A file which contains the buffer value. The first 16 bits contain the type. The next 32 bits gives the length of the data. The remaining value in the data.
How can I find the type from the 16 bits (find if it is int or char...)
I'm super stuck in my though process here. Not able to find a way to convert bits to types.
Say you have the homework assignment:
You are given a file where the first bit encodes the type, the
next 7 bits encode the length, and the rest is the data.
The types are encoded in the following way:
0 is for int
1 is for char
Print the ints or chars separated by newlines.
You just use the given information! Since 1 bit is used to encode the type there are two possible types. So you just read the first bit, then do:
if (bit == 0) {
int *i = ...
}
else if (bit == 1) {
char *c = ...
}
Related
I've got to program a function that receives
a binary number like 10001, and
a decimal number that indicates how many shifts I should perform.
The problem is that if I use the C++ operator <<, the zeroes are pushed from behind but the first numbers aren't dropped... For example
shifLeftAddingZeroes(10001,1)
returns 100010 instead of 00010 that is what I want.
I hope I've made myself clear =P
I assume you are storing that information in int. Take into consideration, that this number actually has more leading zeroes than what you see, ergo your number is most likely 16 bits, meaning 00000000 00000001 . Maybe try AND-ing it with number having as many 1 as the number you want to have after shifting? (Assuming you want to stick to bitwise operations).
What you want is to bit shift and then limit the number of output bits which can be active (hold a value of 1). One way to do this is to create a mask for the number of bits you want, then AND the bitshifted value with that mask. Below is a code sample for doing that, just replace int_type with the type of value your using -- or make it a template type.
int_type shiftLeftLimitingBitSize(int_type value, int numshift, int_type numbits=some_default) {
int_type mask = 0;
for (unsigned int bit=0; bit < numbits; bit++) {
mask += 1 << bit;
}
return (value << numshift) & mask;
}
Your output for 10001,1 would now be shiftLeftLimitingBitSize(0b10001, 1, 5) == 0b00010.
Realize that unless your numbits is exactly the length of your integer type, you will always have excess 0 bits on the 'front' of your number.
I am writing a program and using memcpy to copy some bytes of data, using the following code;
#define ETH_ALEN 6
unsigned char sourceMAC[6];
unsigned char destMAC[6];
char* txBuffer;
....
memcpy((void*)txBuffer, (void*)destMAC, ETH_ALEN);
memcpy((void*)(txBuffer+ETH_ALEN), (void*)sourceMAC, ETH_ALEN);
Now I want to copy some data on to the end of this buffer (txBuffer) that is less than a single byte or greater than one byte, so it is not a multiple of 8 (doesn't finish on a whole byte boundary), so memcpy() can't be used (I don't believe?).
I want to add 16 more bits worth of data which is a round 4 bytes. First I need to add a value into the next 3 bits of txtBuffer which I have stored in an int, and a fourth bit which is always 0. Next I need to copy another 12 bit value, again I have this in an int.
So the first decimal value stored in an int is between 0 and 7 inclusively, the same is true for the second number I mention to go into the final 12 bits. The stored value is within the rang of 2^12. Should I for example 'bit-copy' the last three bits of the int into memory, or merge all these values together some how?
Is there a way I can compile these three values into 4 bytes to copy with memcpy, or should I use something like bitset to copy them in, bit at a time?
How should I solve this issue?
Thank you.
Assuming int is 4 bytes on your platform
int composed = 0;
int three_bits = something;
int twelve_bits = something_else;
composed = (three_bits & 0x07) | (1 << 3) | ((twelve_bits << 4) & 0xFFFFFF0);
GIven the fact that I generate a string containing "0" and "1" of a random length, how can I write the data to a file as bits instead of ascii text ?
Given my random string has 12 bits, I know that I should write 2 bytes (or add 4 more 0 bits to make 16 bits) in order to write the 1st byte and the 2nd byte.
Regardless of the size, given I have an array of char[8] or int[8] or a string, how can I write each individual group of bits as one byte in the output file?
I've googled a lot everywhere (it's my 3rd day looking for an answer) and didn't understand how to do it.
Thank you.
You don't do I/O with an array of bits.
Instead, you do two separate steps. First, convert your array of bits to a number. Then, do binary file I/O using that number.
For the first step, the types uint8_t and uint16_t found in <stdint.h> and the bit manipulation operators << (shift left) and | (or) will be useful.
You haven't said what API you're using, so I'm going to assume you're using I/O streams. To write data to the stream just do this:
f.write(buf, len);
You can't write single bits, the best granularity you are going to get is bytes. If you want bits you will have to do some bitwise work to your byte buffer before you write it.
