I am trying to write a C++ application to send a 64bit word to an Arduino.
I used termios using the method described here
The problem i am having is the byes are arriving at the arduino in least significant byte first.
ie
if a use (where serialword is a uint64_t)
write(fp,(const void*)&serialWord, 8);
the least significant bytes arrive at the arduino first.
this is not the behavior i was wanted, is there a way to get the most significant byes to arrive first? Or is it best to brake the serialword into bytes and send byte by byte?
Thanks
Since the endianess of the CPU's involved are different you will need to reverse the order of bytes before you send them or after your receive them. In this case I would recommend reversing them before you send them just to save CPU cycles on the Arduino. The simplest way using the C++ Standard Library is with std::reverse as shown in the following example
#include <cstdint> // uint64_t (example only)
#include <iostream> // cout (example only)
#include <algorithm> // std::reverse
int main()
{
uint64_t value = 0x1122334455667788;
std::cout << "Before: " << std::hex << value << std::endl;
// swap the bytes
std::reverse(
reinterpret_cast<char*>(&value),
reinterpret_cast<char*>(&value) + sizeof(value));
std::cout << "After: " << std::hex << value << std::endl;
}
This outputs the following:
Before: 1122334455667788
After: 8877665544332211
Related
I am trying to use functions of a library that reads and writes from/to streams. To adapt my data to that library, I wanted to use boost::iostreams from boost 1.77.0. Still, a first very simple example does not work as expected. Why?
#include <boost/iostreams/device/array.hpp>
#include <boost/iostreams/device/back_inserter.hpp>
#include <boost/iostreams/stream.hpp>
#include <iostream>
int main(int, char*[])
{
// Create container
std::vector<char> bytes;
// Set up stream to write three chars to container
boost::iostreams::back_insert_device<std::vector<char>> inserter =
boost::iostreams::back_inserter(bytes);
boost::iostreams::stream stream(inserter);
// Write chars
stream << 1;
stream << 2;
stream << 3;
stream.close();
// Check container
for (char entry : bytes)
{
std::cout << "Entry: " << entry << std::endl;
}
std::cout << "There are " << bytes.size() << " bytes." << std::endl;
// Set up stream to read chars from container
boost::iostreams::basic_array_source<char> source(bytes.data(), bytes.size());
boost::iostreams::stream stream2(source);
// Read chars from container
while (!stream2.eof())
{
std::cout << "Read entry " << stream2.get() << std::endl;
}
return 0;
}
The output is:
Entry: 1
Entry: 2
Entry: 3
There are 3 bytes.
Read entry 49
Read entry 50
Read entry 51
Read entry -1
Why does it read 49, 50 and 51 instead of 1, 2 and 3? The -1 doesn't surprise me that much, it might denote the end of the container. Am I using the classes in a wrong way?
It works correct to me, but not in the intuitive way though. You are putting integers 1 2 and 3 into the vector of chars via stream, so they land there as their ASCII codes, 49, 50 and 51 respectively. In the initial loop you are actually printing characters, not their integer representations, thus. I suggest you should try std::cout << "Entry: " << +entry << std::endl; (note the + sign) and it will become clear.
Currently when I print an address of a variable,
cout << &a, its displayed as a simple hexadecimal number, let's say: 008FFBB8.
How can I print an address with the 0x notation? e.g. 0x6FFDF4.
I tried doing:
cout << hex << &a;
but it doesn't seem to work.
There are many different solutions, the first two that came into my mind are the following:
#include <iostream>
#include <iomanip>
#include <string>
#include <sstream>
int main()
{
std::cout << std::hex
<< "showbase: " << std::showbase<< 42 << '\n'
<< "noshowbase: " << std::noshowbase << 42 << '\n';
//2nd solution
std::string a{"008FFBB8"};
std::cout << "Given: " << a << '\n';
a.erase(0,2);
a = "0x"+a;
std::cout << "Edited: "<< a <<'\n';
//3rd string to int so you can use standard arithmetics
int b = std::stoul(a, nullptr, 16);
std::cout << "Represented as int: " << b << '\n';
std::cin.get();
}
Edit:
This is how it is printed after compilation with GNU GCC(g++) compiler.
The reason visual studio isn't printing it as shown on the screenshot is because visual studio tend not to use GNU GCC but its Microsoft compiler MSVC.(There are other things MSVC isn't doing as you may expect btw.) But good news: you can make it! or here:)
It is just how hex is presented in your configuration with "MSVC"
I hope that answers your question, else leave a comment :)
A simple approach
cout << "0x" << &a;
That said, given that other systems do inlude 0x in &a, in order to make it portable, you should make cout << "0x" conditional based on predefined macro that detects the systems where 0x isn't included.
