使用gcc 4.9,使用Linaro工具链对ARM进行交叉编译,我发现在添加vector.assign()
时-std=c++14
的编译结果会以一种会产生严重性能问题的方式发生变化。
我已经尝试了几种不同的方法来完成这个分配+复制,但只要我使用std::vector
来完成它,所有这些都有这个性能问题。
我可以通过这个玩具示例重现问题:
VectorTest.h
#include <stdint.h>
#include <stddef.h>
#include <vector>
struct VectorWrapper_t
{
VectorWrapper_t(uint8_t const* pData, size_t length);
std::vector<uint8_t> data;
};
VectorTest.cpp
#include "VectorTest.h"
VectorWrapper_t::VectorWrapper_t(uint8_t const* pData, size_t length)
{
data.assign(pData, pData + length);
}
gcc标志:
-std=c++14 \
-mthumb -march=armv7-a -mtune=cortex-a9 \
-mlittle-endian -mfloat-abi=hard -mfpu=neon -Wa,-mimplicit-it=thumb \
-O2 -g
查看程序集,我可以看到原因:原始版本(C ++ 03,我假设?)调用memmove
,而C ++ 14版本则添加了一个额外的循环,它看起来像手动复制数据。看看.loc
标签gcc加上-fverbose-asm
,这个循环中的指令来自stl_construct.h
和stl_uninitialized.h
。
更改为gcc 5.2.1(使用C ++ 14),它的编译几乎与C ++ 03示例完全相同,除了使用memcpy
而不是memmove
。
我可以通过在这里使用std::unique_ptr<uint8_t[]>
而不是vector
解决这个问题。但是,我想深入研究这个问题的底部,以确定使用vector
s的其他地方是否存在性能问题以及如何对它们进行修复(更新到gcc 5.2是不切实际的)。
所以我的问题是:为什么它在C ++ 11/14下的编译方式不同?
作为参考,gcc --version
报道:
arm-linux-gnueabihf-gcc (Linaro GCC 4.9-2014.12) 4.9.3 20141205 (prerelease)
。
这是生成的程序集gcc:
# C++03, gcc 4.9
push {r3, r4, r5, r6, r7, lr} @
movs r3, #0 @ tmp118,
mov r4, r0 @ this, this
str r3, [r0] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_start
mov r5, r2 @ length, length
str r3, [r0, #4] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_finish
str r3, [r0, #8] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_end_of_storage
cbnz r2, .L19 @ length,
mov r0, r4 @, this
pop {r3, r4, r5, r6, r7, pc} @
.L19:
mov r0, r2 @, length
mov r6, r1 @ pData, pData
bl _Znwj @
mov r2, r5 @, length
mov r1, r6 @, pData
mov r7, r0 @ D.13516,
bl memmove @
ldr r0, [r4] @ D.13515, MEM[(struct vector *)this_1(D)].D.11902._M_impl._M_start
cbz r0, .L3 @ D.13515,
bl _ZdlPv @
.L3:
add r5, r5, r7 @ D.13515, D.13516
str r7, [r4] @ D.13516, MEM[(struct vector *)this_1(D)].D.11902._M_impl._M_start
str r5, [r4, #4] @ D.13515, MEM[(struct vector *)this_1(D)].D.11902._M_impl._M_finish
mov r0, r4 @, this
str r5, [r4, #8] @ D.13515, MEM[(struct vector *)this_1(D)].D.11902._M_impl._M_end_of_storage
pop {r3, r4, r5, r6, r7, pc} @
.L6:
ldr r0, [r4] @ D.13515, MEM[(struct _Vector_base *)this_1(D)]._M_impl._M_start
cbz r0, .L5 @ D.13515,
bl _ZdlPv @
.L5:
bl __cxa_end_cleanup @
# C++14, gcc 4.9
push {r3, r4, r5, r6, r7, lr} @
movs r3, #0 @ tmp157,
mov r6, r0 @ this, this
str r3, [r0] @ tmp157, MEM[(struct _Vector_impl *)this_1(D)]._M_start
mov r5, r2 @ length, length
str r3, [r0, #4] @ tmp157, MEM[(struct _Vector_impl *)this_1(D)]._M_finish
str r3, [r0, #8] @ tmp157, MEM[(struct _Vector_impl *)this_1(D)]._M_end_of_storage
cbnz r2, .L25 @ length,
mov r0, r6 @, this
