从SDK中可以获得打包为BGR像素格式的图像,即BGRBGRBGR
。对于另一个应用程序,我需要将此格式转换为RGB plane RRRGGGBBB
。我不想仅为此任务使用额外的库,因此我必须使用自己的代码在格式之间进行转换。
我正在使用C#.NET 4.5 32位,并且数据位于具有相同大小的字节数组中。
现在,我正在遍历数组源并将BGR值分配给它们在目标数组中的适当位置,但这花费的时间太长(对于1.3兆像素的图像需要250毫秒)。代码运行的处理器是Intel Atom E680,可以访问MMX,SSE,SSE2,SSE3,SSSE3。
不幸的是,我不了解内在函数,也无法为类似Fast method to copy memory with translation - ARGB to BGR这样的类似问题转换代码以满足我的需求。
我当前用于在像素格式之间进行转换的代码是:
// the array with the BGRBGRBGR pixel data
byte[] source;
// the array with the RRRGGGBBB pixel data
byte[] result;
// the amount of pixels in one channel, width*height
int imageSize;
for (int i = 0; i < source.Length; i += 3)
{
result[i/3] = source[i + 2]; // R
result[i/3 + imageSize] = source[i + 1]; // G
result[i/3 + imageSize * 2] = source[i]; // B
}
我试图将对源数组的访问分为三个循环,每个通道一个循环,但这并没有真正的帮助。因此,我愿意提出建议。
for (int i = 0; i < source.Length; i += 3)
{
result[i/3] = source[i + 2]; // R
}
for (int i = 0; i < source.Length; i += 3)
{
result[i/3 + imageSize] = source[i + 1]; // G
}
for (int i = 0; i < source.Length; i += 3)
{
result[i/3 + imageSize * 2] = source[i]; // B
}
编辑:我通过除去这样的除法和乘法将其降低到180ms,但是有没有办法使其更快?我仍然认为它仍然很慢,因为内存的读/写不是很理想。
int targetPosition = 0;
int imageSize2 = imageSize * 2;
for (int i = 0; i < source.Length; i += 3)
{
result[targetPosition] = source[i + 2]; // R
targetPosition++;
}
targetPosition = 0;
for (int i = 0; i < source.Length; i += 3)
{
result[targetPosition + imageSize] = source[i + 1]; // G
targetPosition++;
}
targetPosition = 0;
for (int i = 0; i < source.Length; i += 3)
{
result[targetPosition + imageSize2] = source[i]; // B
targetPosition++;
}
多亏了MBo的回答,我才能够将时间从180ms减少到90ms!这是代码:
Converter.cpp:
#include "stdafx.h"
BOOL __stdcall DllMain(HINSTANCE hInst, DWORD dwReason, LPVOID lpReserved) {
return TRUE;
}
const unsigned char Mask[] = { 0, 3, 6, 9,
1, 4, 7, 10,
2, 5, 8, 11,
12, 13, 14, 15};
extern "C" __declspec(dllexport) char* __stdcall ConvertPixelFormat(unsigned char* source, unsigned char *target, int imgSize) {
_asm {
//interleave r1g1b1 r2g2b2 r3g3b3 r4b4g4 r5b5g5 r6... to planar
// r1r2r3r4r5..... g1g2g3g4g5... b1b2b3b4b5...