A portable solution is to insert into a string stream first, then check whether the prefix was added, and concatenate if it wasn't. this also prevents possibility of 0x and the address being separated by concurrent output.
I just wanted to know why Visual Studio is not displaying this notation by default,
Because the authors of their standard library chose to do so.
so I don't have to manually do it.
I'm not sure if there is a way to change their implementation. Before asking how, I recommend considering why you think that you have to do it.
Taking inspiration ;) from MSDN and trying it on VS 2019:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
ios state(nullptr);
int a;
state.copyfmt(cout); // save current formatting
cout << "In hex: 0x" // now load up a bunch of formatting modifiers
<< hex
<< uppercase
<< setw(8)
<< setfill('0')
<< &a // the actual value we wanted to print out
<< endl;
cout.copyfmt(state); // restore previous formatting
}
Output:
In hex: 0x00AFFB44
And as for why, well...you could raise an issue on MSVC and claim parity with all other compilers and platforms if you have the data. But given that they themselves manually slipped in the 0x I'm guessing they are already comfortable with the behavior.
I'm a C# developer, writing a client for a server written in C++. The server streams some arbitrary data over TCP/IP to the client, and we have to reassemble it on the other end. The server sends us first a description of the data, then the data itself.
Problematic Structure:
struct Inner_S
{
double a;
double b[4][4];
};
#pragma pack(1)
struct Packed_S
{
uint8_t c;
Inner_S d;
};
The server is telling the client that the outer structure has an alignment of 1, and the inner structure has an alignment of 8. The protocol spec says:
Alignment of fields within a streamed structure is done according to the Itanium 64-bit C++ Application Binary Interface specification (i.e. the same as a typical GNU compiler on a typical 64-bit platform).
I found the Itanium 64-bit C++ Application Binary Interface specification. I think the part I am looking for is in "Allocation of Members Other Than Virtual Bases" but I get lost in there.
On the C# side I'm reading the data stream and packing my own class with the values extracted from the structure. I need to know where exactly in the stream to look for each element of the structure.
I'm currently handling the structure this way which is wrong according to my users:
(begin structure with alignment 1)(no padding needed)(read simple value)c(begin inner structure with alignment 8)(add padding to alignment 8)0000000(read field)aaaaaaaa(begin array)(read simple value)bbbbbbbb.....
That method is supported by at least one site.
So, when I'm parsing this data, how do I handle alignment in Inner_S?
caaaaaaaabbbbbbbb.... (I think?)
caaaaaaaa0000000bbbbbbbb.... (looks wrong)
As suggested by #Cameron I tried this with offsetof since this involves POD types.
#include <iostream>
#include <cstddef>
#include <cstdint>
using namespace std;
struct Inner_S
{
double a;
double b[4][4];
};
#pragma pack(1)
struct Packed_S
{
uint8_t c;
Inner_S d;
};
int main() {
cout << "Size: " << sizeof(Packed_S) << endl;
cout << "c offset: " << offsetof(Packed_S, c) << endl;
cout << "d offset: " << offsetof(Packed_S, d) << endl;
cout << "a offset: " << offsetof(Packed_S, d.a) << endl;
cout << "b offset: " << offsetof(Packed_S, d.b) << endl;
return 0;
}
Output
Size: 137
c offset: 0
d offset: 1
a offset: 1
b offset: 9
So using your notation the structure is packed as caaaaaaaabbbbbbbb..... Note that if you take out the #pragma pack(1) directive the compiler adds 3 bytes of padding after c.
See here
Question
What is the best way to convert binary to it's integral representation?
Context
Let's imagine that we have a buffer containing binary data obtained from an external source such as a socket connection or a binary file. The data is organised in a well defined format and we know that the first four octets represent a single unsigned 32 bit integer (which could be the size of following data). What would be the more efficient way to covert those octets to a usable format (such as std::uint32_t)?
