We are writing an embedded application code and validating a string for a valid IPv4 format. I am successfully able to do so using string tokenizer but now I need to convert the integers to Host-To-Network order using htonl() function.
Since it an embedded application I cannot include network header and library just to make use of htonl() function.
Is there any way / non-network header in C++ by which I can avail htonl() functionality?
From htonl()'s man page:
The htonl() function converts the unsigned integer hostlong from host byte order to network byte order.
Network byte order is actually just big endian.
All you need to do is write (or find) a function that converts an unsigned integer to big endian and use it in place of htonl. If your system is already in big endian than you don't need to do anything at all.
You can use the following to determine the endianness of your system:
int n = 1;
// little endian if true
if(*(char *)&n == 1) {...}
Source
And you can convert a little endian uint32_t to big endian using the following:
uint32_t htonl(uint32_t x) {
unsigned char *s = (unsigned char *)&x;
return (uint32_t)(s[0] << 24 | s[1] << 16 | s[2] << 8 | s[3]);
}
Source
You don't strictly need htonl. If you have the IP address as individual bytes like this:
uint8_t a [4] = { 192, 168, 2, 1 };
You can just send these 4 bytes, in that exact order, over the network. That is unless you specifically need it as a 4 byte Integer, which you probably don't, since you presumably are not using sockaddr_in & friends.
If you already have the address as a 32 bit integer in host byte order, you can obtain a like this:
uint32_t ip = getIPHostOrder();
uint8_t a [4] = { (ip >> 24) & 0xFF, (ip >> 16) & 0xFF, (ip >> 8) & 0xFF, ip & 0xFF };
This has the advantage of not relying on implementation defined behaviour and being portable.
Related
my task is to read metadata values from a unsigned char array, which contains the bytes of a binary .shp file (Shapefile)
unsigned char* bytes;
The header of the file which is stored in the array and the order of the information stored in it looks like this:
int32_t filecode // BigEndian
int32_t skip[5] // Uninteresting stuff
int32_t filelength // BigEndian
int32_t version // LitteEndian
int32_t shapetype // LitteEndian
// Rest of the header and of the filecontent which I don't need
So my question would be how can I extract this information (except the skip part of course) under consideration of the endianness and read it into the according variables.
I thought about using ifstream, but I couldnt figure out how to do it properly.
Example:
Read the first four bytes of the binary, ensure big endian byte order, store it in a int32_t. Then skip 5* 4 Bytes (5 * int32). Then read four bytes, ensure big endian byte order, and store it in a int32_t. Then read four bytes, ensure little endian byte order, and again store it in a int32_t and so on.
Thanks for your help guys!
So 'reading' a byte array just means extracting the bytes from the positions in the byte array where you know your data is stored. Then you just need to do the appropriate bit manipulations to handle the endianess. So for example, filecode would be this
filecode = (bytes[0] << 24) | (bytes[1] << 16) | (bytes[2] << 8) | bytes[3];
and version would be this
version = bytes[13] | (bytes[14] << 8) | (bytes[15] << 16) | (bytes[16] << 24);
(An offset of 13 for the version seems a bit odd, I'm just going on what you stated above).
I'm writing a program that creates MIDI files, and I'm trying to write the midi messages on a file.
I tested first all the way to create file from zero using the function fputc() and inputting byte per byte all the file, and it went well.
The problem came when I tried to write more than one byte at the same time (e.g. writing a short int or an int into the file), because the function fwrite() put the bytes backwards.
For example:
FILE* midiFile;
midiFile = fopen("test.mid", "wb");
short msg = 0x0006;
fwrite(msg, sizeof(msg), 1, midiFile);
fclose(midifile);
The output written int the file its 0x06 the 0x00, and not the expected: 0x00,0x06.
I read about that, and find that it's caused by the endianness; my Intel processor uses little endian so it writes variables bigger than 1 byte backwards (compared to a big endian machine).
I still need to correct that and write the bytes the way I want to develop my program.
My compiler doesn't identify functions like htonl() or similar (I don't know why) but I'm asking a way to do it, or how to write short's and int's on char arrays (especially short's).
