N records of k bits each

Question

Starting with the address A in the memory of the MK there are N records with the length of k bits each. Records are aligned to the word boundary. Write functions for reading / writing individual bits for any record from this set.

1-word = 4-bytes = 32 bits The question is how to make a record length, for example, not standard 32, but let's say 33 bits or 37 bits? and what operations to use for this?

That's what happened with me, but this is not entirely correct:

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <cmath> //extern char tim(char s9[100], int OldSize, int NewSize); extern void _WRITE_MASK1(); extern void _WRITE_MASK2(); extern void _WRITE_BIT(long *j, int num_z, int num_byte, int num_bite, int _vol); extern int _READ_BIT(long *j, int num_z, int num_byte, int num_bite); #define N 1 // Ввод кол-ва записей по 4 байта. int main() { int K=32; // Ввод кол-ва битов в каждой записи( от 1 до 32 ) int Z=K-1; if((K<1)||(K>32)) { return 0; } unsigned long mass [N]; for (int i=0; i<N; i++) // формирование битов длиной К. { if(Z<2) mass[i]=pow(2,Z); if ((Z>=3)&(Z<10)) mass[i]=pow(2,Z)+pow(2,Z-1)+pow(2,Z-2)+1; if ((Z>=10)&(Z<20)) mass[i]=pow(2,Z)+pow(2,Z-2)+pow(2,Z-4)+pow(2,Z-6)+pow(2,Z-8)+i; if ((Z>=20)&(Z<32)) mass[i]=pow(2,Z)+pow(2,Z-4)+pow(2,Z-8)+pow(2,Z-12)+pow(2,Z-16)+i; } _WRITE_MASK1(); _WRITE_MASK2(); int rtf =1; // 0 или 1 int bit=1; // от 0 до 7 int byte=3; ; // от 0 до 3 int zap=4; // от 0 до кол-ва записей -1 int k=(sizeof(mass))/sizeof(mass[0]); // определение кол-ва элементов if(((bit>7)||(bit<0)||(byte>3)||(byte<0)||(zap<0)||(zap>k-1))||((rtf!=0)&(rtf!=1))) { return 0; } _WRITE_BIT(&mass[0],zap,byte,bit,rtf); //zap от 0 до размера массива-1; byte от 0 до 4, бит от 0 до 7. Устанавливаем бит в памяти равный rtf int bitRead=1; // от 0 до 7 int byteRead=3; // от 0 до 3 int zapRead=4; // от 0 до кол-ва записей -1 if((bitRead>7)||(bitRead<0)||(byteRead>3)||(byteRead<0)||(za pRead<0)||(zapRead>k-1)) { return 0; } int b = _READ_BIT(&mass[0],zapRead,byteRead,bitRead); int c=10; }

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Answer 1 · 2014-04-14T07:53:41

You need to "finish off" the end of the record to the length of the word. In your case (C-style), the easiest way is to work with structures where you can define alignment in advance. Suppose there is such data:

 typedef struct tagRecord { float data1; // word (4 байта) double data2; // double word (8 байт) int data3; // word (4 байта) unsigned char data4; // 1 байт } Record; // sizeof(Record) == 17, //нужно "добить" размер до кратности размеру слова (4), т.е. до 20. // Напишем обертку с необходимым выравниванием typedef struct tagAlignedWrapper { Record data; char align[3]; } AlignedWrapper;

Extreme 3 bytes can be simply ignored (not filled in and not processed in functions), but they will automatically align our structure.

This is useful for mixed data. In your case, 33 bits will be represented as:

 typedef struct tagRecord { char data[5]; // 7 бит в конце игнорируем char align[3]; } Record_33bit_aligned;

And 37 bits , oddly enough, in the same way =)

 typedef struct tagRecord { char data[5]; // 3 бита в конце игнорируем char align[3]; } Record_37bit_aligned;

N records of k bits each

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