If you want to pack your 8 element array of chars into one byte you can do something like this:
char data[8] = ...;
char byte = 0;
for (unsigned i = 0; i != 8; ++i)
{
byte |= (data[i] & 1) << i;
}
f.put(byte);
If data contains ASCII '0' or '1' characters rather than actual 0 or 1 bits replace the |= line with this:
byte |= (data[i] == '1') << i;
Make an unsigned char out of the bits in an array:
unsigned char make_byte(char input[8]) {
unsigned char result = 0;
for (int i=0; i<8; i++)
if (input[i] != '0')
result |= (1 << i);
return result;
}
This assumes input[0] should become the least significant bit in the byte, and input[7] the most significant.
I'm writing a program in C++ to listen to a stream of tcp messages from another program to give tracking data from a webcam. I have the socket connected and I'm getting all the information in but having difficulty splitting it up into the data I want.
Here's the format of the data coming in:
8 byte header:
4 character string,
integer
32 byte message:
integer,
float,
float,
float,
float,
float
This is all being stuck into a char array called buffer. I need to be able to parse out the different bytes into the primitives I need. I have tried making smaller sub arrays such as headerString that was filled by looping through and copying the first 4 elements of the buffer array and I do get the the correct hear ('CCV ') printed out. But when I try the same thing with the next for elements (to get the integer) and try to print it out I get weird ascii characters being printed out. I've tried converting the headerInt array to an integer with the atoi method from stdlib.h but it always prints out zero.
I've already done this in python using the excellent unpack method, is their any alternative in C++?
Any help greatly appreciated,
Jordan
Links
CCV packet structure
Python unpack method
The buffer only contains the raw image of what you read over the
network. You'll have to convert the bytes in the buffer to whatever
format you want. The string is easy:
std::string s(buffer + sOffset, 4);
(Assuming, of course, that the internal character encoding is the same
as in the file—probably an extension of ASCII.)
The others are more complicated, and depend on the format of the
external data. From the description of the header, I gather than the
integers are four bytes, but that still doesn't tell me anything about
their representation. Depending on the case, either:
int getInt(unsigned char* buffer, int offset)
{
return (buffer[offset ] << 24)
| (buffer[offset + 1] << 16)
| (buffer[offset + 2] << 8)
| (buffer[offset + 3] );
}
or
int getInt(unsigned char* buffer, int offset)
{
return (buffer[offset + 3] << 24)
| (buffer[offset + 2] << 16)
| (buffer[offset + 1] << 8)
| (buffer[offset ] );
}
will probably do the trick. (Other four byte representations of
integers are possible, but they are exceedingly rare. Similarly, the
conversion of the unsigned results of the shifts and or's into a int
is implementation defined, but in practice, the above will work almost
everywhere.)
The only hint you give concerning the representation of the floats is in
the message format: 32 bytes, minus a 4 byte integer, leave 28 bytes for
5 floats; but 28 doesn't go into five, so I cannot even guess as to the
length of the floats (except that there must be some padding in there
somewhere). But converting floating point can be more or less
complicated if the external format isn't exactly like the internal
format.
Something like this may work:
struct {
char string[4];
int integers[2];
float floats[5];
} Header;
Header* header = (Header*)buffer;
You should check that sizeof(Header) == 32.
I am learning C/C++ programming & have encountered the usage of 'Bit arrays' or 'Bit Vectors'. Am not able to understand their purpose? here are my doubts -
Are they used as boolean flags?
Can one use int arrays instead? (more memory of course, but..)
What's this concept of Bit-Masking?
If bit-masking is simple bit operations to get an appropriate flag, how do one program for them? is it not difficult to do this operation in head to see what the flag would be, as apposed to decimal numbers?
I am looking for applications, so that I can understand better. for Eg -
Q. You are given a file containing integers in the range (1 to 1 million). There are some duplicates and hence some numbers are missing. Find the fastest way of finding missing
numbers?
For the above question, I have read solutions telling me to use bit arrays. How would one store each integer in a bit?
I think you've got yourself confused between arrays and numbers, specifically what it means to manipulate binary numbers.
I'll go about this by example. Say you have a number of error messages and you want to return them in a return value from a function. Now, you might label your errors 1,2,3,4... which makes sense to your mind, but then how do you, given just one number, work out which errors have occured?
Now, try labelling the errors 1,2,4,8,16... increasing powers of two, basically. Why does this work? Well, when you work base 2 you are manipulating a number like 00000000 where each digit corresponds to a power of 2 multiplied by its position from the right. So let's say errors 1, 4 and 8 occur. Well, then that could be represented as 00001101. In reverse, the first digit = 1*2^0, the third digit 1*2^2 and the fourth digit 1*2^3. Adding them all up gives you 13.
Now, we are able to test if such an error has occured by applying a bitmask. By example, if you wanted to work out if error 8 has occured, use the bit representation of 8 = 00001000. Now, in order to extract whether or not that error has occured, use a binary and like so:
00001101
& 00001000
= 00001000
I'm sure you know how an and works or can deduce it from the above - working digit-wise, if any two digits are both 1, the result is 1, else it is 0.
Now, in C:
int func(...)
