I just started learning socket programming, and I'm trying to implement TCP/UDP protocols using raw sockets.
IP Header
0 7 8 15 16 23 24 31
+--------+--------+--------+--------+
|Ver.|IHL|DSCP|ECN| Total length |
+--------+--------+--------+--------+
| Identification |Flags| Offset |
+--------+--------+--------+--------+
| TTL |Protocol| Header Checksum |
+--------+--------+--------+--------+
| Source IP address |
+--------+--------+--------+--------+
| Destination IP address |
+--------+--------+--------+--------+
When writing IP header, the Flags and Offset part, the length of Offset is not multiple of 8 bit. So I take Flags and Offset together as a whole.
uint8 flags = 0;
uint16 offset = htons(6000); // more than 1 byte, so we need to use htons
// in c, we can left(since it's in big endianness) shift offset 3 bit,
// and then convert flags to uint16, and then merge them together
// in some other languages, for example, Haskell,
// htons like functions may return a bytestring which is not an instance of Bit,
// we need to unpack it back into a list of uint8 in order to use bitwise operations.
This method is not very clean, I'm wondering what's the usual way to construct bytestring when its components are of length more than 1 byte and their endianness also needs to be considered.
In C, the usual way would be to declare a uint16_t, uint32_t, or uint64_t temporary variable, use bitwise operators to assemble the bits within that variable, and then use htons() or htonl() to convert the bits into network (aka big-endian) order.
For example, the Flags and Offset fields, taken together, constitute a 16-bit word. So:
uint8_t flags = /* some 3-bit value */;
uint16_t offset = /* some 13-bit value */;
uint16_t flagsAndOffsetBigEndian = htons(flags | (offset << 3));
memcpy(&header[16], &flagsAndOffsetBigEndian, sizeof(uint16_t));
Related
#include<stdio.h>
int main()
{
typedef unsigned char *byte_pointer;
void show_bytes(byte_pointer start, size_t len)
{
int i;
for (i = 0; i < len; i++)
{
printf(" %.2x", start[i]);
printf("\n");
}
}
void show_int(int x)
{
show_bytes((byte_pointer) &x, sizeof(int));
}
void show_float(int x)
{
show_bytes((byte_pointer) &x, sizeof(float));
}
void show_pointer(int x)
{
show_bytes((byte_pointer) &x, sizeof(void *));
}
int a = 0x12345678;
byte_pointer ap = (byte_pointer) &a;
show_bytes(ap, 3);
return 0;
}
(Solutions according to the CS:APP book)
Big endian: 12 34 56
Little endian: 78 56 34
I know systems have different conventions for storage allocation but if two systems use the same convention but are different endian why are the hex values different?
Endian-ness is an issue that arises when we use more than one storage location for a value/type, which we do because somethings won't fit in a single storage location.
As soon as we use multiple storage locations for a single value that gives rise to the question of: What part of the value will we store in each storage location?
The first byte of a two-byte item will have a lower address than the second byte, and in particular, the address of the second byte will be at +1 from the address of the lower byte.
Storing a two-byte item in two bytes of storage, do we store the most significant byte first and the least significant byte second, or vice versa?
We choose to use directly consecutive bytes for the two bytes of the two-byte item, so no matter which (endian) way we choose to store such an item, we refer to the whole two-byte item by the lower address (the address of its first byte).
We can express these storage choices with a formula, here item[0] refer to the first byte while item[1] refers to the second byte.
item[0] = value >> 8 // also value / 256
item[1] = value & 0xFF // also value % 256
value = (item[0]<<8) | item[1] // also item[0]*256 | item[1]
--vs--
item[0] = value & 0xFF // also value % 256
item[1] = value >> 8 // also value / 256
value = item[0] | (item[1]<<8) // also item[0] | item[1]*256
The first set of formulas is for big endian, and the second for little endian.
By these formulas, it doesn't matter what order we access memory as to whether item[0] first, then item[1], or vice versa, or both at the same time (common in hardware), as long as the formulas for one endian are consistently used.
If the item in question is a four-byte value, then there are 4 possible orderings(!) — though only two of them are truly sensible.
