问题
So I have this question from my professor, and I can not figure out why vector2 is faster and has less cache misses than vector1.
Assume that the code below is a valid compilable C code.
Vector2:
void incrementVector2(INT4* v, int n) {
for (int k = 0; k < 100; ++k) {
for (int i = 0; i < n; ++i) {
v[i] = v[i] + 1;
}
}
}
Vector1:
void incrementVector1(INT4* v, int n) {
for (int i = 0; i < n; ++i) {
for (int k = 0; k < 100; ++k) {
v[i] = v[i] + 1;
}
}
}
NOTE: INT4 means the integer is 4 Bytes in size.
In terms of CPU specs: cache size = 12288KB, line size=64B and only considering this single cache interacting with main memory.
Question
Why does vector2 have a faster runtime than vector1? And why vector1 will have more cache misses than vector2?
Me and a few classmates worked on this for a while and couldn't figure it out. We thought it could be due to the fact that vector2 has better spatial locatlity, since the array is been accessed by the inner loop constantly, while in vector1, only one element is accessed at a time?
we are not sure if this is correct, and also not sure how to bring cache lines in to this either.
回答1:
We thought it could be due to the fact that vector2 has better spatial locatlity, since the array is been accessed by the inner loop constantly, while in vector1, only one element is accessed at a time?
Well, both codes have the same accessing pattern, iterating over the array v with a stride of 1. Cache spacial-locality-wise both codes are the same. However, the second code:
void incrementVector1(INT4* v, int n) {
for (int i = 0; i < n; ++i) {
for (int k = 0; k < 100; ++k) {
v[i] = v[i] + 1;
}
}
}
Has a better temporal-locality because you access the same element 100 times, whereas in:
void incrementVector2(INT4* v, int n) {
for (int k = 0; k < 100; ++k) {
for (int i = 0; i < n; ++i) {
v[i] = v[i] + 1;
}
}
}
you only access it once on every 'n' iterations.
So either you did a mistake, your teacher is playing some kind of strange game or I am missing something obvious.
回答2:
Are you sure that you mean that incrementVector2 is faster?
As dreamcast pointed out, incrementVector1 has better temporal locality, so it should be faster.
And, from benchmarks, 1 is faster than 2
50x faster for an array size of 1000.
This is surprising because the array is small enough to become [and stay] cache hot.
And, incrementVector1 has a much smaller disassembled code size than incrementVector2 (at -O3).
incrementVector1 is 105 bytes and incrementVector2 is 203 bytes.
This is somewhat surprising since I would think that although there is some variation, I would not expect such a large discrepancy [See below].
Here is the program output:
0.000002238 incrementVector1
0.000019535 incrementVector2
8.729x slower
0.000000280 incrementVector1
0.000022454 incrementVector2
80.193x slower
0.000000452 incrementVector1
0.000019617 incrementVector2
43.400x slower
0.000000377 incrementVector1
0.000020632 incrementVector2
54.727x slower
0.000000361 incrementVector1
0.000022612 incrementVector2