Either write the bytes you want in order, one at a time...
or swap the bytes before you write them.
uint8_t msbyte = msg >> 8;
uint8_t lsbyte = msg & 0xFF;
uint8_t buffer[2];
// Big Endian
buffer[0] = msbyte;
buffer[1] = lsbyte;
/* Little endian
buffer[0] = lsbyte;
buffer[1] = msbyte;
*/
fwrite(&buffer[0], 1, sizeof(buffer), midiFile);
Swapping bytes:
uint16_t swap_bytes(const uint16_t value)
{
uint16_t result;
result = value >> 8;
result += (value & 0xFF) << 8;
return result;
}
I'm following this tutorial for using OpenAL in C++: http://enigma-dev.org/forums/index.php?topic=730.0
As you can see in the tutorial, they leave a few methods unimplemented, and I am having trouble implementing file_read_int32_le(char*, FILE*) and file_read_int16_le(char*, FILE*). Apparently what it should do is load 4 bytes from the file (or 2 in the case of int16 I guess..), convert it from little-endian to big endian and then return it as an unsigned integer. Here's the code:
static unsigned int file_read_int32_le(char* buffer, FILE* file) {
size_t bytesRead = fread(buffer, 1, 4, file);
printf("%x\n",(unsigned int)*buffer);
unsigned int* newBuffer = (unsigned int*)malloc(4);
*newBuffer = ((*buffer << 24) & 0xFF000000U) | ((*buffer << 8) & 0x00FF0000U) | ((*buffer >> 8) & 0x0000FF00U) | ((*buffer >> 24) & 0x000000FFU);
printf("%x\n", *newBuffer);
return (unsigned int)*newBuffer;
}
When debugging (in XCode) it says that the hexadecimal value of *buffer is 0x72, which is only one byte. When I create newBuffer using malloc(4), I get a 4-byte buffer (*newBuffer is something like 0xC0000003) which then, after the operations, becomes 0x72000000. I assume the result I'm looking for is 0x00000027 (edit: actually 0x00000072), but how would I achieve this? Is it something to do with converting between the char* buffer and the unsigned int* newBuffer?
Yes, *buffer will read in Xcode's debugger as 0x72, because buffer is a pointer to a char.
If the first four bytes in the memory block pointed to by buffer are (hex) 72 00 00 00, then the return value should be 0x00000072, not 0x00000027. The bytes should get swapped, but not the two "nybbles" that make up each byte.
This code leaks the memory you malloc'd, and you don't need to malloc here anyway.
Your byte-swapping is correct on a PowerPC or 68K Mac, but not on an Intel Mac or ARM-based iOS. On those platforms, you don't have to do any byte-swapping because they're natively little-endian.
Core Foundation provides a way to do this all much more easily:
static uint32_t file_read_int32_le(char* buffer, FILE* file) {
fread(buffer, 1, 4, file); // Get four bytes from the file
uint32_t val = *(uint32_t*)buffer; // Turn them into a 32-bit integer
// Swap on a big-endian Mac, do nothing on a little-endian Mac or iOS
return CFSwapInt32LittleToHost(val);
}
there's a whole range of functions called "htons/htonl/hton" whose sole purpose in life is to convert from "host" to "network" byte order.
http://beej.us/guide/bgnet/output/html/multipage/htonsman.html
Each function has a reciprocal that does the opposite.
Now, these functions won't help you necessarily because they intrinsically convert from your hosts specific byte order, so please just use this answer as a starting point to find what you need. Generally code should never make assumptions about what architecture it's on.
Intel == "Little Endian".
Network == "Big Endian".
Hope this starts you out on the right track.
I've used the following for integral types. On some platforms, it's not safe for non-integral types.
template <typename T> T byte_reverse(T in) {
T out;
char* in_c = reinterpret_cast<char *>(&in);
char* out_c = reinterpret_cast<char *>(&out);
std::reverse_copy(in_c, in_c+sizeof(T), out_c);
return out;
};
So, to put that in your file reader (why are you passing the buffer in, since it appears that it could be a temporary)
static unsigned int file_read_int32_le(FILE* file) {
unsigned int int_buffer;
size_t bytesRead = fread(&int_buffer, 1, sizeof(int_buffer), file);
/* Error or less than 4 bytes should be checked */
return byte_reverse(int_buffer);
}
What's the best way to send float, double, and int16 over serial on Arduino?
The Serial.print() only sends values as ASCII encoded. But I want to send the values as bytes. Serial.write() accepts byte and bytearrays, but what's the best way to convert the values to bytes?