push edi
push esi
mov eax, source //A address
mov edx, target //B address
mov ecx, imgSize
movdqu xmm5, Mask //load shuffling mask
mov edi, imgSize //load interleave step
mov esi, eax
add esi, edi
add esi, edi
add esi, edi
shr ecx, 2 //divide count by 4
dec ecx //exclude last array chunk
jle Rest
Cycle:
movdqu xmm0, [eax] //load 16 bytes
pshufb xmm0, xmm5 //shuffle bytes, we are interested in 12 ones
movd [edx], xmm0 //store 4 bytes of R
psrldq xmm0, 4 //shift right register, now G is on the end
movd [edx + edi], xmm0 //store 4 bytes of G to proper place
psrldq xmm0, 4 //do the same for B
movd [edx + 2 * edi], xmm0
add eax, 12 //shift source index to the next portion
add edx, 4 //shift destination index
loop Cycle
Rest: //treat the rest of array
cmp eax, esi
jae Finish
mov ecx, [eax]
mov [edx], cl //R
mov [edx + edi], ch //G
shr ecx, 16
mov [edx + 2 * edi], cl //B
add eax, 3
add edx, 1
jmp Rest
Finish:
pop esi
pop edi
}
}
C#文件:
// Code to define the method
[DllImport("Converter.dll")]
unsafe static extern void ConvertPixelFormat(byte* source, byte* target, int imgSize);
// Code to execute the conversion
unsafe
{
fixed (byte* sourcePointer = &source[0])
{
fixed (byte* resultPointer = &result[0])
{
ConvertPixelFormat(sourcePointer, resultPointer, imageSize);
}
}
}
最佳答案
我已经在Delphi中实现了这个交错问题,并检查了内置的asm。我没有内部函数,因此使用普通汇编器。
pshufb
等于_mm_shuffle_epi8
(SSSE3 intrinsic)
在每个循环步骤中,我将16个字节的(r1g1b1 r2g2b2 r3g3b3 r4b4g4 r5b5g5 r6)
加载到128位XMM寄存器中,将它们按(r1r2r3r4 g1g2g3g4 b1b2b3b4 xxxx)
顺序洗牌,然后将r,g,b块保存到目标内存中(忽略最后4个字节)。下一步加载(r5b5g5 r6g6b6 r7g7b7 ...)
,依此类推。
请注意,为了简化代码,我没有在第一个代码版本中正确处理数组的尾部。由于您可以使用此代码,因此我已进行了必要的更正。
第一版发行示例:
imgSize = 32
数组大小= 96字节
32/4 = 8个周期
最后一个周期从第84个字节开始,读取16个字节,直到第99个字节-因此我们超出了数组范围!
我只是在此处添加了保护字节:GetMem(A, Size * 3 + 15);
,但是对于实际任务来说,它可能不适用,因此值得对最后一个数组块进行特殊处理。
对于i5-4670机器,此代码在pascal变体上花费967 ms,在asm变体上花费140 ms,以在i5-4670机器上转换200个1.3MP cadr(对于单线程,处理器本身比Atom 680快6-8倍)。速度约为0.75 GB /秒(pas)和5.4 GB /秒(asm)
const
Mask: array[0..15] of Byte = ( 0, 3, 6, 9,
1, 4, 7, 10,
2, 5, 8, 11,
12, 13, 14, 15);
var
A, B: PByteArray;
i, N, Size: Integer;
t1, t2: DWord;
begin
Size := 1280 * 960 * 200;
GetMem(A, Size * 3);
GetMem(B, Size * 3);
for i := 0 to Size - 1 do begin
A[3 * i] := 1;
A[3 * i + 1] := 2;
A[3 * i + 2] := 3;
end;
t1 := GetTickCount;
for i := 0 to Size - 1 do begin
B[i] := A[3 * i];
B[i + Size] := A[3 * i + 1];
B[i + 2 * Size] := A[3 * i + 2];
end;
t2:= GetTickCount;
//interleave r1g1b1 r2g2b2 r3g3b3 r4b4g4 r5b5g5 r6... to planar
//r1r2r3r4r5..... g1g2g3g4g5... b1b2b3b4b5...
asm
push edi
push esi
mov eax, A //A address
mov edx, B //B address
mov ecx, Size
movdqu xmm5, Mask //load shuffling mask
mov edi, Size //load interleave step
mov esi, eax
add esi, edi
add esi, edi
add esi, edi
shr ecx, 2 //divide count by 4
dec ecx //exclude last array chunk
jle @@Rest
@@Cycle:
movdqu xmm0, [eax] //load 16 bytes
pshufb xmm0, xmm5 //shuffle bytes, we are interested in 12 ones
movd [edx], xmm0 //store 4 bytes of R
psrldq xmm0, 4 //shift right register, now G is on the end
movd [edx + edi], xmm0 //store 4 bytes of G to proper place
psrldq xmm0, 4 //do the same for B
movd [edx + 2 * edi], xmm0
add eax, 12 //shift source index to the next portion
add edx, 4 //shift destination index
loop @@Cycle
@@Rest: //treat the rest of array
cmp eax, esi
jae @@Finish
mov ecx, [eax]
mov [edx], cl //R
mov [edx + edi], ch //G
shr ecx, 16
mov [edx + 2 * edi], cl //B
add eax, 3
add edx, 1
jmp @@Rest
@@Finish:
pop esi
pop edi
end;
Memo1.Lines.Add(Format('pas %d asm %d', [t2-t1, GetTickCount - t2]));
FreeMem(A);
FreeMem(B);