62.637x slower
Here's the program I used:
#include <stdio.h>
#include <time.h>
typedef long long tsc_t;
typedef int INT4;
tsc_t
tscget(void)
{
struct timespec ts;
tsc_t tsc;
clock_gettime(CLOCK_MONOTONIC,&ts);
tsc = ts.tv_sec;
tsc *= 1000000000;
tsc += ts.tv_nsec;
return tsc;
}
double
tscsec(tsc_t tsc)
{
double sec;
sec = tsc;
sec /= 1e9;
return sec;
}
typedef void (*incfnc_p)(INT4 *v, int n);
void
incrementVector2(INT4 *v, int n)
{
for (int k = 0; k < 100; ++k) {
for (int i = 0; i < n; ++i) {
v[i] = v[i] + 1;
}
}
}
void
incrementVector1(INT4 *v, int n)
{
for (int i = 0; i < n; ++i) {
for (int k = 0; k < 100; ++k) {
v[i] = v[i] + 1;
}
}
}
INT4 v[1000] = { 0 };
#define DOFNC(_fnc) \
dofnc(_fnc,#_fnc)
tsc_t
dofnc(incfnc_p fnc,const char *sym)
{
tsc_t tscbeg;
tsc_t tscend;
tscbeg = tscget();
fnc(v,sizeof(v) / sizeof(v[0]));
tscend = tscget();
tscend -= tscbeg;
printf("%.9f %s\n",tscsec(tscend),sym);
return tscend;
}
void
dotest(void)
{
tsc_t tsc1 = DOFNC(incrementVector1);
tsc_t tsc2 = DOFNC(incrementVector2);
double ratio;
const char *tag;
if (tsc1 > tsc2) {
tag = "faster";
ratio = tsc1;
ratio /= tsc2;
}
else {
tag = "slower";
ratio = tsc2;
ratio /= tsc1;
}
printf("%.3fx %s\n",ratio,tag);
}
int
main(void)
{
for (int testno = 1; testno <= 5; ++testno) {
printf("\n");
dotest();
}
return 0;
}
Here is the disassembly:
00000000004011c0 <incrementVector2>:
4011c0: 85 f6 test %esi,%esi
4011c2: 0f 8e be 00 00 00 jle 401286 <L06>
4011c8: 89 f2 mov %esi,%edx
4011ca: 41 89 f3 mov %esi,%r11d
4011cd: 66 0f 6f 0d 8b 0e 00 movdqa 0xe8b(%rip),%xmm1 # 402060 <__dso_handle+0x58>
4011d4: 00
4011d5: 49 89 f8 mov %rdi,%r8
4011d8: c1 ea 02 shr $0x2,%edx
4011db: 44 8d 56 ff lea -0x1(%rsi),%r10d
4011df: 41 83 e3 fc and $0xfffffffc,%r11d
4011e3: 41 b9 01 00 00 00 mov $0x1,%r9d
4011e9: 48 c1 e2 04 shl $0x4,%rdx
4011ed: 48 01 fa add %rdi,%rdx
4011f0:L00 41 83 fa 02 cmp $0x2,%r10d
4011f4: 0f 86 8d 00 00 00 jbe 401287 <L07>
4011fa: 48 89 f8 mov %rdi,%rax
4011fd:L01 f3 0f 6f 00 movdqu (%rax),%xmm0
401201: 48 83 c0 10 add $0x10,%rax
401205: 66 0f fe c1 paddd %xmm1,%xmm0
401209: 0f 11 40 f0 movups %xmm0,-0x10(%rax)
40120d: 48 39 d0 cmp %rdx,%rax
401210: 75 eb jne 4011fd <L01>
401212: 44 89 d8 mov %r11d,%eax
401215: 44 39 de cmp %r11d,%esi
401218: 74 22 je 40123c <L03>
40121a:L02 48 63 c8 movslq %eax,%rcx
40121d: 83 04 8f 02 addl $0x2,(%rdi,%rcx,4)
401221: 8d 48 01 lea 0x1(%rax),%ecx
401224: 39 ce cmp %ecx,%esi
401226: 7e 14 jle 40123c <L03>
401228: 48 63 c9 movslq %ecx,%rcx
40122b: 83 c0 02 add $0x2,%eax
40122e: 83 04 8f 02 addl $0x2,(%rdi,%rcx,4)
401232: 39 c6 cmp %eax,%esi
401234: 7e 06 jle 40123c <L03>
401236: 48 98 cltq
401238: 83 04 87 02 addl $0x2,(%rdi,%rax,4)
40123c:L03 41 8d 41 01 lea 0x1(%r9),%eax
401240: 41 83 c1 02 add $0x2,%r9d
401244: 41 83 f9 63 cmp $0x63,%r9d
401248: 75 a6 jne 4011f0 <L00>
40124a: be 65 00 00 00 mov $0x65,%esi
40124f: 4a 8d 7c 97 04 lea 0x4(%rdi,%r10,4),%rdi
401254: 29 c6 sub %eax,%esi
401256: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