For efficiency, the hardware offers us multibyte memory access in one instruction (and with one reference, namely to the lowest address of the multibyte item), and therefore, the hardware itself needs to define and consistently use one of the two possible/reasonable orderings.
If the hardware did not offer multibyte memory access, then the ordering would be entirely up to the software program itself to define (accessing memory one byte at a time), and the program could choose big or little endian, even differently for each variable, as long as it consistently accesses the multiple bytes of memory in the same manner to reassemble the values stored there.
In a similar manner, when we define a structure of multiple items (e.g. struct point { int x; int y; }, software chooses whether x comes first or y comes first in memory ordering. However, since programmers (and compilers) will still choose to use hardware instructions to access individual fields such as x in one go, the hardware's endian configuration remains necessary.
Here is the structure of the packet header in pcap:
struct pcap_pkthdr {
struct timeval ts; /* time stamp */
bpf_u_int32 caplen; /* length of portion present */
bpf_u_int32 len; /* length this packet (off wire)*/
};
I wonder what is the real difference between caplen and len? And where are they used?
len is the actual length of the packet on the wire. caplen is the length which is captured and thus present in the pcap file. caplen can be the same but also smaller than len.
How many bytes of a packet will be captured can be specified for example in tcpdump with -s size. While on many system tcpdump will capture up to 64k by default for example on OpenBSD it will only capture 116 bytes by default.
I am receiving EEG data from a 24 bit ADC over serial. The ADC data is transmitting in 3 bytes from MSB to LSB. The full packet is 21 bytes:
The first byte is the start byte - 0xFF (255 in decimal)
Then packet number byte.
Then the next 3 bytes are the 24 bit ADC value broken into MSB LSB2 LSB1
I can parse the data fine, but re-constructing a 2's complement signed int32 number is causing issues. The values I am getting out certainly don't reflect what the ADC should be giving out.
Below are the lines to read and parse the 504 samples (which gives me 24 ADC values (504samples/21bytes = 24 values)). I have tried uint8 instead of uchar with similar results (when I try int8 I get a invalid specified precision error).
comEEGSMT = serial(com,'BaudRate',3000000);
fopen(comEEGSMT);
rawData(1:504) = fread(comEEGSMT, 504, 'uchar');
fclose(comEEGSMT);
startPackets = find(rawData == 255);
bytes = rawData([startpackets+2 startpackets+3 startpackets+4]);
I have tried the following method to reconstruct the value:
ADC_value = bytes(:,1)*256^2 + bytes(:,2)*256 + bytes(:,3);
and the following line is the formula to convert the above number to volts:
ADC_value_volts = ADC_value*(5/3)*(1/(2^32));
The values are in the range of 4000 - 8000 microvolts with large jumps in value. The values SHOULD be in the range of 200 - 600 microvolts with small changes.
I have found other questions relating to similar issues, but have had no success trying the proposed solutions such as in the link below:
https://uk.mathworks.com/matlabcentral/answers/137965-concatenate-3-bytes-array-of-real-time-serial-data-into-single-precision
Any help would be very much appreciated as I've been stuck on this for quite long.
Thanks Mark
Starting with ADC_value as int32 with value 0, then:
ADC_value |= MSB << 16;
ADC_value |= LSB2 << 8;
ADC_value |= LSB1;
And then, to find out the corresponding volts value, supposing your ADC has a reference voltage VREF, in volts (e.g. 5.0V):
ADC_value_volts = (ADC_value * VREF)/2^24
since your converter is 24 bits, not 32.
Note the above expressions are in C language equivalent, not Matlab.
EDIT:
The ADC data sheet tell us the PGA gain can be set for the following values:
1, 2, 4, 6, 8, 12, 24, one value at the time for each channel.
The FSR (full scale range) of measurement is: (2*VREF)/Gain = 5/3, for Gain=6,
(eq.(5) page 23) so this must be accounted for in expression computing the volts
values. (these can be verified if you have access to the hardware and can make some
measurements).
Data resulted from ADC is already in two's complement, binary form, 24 bits.
The weird thing is the data sheet counts bits starting with 1, not 0, so this
is why shifting with "17" instead 16 - this is in fact 16 for coding.