pop {r3, r4, r5, r6, r7, pc} @
.L25:
mov r0, r2 @, length
mov r4, r1 @ pData, pData
bl _Znwj @
adds r3, r4, r5 @ D.20345, pData, length
mov r7, r0 @ __result,
cmp r4, r3 @ pData, D.20345
ittt ne
addne r1, r4, #-1 @ ivtmp.76, pData,
movne r3, r0 @ __result, __result
addne r4, r0, r5 @ D.20346, __result, length
beq .L26 @,
.L7:
ldrb r2, [r1, #1]! @ zero_extendqisi2 @ D.20348, MEM[base: _48, offset: 0]
cbz r3, .L6 @ __result,
strb r2, [r3] @ D.20348, MEM[base: __result_23, offset: 0B]
.L6:
adds r3, r3, #1 @ __result, __result,
cmp r3, r4 @ __result, D.20346
bne .L7 @,
.L8:
ldr r0, [r6] @ D.20346, MEM[(struct vector *)this_1(D)].D.18218._M_impl._M_start
cbz r0, .L5 @ D.20346,
bl _ZdlPv @
.L5:
str r7, [r6] @ __result, MEM[(struct vector *)this_1(D)].D.18218._M_impl._M_start
mov r0, r6 @, this
str r4, [r6, #4] @ D.20346, MEM[(struct vector *)this_1(D)].D.18218._M_impl._M_finish
str r4, [r6, #8] @ D.20346, MEM[(struct vector *)this_1(D)].D.18218._M_impl._M_end_of_storage
pop {r3, r4, r5, r6, r7, pc} @
.L26:
adds r4, r0, r5 @ D.20346, __result, length
b .L8 @
.L11:
ldr r0, [r6] @ D.20346, MEM[(struct _Vector_base *)this_1(D)]._M_impl._M_start
cbz r0, .L10 @ D.20346,
bl _ZdlPv @
.L10:
bl __cxa_end_cleanup @
# C++14, gcc 5.2
push {r3, r4, r5, r6, r7, lr} @
movs r3, #0 @ tmp118,
mov r4, r0 @ this, this
str r3, [r0] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_start
str r3, [r0, #4] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_finish
str r3, [r0, #8] @ tmp118, MEM[(struct _Vector_impl *)this_1(D)]._M_end_of_storage
cbnz r2, .L19 @ length,
mov r0, r4 @, this
pop {r3, r4, r5, r6, r7, pc} @
.L19:
mov r0, r2 @, length
mov r6, r1 @ pData, pData
mov r5, r2 @ length, length
bl _Znwj @
mov r2, r5 @, length
mov r1, r6 @, pData
mov r7, r0 @ D.20824,
bl memcpy @
ldr r0, [r4] @ D.20823, MEM[(struct vector *)this_1(D)].D.18751._M_impl._M_start
cbz r0, .L3 @ D.20823,
bl _ZdlPv @
.L3:
add r5, r5, r7 @ D.20823, D.20824
str r7, [r4] @ D.20824, MEM[(struct vector *)this_1(D)].D.18751._M_impl._M_start
str r5, [r4, #4] @ D.20823, MEM[(struct vector *)this_1(D)].D.18751._M_impl._M_finish
mov r0, r4 @, this
str r5, [r4, #8] @ D.20823, MEM[(struct vector *)this_1(D)].D.18751._M_impl._M_end_of_storage
pop {r3, r4, r5, r6, r7, pc} @
.L6:
ldr r0, [r4] @ D.20823, MEM[(struct _Vector_base *)this_1(D)]._M_impl._M_start
cbz r0, .L5 @ D.20823,
bl _ZdlPv @
.L5:
bl __cxa_end_cleanup @
这是4.9.2版本中的GCC错误,请参阅PR 64476。默认的-std=gnu++03
模式和-std=c++14
之间的区别在于,对于C ++ 11及更高版本,它可能具有不可分配的普通类型(因为它们可以具有已删除的赋值运算符),这导致std::uninitialized_copy
的实现采取不同的(慢的)代码路径。对可转让性的检查是错误的,这意味着我们在不需要时采取了慢速路径。
两年前我为GCC 4.9.3修复了它,但是你的编译器基于4.9.2和4.9.3版本之间的快照,并且有几周的时间来修复它。
你可以让Linaro将他们的GCC 4.9编译器更新到4.9.4,或者至少应用修补此bug的补丁。