我正在开发一个iOS应用程序,需要转换图像从RGB -> BGRA相当快。如果可能的话,我想使用霓虹灯本征。有没有比简单地分配组件更快的方法?
void neonPermuteRGBtoBGRA(unsigned char* src, unsigned char* dst, int numPix)
{
numPix /= 8; //process 8 pixels at a time
uint8x8_t alpha = vdup_n_u8 (0xff);
for (int i=0; i<numPix; i++)
{
uint8x8x3_t rgb = vld3_u8 (src);
uint8x8x4_t bgra;
bgra.val[0] = rgb.val[2]; //these lines are slow
bgra.val[1] = rgb.val[1]; //these lines are slow
bgra.val[2] = rgb.val[0]; //these lines are slow
bgra.val[3] = alpha;
vst4_u8(dst, bgra);
src += 8*3;
dst += 8*4;
}
}发布于 2012-08-14 14:04:18
ARMCC的反汇编也不是那么快:
它没有使用最合适的instructions
时都会导致很大的问题。
试试这个:
mov r2, r2, lsr #3
vmov.u8, d3, #0xff
loop:
vld3.8 {d0-d2}, [r0]!
subs r2, r2, #1
vswp d0, d2
vst4.8 {d0-d3}, [r1]!
bgt loop
bx lr我建议的代码也没有完全优化,但进一步的“真正”优化将严重损害可读性。所以我就到此为止了。
发布于 2012-08-10 22:31:23
这取决于编译器。例如,当我使用armcc (5.01)编译上面的代码并反汇编它时,我得到的结果是这样的(我只是将循环放在循环中,并将alpha赋值移到了循环之外)。
18: f420440d vld3.8 {d4-d6}, [r0]!
1c: e2822001 add r2, r2, #1 ; 0x1
20: eeb01b45 fcpyd d1, d5
24: eeb00b46 fcpyd d0, d6
28: eeb02b44 fcpyd d2, d4
2c: f401000d vst4.8 {d0-d3}, [r1]!
30: e1520003 cmp r2, r3
34: bafffff7 blt 18 <neonPermuteRGBtoBGRA_armcc+0x18>如果我用gcc (4.4.3)编译代码,然后再反汇编,我会得到:
40: f967 040f vld3.8 {d16-d18}, [r7]
44: 46d6 mov lr, sl
46: ecca 0b06 vstmia sl, {d16-d18}
4a: 9d02 ldr r5, [sp, #8]
4c: ed8d 8b1a vstr d8, [sp, #104]
50: 3718 adds r7, #24
52: e8be 000f ldmia.w lr!, {r0, r1, r2, r3}
56: f108 0801 add.w r8, r8, #1 ; 0x1
5a: c50f stmia r5!, {r0, r1, r2, r3}
5c: eddd 0b24 vldr d16, [sp, #144]
60: e89e 0003 ldmia.w lr, {r0, r1}
64: edcd 0b16 vstr d16, [sp, #88]
68: eddd 0b22 vldr d16, [sp, #136]
6c: edcd 0b18 vstr d16, [sp, #96]
70: e885 0003 stmia.w r5, {r0, r1}
74: ed9d 0b26 vldr d0, [sp, #152]
78: 9d03 ldr r5, [sp, #12]
7a: ed8d 0b14 vstr d0, [sp, #80]
7e: cd0f ldmia r5!, {r0, r1, r2, r3}
80: 46ae mov lr, r5
82: 465d mov r5, fp
84: c50f stmia r5!, {r0, r1, r2, r3}
86: e89e 000f ldmia.w lr, {r0, r1, r2, r3}
8a: e885 000f stmia.w r5, {r0, r1, r2, r3}
8e: 9501 str r5, [sp, #4]
90: 465d mov r5, fp
92: 2100 movs r1, #0
94: 2220 movs r2, #32
96: 4620 mov r0, r4
98: f7ff fffe bl 0 <memset>
9c: cd0f ldmia r5!, {r0, r1, r2, r3}
9e: 4625 mov r5, r4
a0: c50f stmia r5!, {r0, r1, r2, r3}
a2: f8dd c004 ldr.w ip, [sp, #4]
a6: e89c 000f ldmia.w ip, {r0, r1, r2, r3}
aa: e885 000f stmia.w r5, {r0, r1, r2, r3}
ae: ecd4 0b08 vldmia r4, {d16-d19}
b2: f946 000f vst4.8 {d16-d19}, [r6]
b6: 3620 adds r6, #32
b8: 45c8 cmp r8, r9
ba: dbc1 blt.n 40 <memset+0x40>并且使用armcc的执行时间快了10倍。
如果我用gcc (内联汇编)编译armcc为函数生成的汇编代码(现在看起来alpha又回到了循环中:)
void neonPermuteRGBtoBGRA_gas(unsigned char* src, unsigned char* dst,
int numPix) {
asm(
" ASR r3,r2,#31\n"
" VMOV.I8 d1,#0xff\n"
" ADD r2,r2,r3,LSR #29\n"
" ASR r3,r2,#3\n"
" MOV r2,#0\n"
" CMP r3,#0\n"
" BLE end\n"
"loop:\n"
" VLD3.8 {d4,d5,d6},[r0]!\n"
" ADD r2,r2,#1\n"
" CMP r3,r2\n"
" VMOV.F64 d3,d5\n"
" VMOV.F64 d2,d6\n"
" VMOV.F64 d5,d1\n"
" VMOV.F64 d0,d4\n"
" VST4.8 {d2,d3,d4,d5},[r1]!\n"
" BGT loop\n"
"end:\n"
);
}我和gcc也得到了相同的执行时间。
最后,我建议你要么反汇编你的二进制文件,然后检查编译器是否产生了你想要的东西,要么使用汇编。
顺便说一句,如果你想进一步改进这个函数的执行时间,我建议你去看看
https://stackoverflow.com/questions/11890997
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