth benchmark:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
int main()
{
unsigned long n, r, i;
unsigned char *p;
clock_t c0, c1;
double elapsed;
n = 1000 * 1000 * 1000; /* GB */
r = 100; /* repeat */
p = calloc(n, 1);
c0 = clock();
for(i = 0; i < r; ++i) {
memset(p, (int)i, n);
printf("%4d/%4ld\r", p[0], r); /* "use" the result */
fflush(stdout);
}
c1 = clock();
elapsed = (c1 - c0) / (double)CLOCKS_PER_SEC;
printf("Bandwidth = %6.3f GB/s (Giga = 10^9)\n", (double)n * r / elapsed / 1e9);
free(p);
}
On my system (details below) with a single DDR3-1600 memory module, it outputs:
Bandwidth = 4.751 GB/s (Giga = 10^9)
This is 37% of the theoretical RAM speed: 1.6 GHz * 8 bytes = 12.8 GB/s
On the other hand, here's a similar "read" test:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
unsigned long do_xor(const unsigned long* p, unsigned long n)
{
unsigned long i, x = 0;
for(i = 0; i < n; ++i)
x ^= p[i];
return x;
}
int main()
{
unsigned long n, r, i;
unsigned long *p;
clock_t c0, c1;
double elapsed;
n = 1000 * 1000 * 1000; /* GB */
r = 100; /* repeat */
p = calloc(n/sizeof(unsigned long), sizeof(unsigned long));
c0 = clock();
for(i = 0; i < r; ++i) {
p[0] = do_xor(p, n / sizeof(unsigned long)); /* "use" the result */
printf("%4ld/%4ld\r", i, r);
fflush(stdout);
}
c1 = clock();
elapsed = (c1 - c0) / (double)CLOCKS_PER_SEC;
printf("Bandwidth = %6.3f GB/s (Giga = 10^9)\n", (double)n * r / elapsed / 1e9);
free(p);
}
It outputs:
Bandwidth = 11.516 GB/s (Giga = 10^9)
I can get close to the theoretical limit for read performance, such as XORing a large array, but writing appears to be much slower. Why?
OS Ubuntu 14.04 AMD64 (I compile with gcc -O3. Using -O3 -march=native makes the read performance slightly worse, but does not affect memset)
CPU Xeon E5-2630 v2
RAM A single "16GB PC3-12800 Parity REG CL11 240-Pin DIMM" (What it says on the box) I think that having a single DIMM makes performance more predictable. I'm assuming that with 4 DIMMs, memset will be up to 4 times faster.
Motherboard Supermicro X9DRG-QF (Supports 4-channel memory)
Additional system: A laptop with 2x 4GB of DDR3-1067 RAM: read and write are both about 5.5 GB/s, but note that it uses 2 DIMMs.
P.S. replacing memset with this version results in exactly the same performance
void *my_memset(void *s, int c, size_t n)
{
unsigned long i = 0;
for(i = 0; i < n; ++i)
((char*)s)[i] = (char)c;
return s;
}