40125d: 00 00 00
401260:L04 41 8b 10 mov (%r8),%edx
401263: 8d 42 01 lea 0x1(%rdx),%eax
401266: 01 f2 add %esi,%edx
401268: 0f 1f 84 00 00 00 00 nopl 0x0(%rax,%rax,1)
40126f: 00
401270:L05 89 c1 mov %eax,%ecx
401272: 83 c0 01 add $0x1,%eax
401275: 39 d0 cmp %edx,%eax
401277: 75 f7 jne 401270 <L05>
401279: 41 89 08 mov %ecx,(%r8)
40127c: 49 83 c0 04 add $0x4,%r8
401280: 49 39 f8 cmp %rdi,%r8
401283: 75 db jne 401260 <L04>
401285: c3 retq
401286:L06 c3 retq
401287:L07 31 c0 xor %eax,%eax
401289: eb 8f jmp 40121a <L02>
40128b: 0f 1f 44 00 00 nopl 0x0(%rax,%rax,1)
0000000000401290 <incrementVector1>:
401290: 85 f6 test %esi,%esi
401292: 7e 5f jle 4012f3 <L02>
401294: 8d 46 ff lea -0x1(%rsi),%eax
401297: 83 f8 02 cmp $0x2,%eax
40129a: 76 59 jbe 4012f5 <L04>
40129c: 89 f2 mov %esi,%edx
40129e: 66 0f 6f 0d ca 0d 00 movdqa 0xdca(%rip),%xmm1 # 402070 <__dso_handle+0x68>
4012a5: 00
4012a6: 48 89 f8 mov %rdi,%rax
4012a9: c1 ea 02 shr $0x2,%edx
4012ac: 48 c1 e2 04 shl $0x4,%rdx
4012b0: 48 01 fa add %rdi,%rdx
4012b3:L00 f3 0f 6f 00 movdqu (%rax),%xmm0
4012b7: 48 83 c0 10 add $0x10,%rax
4012bb: 66 0f fe c1 paddd %xmm1,%xmm0
4012bf: 0f 11 40 f0 movups %xmm0,-0x10(%rax)
4012c3: 48 39 d0 cmp %rdx,%rax
4012c6: 75 eb jne 4012b3 <L00>
4012c8: 89 f0 mov %esi,%eax
4012ca: 83 e0 fc and $0xfffffffc,%eax
4012cd: 39 f0 cmp %esi,%eax
4012cf: 74 23 je 4012f4 <L03>
4012d1:L01 48 63 d0 movslq %eax,%rdx
4012d4: 83 04 97 64 addl $0x64,(%rdi,%rdx,4)
4012d8: 8d 50 01 lea 0x1(%rax),%edx
4012db: 39 f2 cmp %esi,%edx
4012dd: 7d 14 jge 4012f3 <L02>
4012df: 48 63 d2 movslq %edx,%rdx
4012e2: 83 c0 02 add $0x2,%eax
4012e5: 83 04 97 64 addl $0x64,(%rdi,%rdx,4)
4012e9: 39 c6 cmp %eax,%esi
4012eb: 7e 06 jle 4012f3 <L02>
4012ed: 48 98 cltq
4012ef: 83 04 87 64 addl $0x64,(%rdi,%rax,4)
4012f3:L02 c3 retq
4012f4:L03 c3 retq
4012f5:L04 31 c0 xor %eax,%eax
4012f7: eb d8 jmp 4012d1 <L01>
4012f9: 0f 1f 80 00 00 00 00 nopl 0x0(%rax)
Here is the output of -S -fverbose-asm:
.globl incrementVector2
.type incrementVector2, @function
incrementVector2:
.LFB13:
.cfi_startproc
testl %esi, %esi # n
jle .L19 #,
movl %esi, %edx # n, bnd.1
movl %esi, %r11d # n, niters_vector_mult_vf.2
movdqa .LC1(%rip), %xmm1 #, tmp152
movq %rdi, %r8 # v, ivtmp.33
shrl $2, %edx #,
leal -1(%rsi), %r10d #,
andl $-4, %r11d #, niters_vector_mult_vf.2
# orig.c:38: for (int i = 0; i < n; ++i) {
movl $1, %r9d #, ivtmp.47
salq $4, %rdx #, tmp150
addq %rdi, %rdx # v, _144
.L10:
cmpl $2, %r10d #, _72
jbe .L14 #,
# orig.c:36: {
movq %rdi, %rax # v, ivtmp.36
.L8:
# orig.c:39: v[i] = v[i] + 1;
movdqu (%rax), %xmm0 # MEM[base: _138, offset: 0B], vect__42.13
addq $16, %rax #, ivtmp.36
paddd %xmm1, %xmm0 # tmp152, vect__42.13
# orig.c:39: v[i] = v[i] + 1;
movups %xmm0, -16(%rax) # vect__42.13, MEM[base: _138, offset: 0B]
cmpq %rdx, %rax # _144, ivtmp.36
jne .L8 #,
# orig.c:38: for (int i = 0; i < n; ++i) {
movl %r11d, %eax # niters_vector_mult_vf.2, i
cmpl %r11d, %esi # niters_vector_mult_vf.2, n
je .L9 #,
.L13:
# orig.c:39: v[i] = v[i] + 1;
movslq %eax, %rcx # i, i
