I was tasked to develop a test software to generate 100Gbps of traffic over 1 TCP socket on Linux (X86-64, kernel 4.15) on a machine with 32GB of RAM.
I developed something like the following code (removed some sanity checks for simplicity) to run on a pair of veth interfaces (one of them is in a different netns).
It generate about 60Gbps on my PC according to bmon,an open source software. To my surprise, if I remove the statement memset(buff, 0, size);, I get about 94Gbps. That's very puzzling.
void test(int sock) {
int size = 500 * 0x100000;
char *buff = malloc(size);
//optional
memset(buff, 0, size);
int offset = 0;
int chunkSize = 0x200000;
while (1) {
offset = 0;
while (offset < size) {
chunkSize = size - offset;
if (chunkSize > CHUNK_SIZE) chunkSize = CHUNK_SIZE;
send(sock, &buff[offset], chunkSize, 0);
offset = chunkSize;
}
}
}
I did some experiments by replacing memset(buff, 0, size); with the following (initialize a portion of the buff),
memset(buff, 0, size * ratio);
if ratio is 0, the throughput is highest at around 94Gbps, as the ratio goes up to 100 percent (1.0), the throughput goes down to around 60Gbps. If the ratio is 0.5 (50%), the throughput goes to about 72Gbps
Appreciate any light you can shed on this.
Edit 1. Here is a relatively complete code that shows the effect that copy on initialized buffer appearing to be slower.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <sys/time.h>
#include <sys/stat.h>
int size = 500 * 0x100000;
char buf[0x200000];
double getTS() {
struct timeval tv;
gettimeofday(&tv, NULL);
return tv.tv_sec tv.tv_usec/1000000.0;
}
void test1(int init) {
char *p = malloc(size);
int offset = 0;
if (init) memset(p, 0, size);
double startTs = getTS();
for (int i = 0; i < 100; i ) {
offset = 0;
while (offset < size) {
memcpy(&buf[0], p offset, 0x200000);
offset = 0x200000;
}
}
printf("took %f secs\n", getTS() - startTs);
}
int main(int argc, char *argv[]) {
test1(argc > 1);
return 0;
}
On my PC (Linux 18.04, Linux 4.15 with 32GB RAM), tried twice without initialization, it took 1.35 seconds. With initialization, it took 3.02 seconds.
Edit 2. Would love to get sendfile (Thanks @marco-bonelli) as fast as sending from the buffer with all 0 (by calloc). I think it's going to be an requirement for my task pretty soon.
CodePudding user response：
I have been running various tests to investigate this surprising result.
I wrote the test program below that combines various operations in the init phase and the loop:
#include <stdio.h>
#include <unistd.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <sys/time.h>
#include <sys/stat.h>
int alloc_size = 500 * 0x100000; // 500 MB
char copy_buf[0x200000]; // 2MB
double getTS() {
struct timeval tv;
gettimeofday(&tv, NULL);
return tv.tv_sec tv.tv_usec/1000000.0;
}
// set a word on each page of a memory area
void memtouch(void *buf, size_t size) {
uint64_t *p = buf;
size /= sizeof(*p);
for (size_t i = 0; i < size; i = 4096 / sizeof(*p))
p[i] = 0;
}
// compute the sum of words on a memory area
uint64_t sum64(const void *buf, size_t size) {
uint64_t sum = 0;
const uint64_t *p = buf;
size /= sizeof(*p);
for (size_t i = 0; i < size; i )
sum = p[i];
return sum;
}
void test1(int n, int init, int run) {
int size = alloc_size;
char msg[80];
int pos = 0;
double times[n 1];
uint64_t sum = 0;
double initTS = getTS();
char *p = malloc(size);
pos = snprintf(msg pos, sizeof msg - pos, "malloc");
