将数据按X位移入和移出文件的最佳方法是什么? [英] What is the best way to shift data by X bits into and out of a file?
问题描述
我有一个数据集,由一个掩码文件和一个数据文件组成.掩码文件告诉解码器,对于数据文件的当前偏移量,每个字段是否存在8位或4位.我需要根据掩码将数据移出,并写入解码的文件,每个字段所有8位.我正在尝试用C语言实现这一目标.
I have a dataset, made up of a mask file, and a data file. the mask file tells the decoder whether there are 8 bits per field present, or 4, for the current offset of the datafile. I need to shift the data out according to the mask, and write the decoded file, with all 8 bits per field. I'm trying to accomplish this in C.
void shift_4bits_left(unsigned char* array, unsigned short size)
{
int i;
unsigned char shifted = 0x00;
unsigned char overflow = (0xF0 & array[0]) >> 4;
for (i = (size - 1); i >= 0; i--)
{
shifted = (array[i] << 4) | overflow;
overflow = (0xF0 & array[i]) >> 4;
array[i] = shifted;
}
}
功能:
while(len>count){
//count=0;
if(bit==0)yy=fread(blockchar,1,len,fp1);
if(bit==0)memcpy(blockchar2,blockchar,len);
count+=yy;
memset(outputbuf,0,64);
for(x=0;x<len;x++){
count2=0;
//count2=1;
if(sometests(blockchar[x])==1)
{
shift_4bits_left(&blockchar[x],(yy-(bittest/2))+1);
count2=1;
total++;
if(total%2==0)len--;
}
//set bit for current position in mask file's buffer
if((x)%8 == 0)outputbuf[x]+=(count2<<7);
if((x)%8 == 1)outputbuf[x]+=(count2<<6);
if((x)%8 == 2)outputbuf[x]+=(count2<<5);
if((x)%8 == 3)outputbuf[x]+=(count2<<4);
if((x)%8 == 4)outputbuf[x]+=(count2<<3);
if((x)%8 == 5)outputbuf[x]+=(count2<<2);
if((x)%8 == 6)outputbuf[x]+=(count2<<1);
if((x)%8 == 7)outputbuf[x]+=count2;
}
问题是,使用大型数据集时,这非常慢,我希望对文件进行迭代,而无需将文件存储在数组中.
The problem is this is very slow when working with large datasets, I would prefer to make this iterative over the file, without storing the files in arrays.
推荐答案
我通过使用memcpy一次查看了256个字节(不移位时)来解决了这个问题.感谢您的帮助.
I solved it by looking at 256 bytes at a time, and when not shifting, using memcpy. Thanks for the help guys.
这篇关于将数据按X位移入和移出文件的最佳方法是什么?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!