I tried to cast an int16 to an byte*, without luck. I also used memcpy, but that uses to many CPU cycles. Arduino uses plain C/C++. It's an ATmega328 microcontroller.
hm. How about this:
void send_float (float arg)
{
// get access to the float as a byte-array:
byte * data = (byte *) &arg;
// write the data to the serial
Serial.write (data, sizeof (arg));
}
Yes, to send these numbers you have to first convert them to ASCII strings. If you are working with C, sprintf() is, IMO, the handiest way to do this conversion:
[Added later: AAAGHH! I forgot that for ints/longs, the function's input argument wants to be unsigned. Likewise for the format string handed to sprintf(). So I changed it below. Sorry about my terrible oversight, which would have been a hard-to-find bug. Also, ulong makes it a little more general.]
char *
int2str( unsigned long num ) {
static char retnum[21]; // Enough for 20 digits plus NUL from a 64-bit uint.
sprintf( retnum, "%ul", num );
return retnum;
}
And similar for floats and doubles. The code doing the conversion has be known in advance. It has to be told - what kind of an entity it's converting, so you might end up with functions char *float2str( float float_num) and char *dbl2str( double dblnum).
You'll get a NUL-terminated left-adjusted (no leading blanks or zeroes) character string out of the conversion.
You can do the conversion anywhere/anyhow you like; these functions are just illustrations.
Use the Firmata protocol. Quote:
Firmata is a generic protocol for communicating with microcontrollers
from software on a host computer. It is intended to work with any host
computer software package. Right now there is a matching object in a
number of languages. It is easy to add objects for other software to
use this protocol. Basically, this firmware establishes a protocol for
talking to the Arduino from the host software. The aim is to allow
people to completely control the Arduino from software on the host
computer.
The jargon word you need to look up is "serialization".
It is an interesting problem over a serial connection which might have restrictions on what characters can go end to end, and might not be able to pass eight bits per character either.
Restrictions on certain character codes are fairly common. Here's a few off the cuff:
If software flow control is in use, then conventionally the control characters DC1 and DC3 (Ctrl-Q and Ctrl-S, also sometimes called XON and XOFF) cannot be transmitted as data because they are sent to start and stop the sender at the other end of the cable.
On some devices, NUL and/or DEL characters (0x00 and 0x7F) may simply vanish from the receiver's FIFO.
If the receiver is a Unix tty, and the termio modes are not set correctly, then the character Ctrl-D (EOT or 0x04) can cause the tty driver to signal an end-of-file to the process that has the tty open.
A serial connection is usually configurable for byte width and possible inclusion of a parity bit. Some connections will require that a 7-bit byte with a parity are used, rather than an 8-bit byte. It is even possible for connection to (seriously old) legacy hardware to configure many serial ports for 5-bit and 6-bit bytes. If less than 8-bits are available per byte, then a more complicated protocol is required to handle binary data.
ASCII85 is a popular technique for working around both 7-bit data and restrictions on control characters. It is a convention for re-writing binary data using only 85 carefully chosen ASCII character codes.
In addition, you certainly have to worry about byte order between sender and receiver. You might also have to worry about floating point format, since not every system uses IEEE-754 floating point.
The bottom line is that often enough choosing a pure ASCII protocol is the better answer. It has the advantage that it can be understood by a human, and is much more resistant to issues with the serial connection. Unless you are sending gobs of floating point data, then inefficiency of representation may be outweighed by ease of implementation.
Just be liberal in what you accept, and conservative about what you emit.
Does size matter? If it does, you can encode each 32 bit group into 5 ASCII characters using ASCII85, see http://en.wikipedia.org/wiki/Ascii85.
This simply works. Use Serial.println() function
void setup() {
Serial.begin(9600);
}
void loop() {
float x = 23.45585888;
Serial.println(x, 10);
delay(1000);
}
And this is the output:
Perhaps that is best Way to convert Float to Byte and Byte to Float,-Hamid Reza.
int breakDown(int index, unsigned char outbox[], float member)
{
unsigned long d = *(unsigned long *)&member;
outbox[index] = d & 0x00FF;
index++;
outbox[index] = (d & 0xFF00) >> 8;
index++;
outbox[index] = (d & 0xFF0000) >> 16;
index++;
outbox[index] = (d & 0xFF000000) >> 24;
index++;
return index;
}
float buildUp(int index, unsigned char outbox[])
{
unsigned long d;
d = (outbox[index+3] << 24) | (outbox[index+2] << 16)
| (outbox[index+1] << 8) | (outbox[index]);
float member = *(float *)&d;
return member;
}
regards.