if (init > 0) {
memset(p, init - 1, size);
pos = snprintf(msg pos, sizeof msg - pos, " memset%.0d", init - 1);
} else
if (init == -1) {
memtouch(p, size);
pos = snprintf(msg pos, sizeof msg - pos, " memtouch");
} else
if (init == -2) {
sum = sum64(p, size);
pos = snprintf(msg pos, sizeof msg - pos, " sum64");
} else {
/* leave p uninitialized */
}
pos = snprintf(msg pos, sizeof msg - pos, " rep(%d, ", n);
if (run > 0) {
pos = snprintf(msg pos, sizeof msg - pos, "memset%.0d)", run - 1);
} else
if (run < 0) {
pos = snprintf(msg pos, sizeof msg - pos, "sum64)");
} else {
pos = snprintf(msg pos, sizeof msg - pos, "memcpy)");
}
double startTS = getTS();
for (int i = 0; i < n; i ) {
if (run > 0) {
memset(p, run - 1, size);
} else
if (run < 0) {
sum = sum64(p, size);
} else {
int offset = 0;
while (offset < size) {
memcpy(copy_buf, p offset, 0x200000);
offset = 0x200000;
}
}
times[i] = getTS();
}
double firstTS = times[0] - startTS;
printf("%f %f", startTS - initTS, firstTS);
if (n > 2) {
double avgTS = (times[n - 2] - times[0]) / (n - 2);
printf(" / %f", avgTS);
}
if (n > 1) {
double lastTS = times[n - 1] - times[n - 2];
printf(" / %f", lastTS);
}
printf(" secs %s", msg);
if (sum != 0) {
printf(" sum=6llx", (unsigned long long)sum);
}
printf("\n");
free(p);
}
int main(int argc, char *argv[]) {
int n = 4;
if (argc < 2) {
test1(n, 0, 0);
test1(n, 0, 1);
test1(n, 0, -1);
test1(n, 1, 0);
test1(n, 1, 1);
test1(n, 1, -1);
test1(n, 2, 0);
test1(n, 2, 1);
test1(n, 2, -1);
test1(n, -1, 0);
test1(n, -1, 1);
test1(n, -1, -1);
test1(n, -2, 0);
test1(n, -2, 1);
test1(n, -2, -1);
} else {
test1(argc > 1 ? strtol(argv[1], NULL, 0) : n,
argc > 2 ? strtol(argv[2], NULL, 0) : 0,
argc > 3 ? strtol(argv[2], NULL, 0) : 0);
}
return 0;
}
Running it on an old linux box running Debian Linux 3.16.0-11-amd64, I got these timings:
The columns are
- init phase
- first iteration of the loop
- average of the second to penultimate iterations
- last iteration of the loop
- sequence of operations
0.000071 0.242601 / 0.113761 / 0.113711 secs malloc rep(4, memcpy)
0.000032 0.349896 / 0.125809 / 0.125681 secs malloc rep(4, memset)
0.000032 0.190461 / 0.049150 / 0.049210 secs malloc rep(4, sum64)
0.350089 0.186691 / 0.186705 / 0.186548 secs malloc memset rep(4, memcpy)
0.350078 0.125603 / 0.125632 / 0.125531 secs malloc memset rep(4, memset)
0.349931 0.105991 / 0.105859 / 0.105788 secs malloc memset rep(4, sum64)
0.349860 0.186950 / 0.187031 / 0.186494 secs malloc memset1 rep(4, memcpy)
0.349584 0.125537 / 0.125525 / 0.125535 secs malloc memset1 rep(4, memset)
0.349620 0.106026 / 0.106114 / 0.105756 secs malloc memset1 rep(4, sum64) sum=ebebebebebe80000
0.339846 0.186593 / 0.186686 / 0.186498 secs malloc memtouch rep(4, memcpy)
0.340156 0.125663 / 0.125716 / 0.125751 secs malloc memtouch rep(4, memset)
0.340141 0.105861 / 0.105806 / 0.105869 secs malloc memtouch rep(4, sum64)
0.190330 0.113774 / 0.113730 / 0.113754 secs malloc sum64 rep(4, memcpy)
0.190149 0.400483 / 0.125638 / 0.125624 secs malloc sum64 rep(4, memset)
0.190214 0.049136 / 0.049170 / 0.049149 secs malloc sum64 rep(4, sum64)
The timings are consistent with the observations of the OP. I found an explanation that is consistent with the observed timings:
if the first access to a page is a read operation, the timings are substantially better than if the first operation is a write access.