{
int retval = 0;
if ( sometestthatmeans an error )
{
retval += 1;
}
if ( sometestthatmeans an error )
{
retval += 2;
}
return retval
}
int anotherfunc(...)
{
uint8_t x = func(...)
/* binary and with 8 and shift 3 plaes to the right
* so that the resultant expression is either 1 or 0 */
if ( ( ( x & 0x08 ) >> 3 ) == 1 )
{
/* that error occurred */
}
}
Now, to practicalities. When memory was sparse and protocols didn't have the luxury of verbose xml etc, it was common to delimit a field as being so many bits wide. In that field, you assign various bits (flags, powers of 2) to a certain meaning and apply binary operations to deduce if they are set, then operate on these.
I should also add that binary operations are close in idea to the underlying electronics of a computer. Imagine if the bit fields corresponded to the output of various circuits (carrying current or not). By using enough combinations of said circuits, you make... a computer.
regarding the usage the bits array :
if you know there are "only" 1 million numbers - you use an array of 1 million bits. in the beginning all bits will be zero and every time you read a number - use this number as index and change the bit in this index to be one (if it's not one already).
after reading all numbers - the missing numbers are the indices of the zeros in the array.
for example, if we had only numbers between 0 - 4 the array would look like this in the beginning: 0 0 0 0 0.
if we read the numbers : 3, 2, 2
the array would look like this: read 3 --> 0 0 0 1 0. read 3 (again) --> 0 0 0 1 0. read 2 --> 0 0 1 1 0. check the indices of the zeroes: 0,1,4 - those are the missing numbers
BTW, of course you can use integers instead of bits but it may take (depends on the system) 32 times memory
Sivan
Bit Arrays or Bit Vectors can be though as an array of boolean values. Normally a boolean variable needs at least one byte storage, but in a bit array/vector only one bit is needed.
This gets handy if you have lots of such data so you save memory at large.
Another usage is if you have numbers which do not exactly fit in standard variables which are 8,16,32 or 64 bit in size. You could this way store into a bit vector of 16 bit a number which consists of 4 bit, one that is 2 bit and one that is 10 bits in size. Normally you would have to use 3 variables with sizes of 8,8 and 16 bit, so you only have 50% of storage wasted.
But all these uses are very rarely used in business aplications, the come to use often when interfacing drivers through pinvoke/interop functions and doing low level programming.
Bit Arrays of Bit Vectors are used as a mapping from position to some bit value. Yes it's basically the same thing as an array of Bool, but typical Bool implementation is one to four bytes long and it uses too much space.
We can store the same amount of data much more efficiently by using arrays of words and binary masking operations and shifts to store and retrieve them (less overall memory used, less accesses to memory, less cache miss, less memory page swap). The code to access individual bits is still quite straightforward.
There is also some bit field support builtin in C language (you write things like int i:1; to say "only consume one bit") , but it is not available for arrays and you have less control of the overall result (details of implementation depends on compiler and alignment issues).
Below is a possible way to answer to your "search missing numbers" question. I fixed int size to 32 bits to keep things simple, but it could be written using sizeof(int) to make it portable. And (depending on the compiler and target processor) the code could only be made faster using >> 5 instead of / 32 and & 31 instead of % 32, but that is just to give the idea.
#include <stdio.h>
#include <errno.h>
#include <stdint.h>
int main(){
/* put all numbers from 1 to 1000000 in a file, except 765 and 777777 */
{
printf("writing test file\n");
int x = 0;
FILE * f = fopen("testfile.txt", "w");
for (x=0; x < 1000000; ++x){
if (x == 765 || x == 777760 || x == 777791){
continue;
}
fprintf(f, "%d\n", x);
}
fprintf(f, "%d\n", 57768); /* this one is a duplicate */
fclose(f);
}
uint32_t bitarray[1000000 / 32];
/* read file containing integers in the range [1,1000000] */
/* any non number is considered as separator */
/* the goal is to find missing numbers */
printf("Reading test file\n");
{
unsigned int x = 0;
FILE * f = fopen("testfile.txt", "r");
while (1 == fscanf(f, " %u",&x)){
bitarray[x / 32] |= 1 << (x % 32);
}
fclose(f);
}
/* find missing number in bitarray */
{
int x = 0;
for (x=0; x < (1000000 / 32) ; ++x){
int n = bitarray[x];
if (n != (uint32_t)-1){
printf("Missing number(s) between %d and %d [%x]\n",
x * 32, (x+1) * 32, bitarray[x]);
int b;
for (b = 0 ; b < 32 ; ++b){
if (0 == (n & (1 << b))){
printf("missing number is %d\n", x*32+b);
}
}
}
}
}
}
That is used for bit flags storage, as well as for parsing different binary protocols fields, where 1 byte is divided into a number of bit-fields. This is widely used, in protocols like TCP/IP, up to ASN.1 encodings, OpenPGP packets, and so on.