(revealed in fig 47, page 42).
So the computing formula of ADC_value_volts should be:
ADC_value_volts = (AC_value * FSR/(2^23))/3 (1LSB=FSR/(2^23), pg.37)
If some other calculations/modifications from original, them these must be explained by provider.
If the provider is not friendly, worth to be changed...
One thing I've never really understood is why in many libraries, constants are defined like this:
public static final int DM_FILL_BACKGROUND = 0x2;
public static final int DM_FILL_PREVIOUS = 0x3;
public static final int TRANSPARENCY_MASK = 1 << 1;
public static final int TRANSPARENCY_PIXEL = 1 << 2;
What's up with the 0x and << stuff? Why aren't people just using ordinary integer values?
The bit shifting of 1 is usually for situations where you have non-exclusive values that you want to store.
For example, say you want to be able to draw lines on any side of a box. You define:
LEFT_SIDE = 1 << 0 # binary 0001 (1)
RIGHT_SIDE = 1 << 1 # binary 0010 (2)
TOP_SIDE = 1 << 2 # binary 0100 (4)
BOTTOM_SIDE = 1 << 3 # binary 1000 (8)
----
0111 (7) = LEFT_SIDE | RIGHT_SIDE | TOP_SIDE
Then you can combine them for multiple sides:
DrawBox (LEFT_SIDE | RIGHT_SIDE | TOP_SIDE) # Don't draw line on bottom.
The fact that they're using totally different bits means that they're independent of each other. By ORing them you get 1 | 2 | 4 which is equal to 7 and you can detect each individual bit with other boolean operations (see here and here for an explanation of these).
If they were defined as 1, 2, 3 and 4 then you'd probably either have to make one call for each side or you'd have to pass four different parameters, one per side. Otherwise you couldn't tell the difference between LEFT and RIGHT (1 + 2 = 3) and TOP (3), since both of them would be the same value (with a simple addition operation).
The 0x stuff is just hexadecimal numbers which are easier to see as binary bitmasks (each hexadecimal digit corresponds exactly with four binary digits. You'll tend to see patterns like 0x01, 0x02, 0x04, 0x08, 0x10, 0x20 and so on, since they're the equivalent of a single 1 bit moving towards the most significant bit position - those values are equivalent to binary 00000001, 00000010, 00000100, 00001000, 00010000, 00100000 and so on.
Aside: Once you get used to hex, you rarely have to worry about the 1 << n stuff. You can instantly recognise 0x4000 as binary 0100 0000 0000 0000. That's less obvious if you see the value 16384 in the code although some of us even recognise that :-)
Regarding << stuff: this in my preferred way.
When I need to define the constant with 1 in the bit 2 position, and 0 in all other bits, I can define it as 4, 0x4 or 1<<2. 1<<2 is more readable, to my opinion, and explains exactly the purpose of this constant.
BTW, all these ways give the same performance, since calculations are done at compile time.
To reduce the download size of an iPhone application I'm compressing some audio files. Specifically I'm using afconvert on the command line to change .wav format to .caf format w/ ima4 compression.
I've read this (wooji-juice.com) awesome post about this exact topic. I'm having trouble w/ the "decoding ima4 packets" step. I've looked at their sample code and I'm stuck. Please help w/ some pseudo code or sample code that can guide me in the right direction.
Thanks!
Additional info:
Here is what I've completed and where I'm having trouble...
I can play .wav files in both the simulator and on the phone.
I can compress .wav files to .caf w/ ima4 compression using afconvert on the command line. I'm using the SoundEngine that came w/ CrashLanding (I fixed one memory leak).
I modified the SoundEngine code to look for the mFormatID 'ima4'.
I don't understand the blog post linked above starting w/ "Calculating the size of the unpacked data". Why do I need to do this? Also, what does the term "packet" refer to? I'm very new to any sort of audio programming.
After gathering all the data from Wooji-Juice, Multimedia Wiki and Apple, here is my proposal (may need some experiment):
File structure
Apple IMA4 file are made of packet of 34 bytes. This is the packet unit used to build the file.