Example
Here is what I have tried so far:
#include <algorithm>
#include <array>
#include <cstdint>
#include <cstring>
#include <iostream>
int main()
{
std::array<char, 4> buffer = { 0x01, 0x02, 0x03, 0x04 };
std::uint32_t n = 0;
n |= static_cast<std::uint32_t>(buffer[0]);
n |= static_cast<std::uint32_t>(buffer[1]) << 8;
n |= static_cast<std::uint32_t>(buffer[2]) << 16;
n |= static_cast<std::uint32_t>(buffer[3]) << 24;
std::cout << "Bit shifting: " << n << "\n";
n = 0;
std::memcpy(&n, buffer.data(), buffer.size());
std::cout << "std::memcpy(): " << n << "\n";
n = 0;
std::copy(buffer.begin(), buffer.end(), reinterpret_cast<char*>(&n));
std::cout << "std::copy(): " << n << "\n";
}
On my system, the result of the following program is
Bit shifting: 67305985
std::memcpy(): 67305985
std::copy(): 67305985
Are they all standard compliant or are they using implementation defined behaviour?
Which one is the more efficient?
Is there an bette way to make that conversion?
You essentially are asking about endianness. While your program might work on one computer, it might not on another. If the "well defined format" is network order, there are a standard set of macros/functions to convert to and from network order to the natural order for your specific machine.
This question already has answers here:
Restore the state of std::cout after manipulating it
(9 answers)
Closed 4 years ago.
If I apply an arbitrary number of manipulators to a stream, is there a way to undo the application of those manipulators in a generic way?
For example, consider the following:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
cout << "Hello" << hex << 42 << "\n";
// now i want to "roll-back" cout to whatever state it was in
// before the code above, *without* having to know
// what modifiers I added to it
// ... MAGIC HAPPENS! ...
cout << "This should not be in hex: " << 42 << "\n";
}
Suppose I want to add code at MAGIC HAPPENS that will revert the state of the stream manipulators to whatever it was before I did cout << hex. But I don't know what manipulators I added. How can I accomplish this?
In other words, I'd like to be able to write something like this (psudocode/fantasy code):
std::something old_state = cout.current_manip_state();
cout << hex;
cout.restore_manip_state(old_state);
Is this possible?
EDIT:
In case you're curious, I'm interested in doing this in a custom operator<<() I'm writing for a complex type. The type is a kind of discriminated union, and different value types will have different manips applied to the stream.
EDIT2:
Restriction: I cannot use Boost or any other 3rd party libraries. Solution must be in standard C++.
Yes.
You can save the state and restore it:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
std::ios state(NULL);
state.copyfmt(std::cout);
cout << "Hello" << hex << 42 << "\n";
// now i want to "roll-back" cout to whatever state it was in
// before the code above, *without* having to know what modifiers I added to it
// ... MAGIC HAPPENS! ...
std::cout.copyfmt(state);
cout << "This should not be in hex: " << 42 << "\n";
}
If you want to get back to the default state you don't even need to save the state you can extract it from a temporary object.
std::cout.copyfmt(std::ios(NULL));
The standard manipulators all manipulate a stream's format flags, precision and width settings. The width setting is reset by most formatted output operations anyway. These can all be retrieved like this:
std::ios_base::fmtflags saveflags = std::cout.flags();
std::streamsize prec = std::cout.precision();
std::streamsize width = std::cout.width();
and restored:
std::cout.flags( saveflags );
std::cout.precision( prec );
std::cout.width( width );
Turning this into an RAII class is an exercise for the reader...
Saving and restoring state is not exception-safe. I would propose to shuffle everything into a stringstream, and finally you put that on the real stream (which has never changed its flags at all).
#include <iostream>
#include <iomanip>
#include <sstream>
int main()
{
std::ostringstream out;
out << "Hello" << std::hex << 42 << "\n";
std::cout << out.str();
// no magic necessary!
std::cout << "This should not be in hex: " << 42 << "\n";
}
Of course this is a little less performant. The perfect solutions depends on your specific needs.
Boost IO State saver might be of help.
http://www.boost.org/doc/libs/1_40_0/libs/io/doc/ios_state.html
I know that is an old question, but for future generations:
You can also write a simple state saver yourself (it will certainly help you avoid leaving the state changed). Just use the solution suggested by #loki and run it from the constructor/destructor of an object (in short: RAII) along these lines:
class stateSaver
{
public:
stateSaver(ostream& os): stream_(os), state_(nullptr) { state_.copyfmt(os); }
~stateSaver() { stream_.copyfmt(state_); }
private:
std::ios state_;
ostream& stream_;
};
Then, you will use it like this:
void myFunc() {
stateSaver state(cout);
cout << hex << 42 << endl; // will be in hex
}
int main() {
cout << 42 << endl; // will be in dec
myFunc();
cout << 42 << endl; // will also be in dec
}