从SDK中我获得了包含像素格式BGR的图像,即BGRBGRBGR
。对于其他应用程序,我需要将此格式转换为RGB平面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。
我使用的是C#.NET 4.5 32位,数据是字节数组,大小相同。
现在我正在迭代数组源并将BGR值分配给目标数组中的适当位置,但这需要太长时间(130万像素图像为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,也无法转换类似问题的代码,例如Fast method to copy memory with translation - ARGB to BGR以满足我的需求。
我目前用于在像素格式之间转换的代码是:
// 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
}
我尝试将对源数组的访问分成三个循环,每个循环一个循环,但它并没有真正帮助。所以我愿意接受建议。
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:我通过删除这样的除法和乘法将其降低到180ms,但有没有办法让它更快?它仍然非常慢,我猜是因为内存读/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++;
}
感谢MBo的回答,我能够将时间从180毫秒减少到90毫秒!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);
}
}
}
答案 0 :(得分:1)
我在Delphi中实现了这个交错问题并检查了内置的asm。我没有内在函数,所以使用普通汇编程序
pshufb
等于_mm_shuffle_epi8
(SSSE3 intrinsic)
在每个循环步骤中,我将16个字节(r1g1b1 r2g2b2 r3g3b3 r4b4g4 r5b5g5 r6)
加载到128位XMM寄存器,将它们拖入(r1r2r3r4 g1g2g3g4 b1b2b3b4 xxxx)
顺序,并将r,g,b块保存到目标内存(忽略最后4个字节) 。下一步加载(r5b5g5 r6g6b6 r7g7b7 ...)
,依此类推。
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,因此我进行了必要的更正。
第一个版本问题示例:
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个字节到第99个字节 - 所以我们用完了数组范围!
我刚刚在这里添加了保护字节:GetMem(A, Size * 3 + 15);
,但对于实际任务,它可能是不适用的,因此值得对最后一个数组块进行特殊处理。
对于pascal变体,此代码需要967 ms,对于asm变体,此代码需要140 ms才能在i5-4670计算机上转换200个130万像素的处理器(单个线程的处理器本身比Atom 680快6-8倍)。速度约为0.75 GB /秒(pas)和5.4 GB / Sec(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);
答案 1 :(得分:0)
您可以尝试向后计数,即int i = source.Length - 1; i >=0 ; i -= 3
,因此每个for循环只读取一次属性source.Length
,而不是每次迭代。
答案 2 :(得分:0)
我遵循了Ivan的建议并想出了这个改进,摆脱了分裂(在C中实现):
int offset = 0;
for (int i = 0; i < ARRAYSIZE(source); i += 3) {
offset++;
result[offset] = source[i + 2]; // R
result[offset + imageSize] = source[i + 1]; // G
result[offset + imageSize * 2] = source[i]; // B
}
这可以节省我机器上大约40%的运行时间。
答案 3 :(得分:0)
第一步:避免多次阅读来源(参见答案https://stackoverflow.com/a/27542680/949044)。这也很适合CPU缓存,目前还没有使用:你正在读取3个字节中的1个字节,因此抛出了2 / 3ds的缓存行。 所以它可以这样:
int targetPositionR = 0;
int targetPositionG = imageSize;
int targetPositionB = imageSize * 2;
for (int i = 0; i < source.Length; i += 3)
{
result[targetPositionB] = source[i]; // B
result[targetPositionG] = source[i + 1]; // G
result[targetPositionR] = source[i + 2]; // R
targetPositionB++;
targetPositionG++;
targetPositionR++;
}
第二步:一次写入4个字节,而不是1个字节。但是,它需要一个额外的缓冲区和副本:
int[] dwPlanar = new int[imageSize*3/4];
int targetPositionR = 0;
int targetPositionG = imageSize / 4;
int targetPositionB = imageSize * 2 / 4;
for (int i = 0; i < source.Length; i += 12)
{
int dwB = (source[i ]) | (source[i+3] << 8) | (source[i+6] << 16) | (source[i+9] << 24);
int dwG = (source[i+1]) | (source[i+4] << 8) | (source[i+7] << 16) | (source[i+10] << 24);
int dwR = (source[i+2]) | (source[i+5] << 8) | (source[i+8] << 16) | (source[i+11] << 24);
dwPlanar[targetPositionB] = dwB; // B
dwPlanar[targetPositionG] = dwG; // G
dwPlanar[targetPositionR] = dwR; // R
targetPositionB++;
targetPositionG++;
targetPositionR++;
}
Buffer.BlockCopy(dwPlanar,0,result,0,imageSize * 3);
我认为它会有所帮助,因为c#会进行更少的数组边界检查,并且通常最好尽可能用更大的块来编写。
(免责声明:我不熟悉c#,我不知道这段代码是否会编译,它只是一种算法)。