Each 34 bytes packet has two parts:
the first 2 bytes contain the preamble: an initial predictor and a step index
the 32 bytes left contain the sound nibbles (a nibble of 4 bits is used to retrieve a 16 bits sample)
Each packet has 32 bytes of compressed data, that represent 64 samples of 16 bits.
If the sound file is stereo, the packets are interleaved (one for the left, one for the right); there must be an even number of packets.
Decoding
Each packet of 34 bytes will lead to the decompression of 64 samples of 16 bits. So the size of the uncompressed data is 128 bytes per packet.
The decoding pseudo code looks like:
int[] ima_index_table = ... // Index table from [Multimedia Wiki][2]
int[] step_table = ... // Step table from [Multimedia Wiki][2]
byte[] packet = ... // A packet of 34 bytes compressed
short[] output = ... // The output buffer of 128 bytes
int preamble = (packet[0] << 8) | packet[1];
int predictor = preamble && 0xFF80; // See [Multimedia Wiki][2]
int step_index = preamble && 0x007F; // See [Multimedia Wiki][2]
int i;
int j = 0;
for(i = 2; i < 34; i++) {
byte data = packet[i];
int lower_nibble = data && 0x0F;
int upper_nibble = (data && 0xF0) >> 4;
// Decode the lower nibble
step_index += ima_index_table[lower_nibble];
diff = ((signed)nibble + 0.5f) * step / 4;
predictor += diff;
step = ima_step_table[step index];
// Clamp the predictor value to stay in range
if (predictor > 65535)
output[j++] = 65535;
else if (predictor < -65536)
output[j++] = -65536;
else
output[j++] = (short) predictor;
// Decode the uppper nibble
step_index += ima_index_table[upper_nibble];
diff = ((signed)nibble + 0.5f) * step / 4;
predictor += diff;
step = ima_step_table[step index];
// Clamp the predictor value to stay in range
if (predictor > 65535)
output[j++] = 65535;
else if (predictor < -65536)
output[j++] = -65536;
else
output[j++] = (short) predictor;
}
The term "packet" refers to a group of compressed audio samples with a header. You need the header to decode the data immediately following. If you consider your ima4 file to be a book, then each packet is a page. At the top are the values needed to decode that page, followed by the compressed audio.
That's why you need to calculate the size of the unpacked data (and then make space for it) -- since it's compressed, you need to convert data from compressed audio to uncompressed audio before you can output it. In order to allocate an output buffer, you need to know how big it has to be (note: you may need to output in chunks that are larger than a single packet at a time).
It looks like the typical structure, per the earlier "Overview" section, is that sets of 64 samples, each 16 bits (so 128 bytes) are translated to a 2-byte header and a 32-byte set of compressed samples (34 bytes in all). So, in the typical case, you can produce your expected output datasize by taking the input data size, dividing by 34 to get the number of packets, then multiplying by 128 bytes for the uncompressed audio per packet.
You shouldn't do that, though. It looks like you should instead query kAudioFilePropertyDataFormat to get the mBytesPerPacket -- this is the "34" value above, and mFramesPerPacket -- this is the 64, above, that gets multiplied by 2 (for 16-byte samples) to make 128 bytes of output.
Then, for each packet, you will need to run through the decoding described in the post. In somewhat longer pseudo C-code, assuming you are getting arrays of bytes, to handle the header:
packet = GetPacket();
Header = (packet[0] << 8) | packet[1]; //Big-endian 16-bit value
step_index = Header & 0x007f; //Lower seven bits
predictor = Header & 0xff80; //Upper nine bits
for (i = 2; i < mBytesPerPacket; i++)
{
nibble = packet[i] & 0x0f; //Low Nibble
process that nibble, per the blogpost -- be careful on sign-extension!
nibble = (packet[i] & 0xf0) >> 4; //High Nibble
process that nibble, per the blogpost -- be careful on sign-extension!
}
The sign-extension above refers to the fact that the post involves handling each nibble both in an unsigned and a signed way. If the high bit of a nibble (bit 3) is a 1, then it is negative; additionally the bit-shift may do sign-extension. This is not handled in the above pseudocode.