`
Structures and unions solve that issue. Use a packed structure with a byte sized union matching the structure. Overlap the pointers to the structure and union (or add the union in the structure). Use Serial.write to send the stream. Have a matching structure/union on receiving end. As long as byte order matches no issue otherwise you can unpack using the "C" hto(s..l) functions. Add "header" info to decode different structures/unions.
For Arduino IDE:
float buildUp(int index, unsigned char outbox[])
{
unsigned long d;
d = (long(outbox[index +3]) << 24) | \
(long(outbox[index +2]) << 16) | \
(long(outbox[index +1]) << 8) | \
(long(outbox[index]));
float member = *(float *)&d;
return member;
}
otherwise not working.
My background is php so entering the world of low-level stuff like char is bytes, which are bits, which is binary values, etc is taking some time to get the hang of.
What I am trying to do here is sent some values from an Ardunio board to openFrameWorks (both are c++).
What this script currently does (and works well for one sensor I might add) when asked for the data to be sent is:
int value_01 = analogRead(0); // which outputs between 0-1024
unsigned char val1;
unsigned char val2;
//some Complicated bitshift operation
val1 = value_01 &0xFF;
val2 = (value_01 >> 8) &0xFF;
//send both bytes
Serial.print(val1, BYTE);
Serial.print(val2, BYTE);
Apparently this is the most reliable way of getting the data across.
So now that it is send via serial port, the bytes are added to a char string and converted back by:
int num = ( (unsigned char)bytesReadString[1] << 8 | (unsigned char)bytesReadString[0] );
So to recap, im trying to get 4 sensors worth of data (which I am assuming will be 8 of those serialprints?) and to have int num_01 - num_04... at the end of it all.
Im assuming this (as with most things) might be quite easy for someone with experience in these concepts.
Write a function to abstract sending the data (I've gotten rid of your temporary variables because they don't add much value):
void send16(int value)
{
//send both bytes
Serial.print(value & 0xFF, BYTE);
Serial.print((value >> 8) & 0xFF, BYTE);
}
Now you can easily send any data you want:
send16(analogRead(0));
send16(analogRead(1));
...
Just send them one after the other.
Note that the serial driver lets you send one byte (8 bits) at a time. A value between 0 and 1023 inclusive (which looks like what you're getting) fits in 10 bits. So 1 byte is not enough. 2 bytes, i.e. 16 bits, are enough (there is some extra space, but unless transfer speed is an issue, you don't need to worry about this wasted space).
So, the first two bytes can carry the data for your first sensor. The next two bytes carry the data for the second sensor, the next two bytes for the third sensor, and the last two bytes for the last sensor.
I suggest you use the function that R Samuel Klatchko suggested on the sending side, and hopefully you can work out what you need to do on the receiving side.
int num = ( (unsigned char)bytesReadString[1] << 8 |
(unsigned char)bytesReadString[0] );
That code will not do what you expect.
When you shift an 8-bit unsigned char, you lose the extra bits.
11111111 << 3 == 11111000
11111111 << 8 == 00000000
i.e. any unsigned char, when shifted 8 bits, must be zero.
You need something more like this:
typedef unsigned uint;
typedef unsigned char uchar;
uint num = (static_cast<uint>(static_cast<uchar>(bytesReadString[1])) << 8 ) |
static_cast<uint>(static_cast<uchar>(bytesReadString[0]));
You might get the same result from:
typedef unsigned short ushort;
uint num = *reinterpret_cast<ushort *>(bytesReadString);
If the byte ordering is OK. Should work on Little Endian (x86 or x64), but not on Big Endian (PPC, Sparc, Alpha, etc.)
To generalise the "Send" code a bit --
void SendBuff(const void *pBuff, size_t nBytes)
{
const char *p = reinterpret_cast<const char *>(pBuff);
for (size_t i=0; i<nBytes; i++)
Serial.print(p[i], BYTE);
}
template <typename T>
void Send(const T &t)
{
SendBuff(&t, sizeof(T));
}