# orig.c:39: v[i] = v[i] + 1;
addl $2, (%rdi,%rcx,4) #, *_40
# orig.c:38: for (int i = 0; i < n; ++i) {
leal 1(%rax), %ecx #, i
# orig.c:38: for (int i = 0; i < n; ++i) {
cmpl %ecx, %esi # i, n
jle .L9 #,
# orig.c:39: v[i] = v[i] + 1;
movslq %ecx, %rcx # i, i
# orig.c:38: for (int i = 0; i < n; ++i) {
addl $2, %eax #, i
# orig.c:39: v[i] = v[i] + 1;
addl $2, (%rdi,%rcx,4) #, *_107
# orig.c:38: for (int i = 0; i < n; ++i) {
cmpl %eax, %esi # i, n
jle .L9 #,
# orig.c:39: v[i] = v[i] + 1;
cltq
# orig.c:39: v[i] = v[i] + 1;
addl $2, (%rdi,%rax,4) #, *_58
.L9:
leal 1(%r9), %eax #, _147
addl $2, %r9d #, ivtmp.47
cmpl $99, %r9d #, ivtmp.47
jne .L10 #,
movl $101, %esi #, tmp147
leaq 4(%rdi,%r10,4), %rdi #, _134
subl %eax, %esi # _147, tmp146
.p2align 4,,10
.p2align 3
.L12:
movl (%r8), %edx # MEM[base: _126, offset: 0B], _2
leal 1(%rdx), %eax #, ivtmp.21
addl %esi, %edx # tmp146, _122
.p2align 4,,10
.p2align 3
.L11:
movl %eax, %ecx # ivtmp.21, _25
addl $1, %eax #, ivtmp.21
# orig.c:38: for (int i = 0; i < n; ++i) {
cmpl %edx, %eax # _122, ivtmp.21
jne .L11 #,
movl %ecx, (%r8) # _25, MEM[base: _126, offset: 0B]
addq $4, %r8 #, ivtmp.33
# orig.c:37: for (int k = 0; k < 100; ++k) {
cmpq %rdi, %r8 # _134, ivtmp.33
jne .L12 #,
ret
.L19:
ret
.L14:
# orig.c:38: for (int i = 0; i < n; ++i) {
xorl %eax, %eax # i
jmp .L13 #
.cfi_endproc
.LFE13:
.size incrementVector2, .-incrementVector2
.p2align 4,,15
.globl incrementVector1
.type incrementVector1, @function
incrementVector1:
.LFB14:
.cfi_startproc
# orig.c:47: for (int i = 0; i < n; ++i) {
testl %esi, %esi # n
jle .L20 #,
leal -1(%rsi), %eax #, tmp118
cmpl $2, %eax #, tmp118
jbe .L27 #,
movl %esi, %edx # n, bnd.51
movdqa .LC2(%rip), %xmm1 #, tmp131
movq %rdi, %rax # v, ivtmp.62
shrl $2, %edx #,
salq $4, %rdx #, tmp121
addq %rdi, %rdx # v, _58
.L23:
movdqu (%rax), %xmm0 # MEM[base: _53, offset: 0B], vect__8.57
addq $16, %rax #, ivtmp.62
paddd %xmm1, %xmm0 # tmp131, vect__8.57
movups %xmm0, -16(%rax) # vect__8.57, MEM[base: _53, offset: 0B]
cmpq %rdx, %rax # _58, ivtmp.62
jne .L23 #,
movl %esi, %eax # n, tmp.53
andl $-4, %eax #, tmp.53
cmpl %esi, %eax # n, tmp.53
je .L29 #,
.L22:
# orig.c:49: v[i] = v[i] + 1;
movslq %eax, %rdx # tmp.53, tmp.53
addl $100, (%rdi,%rdx,4) #, *_3
# orig.c:47: for (int i = 0; i < n; ++i) {
leal 1(%rax), %edx #, i
# orig.c:47: for (int i = 0; i < n; ++i) {
cmpl %esi, %edx # n, i
jge .L20 #,
# orig.c:49: v[i] = v[i] + 1;
movslq %edx, %rdx # i, i
# orig.c:47: for (int i = 0; i < n; ++i) {
addl $2, %eax #, i
addl $100, (%rdi,%rdx,4) #, *_47
# orig.c:47: for (int i = 0; i < n; ++i) {
cmpl %eax, %esi # i, n
jle .L20 #,
# orig.c:49: v[i] = v[i] + 1;
cltq
addl $100, (%rdi,%rax,4) #, *_25
.L20:
# orig.c:52: }
ret
.L29:
ret
.L27:
# orig.c:47: for (int i = 0; i < n; ++i) {
xorl %eax, %eax # tmp.53
jmp .L22 #
.cfi_endproc
.LFE14:
.size incrementVector1, .-incrementVector1
.section .rodata.str1.1,"aMS",@progbits,1
.LC3:
来源：https://stackoverflow.com/questions/65926332/cache-misses-when-accessing-an-array-in-nested-loop