Here are some observations consistent with this explanation:
malloc()for a large 500MB block just makes a system call to map the memory, it does not access this memory andcallocprobably would do exactly the same thing.- if you do not initialize this memory, it still gets mapped in RAM as zero initialized pages for security reasons.
- when you initialize the memory with
memset, the first access to the whole block is a write access, then the timings for the loop are slower - initializing the memory to all bytes
1produces exactly the same timings - if instead I use
memtouch, only writing the first word to zero, I get the same timings in the loop - conversely, if instead of initializing the memory, I compute a checksum, it comes out as zero (which is not guarateed, but expected) and the timings in the loop are faster.
- if no access is performed to the block, then the timings in the loop depend on the operation performed:
memcpyandsum64are faster because the first access is a read access, whilememsetis slower because the first access is a write access.
This seems specific to the linux kernel, I do not observe the same difference on macOS, but with a different processor. This behavior might be specific to older linux kernels and/or CPU and bridge architecture.
FINAL UPDATE: As commented by Peter Cordes, a read from a never-written anonymous page will get every page copy-on-write mapped to the same physical page of zeros, so you can get TLB misses but L1d cache hits when reading it. (Applies to the .bss, and memory from mmap(MAP_ANONYMOUS), like glibc calloc and malloc use for large allocations.) He wrote up some details with perf results for an experiment on Why is iterating though `std::vector` faster than iterating though `std::array`?
This explains why memcpy or just reading from memory that is only implicitly initialized to zero is faster than from memory explicitly written to. A good reason to use calloc() instead of malloc() memset() for sparse data.
CodePudding user response：
Linux uses virtual memory and backs-up that virtual memory with physical memory (allocated as pages) only on demand. When your two example programs request "memory" for a buffer using malloc(), only virtual memory is actually allocated. Only when your program uses that "memory" (e.g. write to that memory buffer) will Linux assign a physical memory page to map to the virtual page. This permits Linux to over-commit memory in a manner very similar to filesystem allocation using sparse files.
When either of your programs initializes the allocated buffer using memset(), that would force assignment of physical pages to correspond to the virtual memory. Perhaps this results in some page swapping during the socket transfer or buffer copying?
But when the memory buffer has not been initialized (and not yet mapped to a physical page), then could there be an optimization of the page fault (due to a read for I/O or copy operation) to just access from a special page (sort of like reading from an unwritten part of a sparse file), and not perform the page assignment?
Based on your question, there does seem to be somekind of page optimization for a page fault. So let's hypothesize that reading virtual memory that has not been written does not trigger an allocation and mapping of physical memory.
To test this hypothesis, we can use the top utility to obtain the virtual versus physical memory usage of your second program. The man page for top describes virtual memory usage is the total of "everything in-use and/or reserved (all quadrants)." Resident memory usage is "anything occupying physical memory which, beginning with Linux-4.5, is the sum of the following three fields:
RSan - quadrant 1 pages, which include any former quadrant 3 pages if modified
RSfd - quadrant 3 and quadrant 4 pages
RSsh - quadrant 2 pages"
When your 2nd program initializes the allocated buffer, this slow version uses 518.564 MB of virtual memory and 515.172 MB of resident memory. The similar memory numbers seems to indicate that the malloc()'d buffer of 500MB is backed-up with physical memory (as it should be).
When your 2nd program does not initialize the allocated buffer, this fast version uses the same 518.564 MB of virtual memory but only 3.192 MB of resident memory. The disparate memory numbers are a good indication that (most if not all of) the malloc()'d buffer of 500 MB is not backed-up with physical memory.
So the hypothesis seems valid.
Peter Cordes' comment confirms that there is such a page-fault optimization: that a "read from a never-written anonymous page will get every page copy-on-write mapped to the same physical page of zeros, so you can get TLB misses but L1d cache hits when reading it."
So your improved transfer rates and copy times seem to be due to reduced overhead from page swapping in the virtual memory subsystem.