I'm having a trouble in trying to send a char (i.e. "R") from my qt5 application on WIN7 to comport which is connected to an Arduino.
I intend to blink a led on Arduino and my arduino part works OK.
Here is my qt code:
#include <QTextStream>
#include <QCoreApplication>
#include <QtSerialPort/QSerialPortInfo>
#include <QSerialPort>
#include <iostream>
#include <QtCore>
QT_USE_NAMESPACE
using namespace std;
QSerialPort serial;
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
QTextStream out(stdout);
QList<QSerialPortInfo> serialPortInfoList = QSerialPortInfo::availablePorts();
out << QObject::tr("Total number of ports available: ") << serialPortInfoList.count() << endl;
foreach (const QSerialPortInfo &serialPortInfo, serialPortInfoList) {
out << endl
<< QObject::tr("Port: ") << serialPortInfo.portName() << endl
<< QObject::tr("Location: ") << serialPortInfo.systemLocation() << endl
<< QObject::tr("Description: ") << serialPortInfo.description() << endl
<< QObject::tr("Manufacturer: ") << serialPortInfo.manufacturer() << endl
<< QObject::tr("Vendor Identifier: ") << (serialPortInfo.hasVendorIdentifier() ? QByteArray::number(serialPortInfo.vendorIdentifier(), 16) : QByteArray()) << endl
<< QObject::tr("Product Identifier: ") << (serialPortInfo.hasProductIdentifier() ? QByteArray::number(serialPortInfo.productIdentifier(), 16) : QByteArray()) << endl
<< QObject::tr("Busy: ") << (serialPortInfo.isBusy() ? QObject::tr("Yes") : QObject::tr("No")) << endl;
}
serial.setPortName("COM5");
serial.open(QIODevice::ReadWrite);
serial.setBaudRate(QSerialPort::Baud9600);
serial.setDataBits(QSerialPort::Data8);
serial.setParity(QSerialPort::NoParity);
serial.setStopBits(QSerialPort::OneStop);
serial.setFlowControl(QSerialPort::NoFlowControl);
if(!serial.isOpen())
{
std::cout<<"port is not open"<<endl;
//serial.open(QIODevice::ReadWrite);
}
if(serial.isWritable()==true)
{
std::cout<<"port writable..."<<endl;
}
QByteArray data("R");
serial.write(data);
serial.flush();
std::cout<<"value sent!!! "<<std::endl;
serial.close();
return 0;
}
My source code consists of two parts,
1- serialportinfolist .... which works just fine
2- opening and writing data... I get no issue when running the code and the display shows the result as if nothing has gone wrong!
HOWEVER, the led on the board does not turn on when I run this code.
I test this with Arduino Serial Monitor and it turns on but cant turn on from Qt.
Are you waiting for cr lf (0x0D 0x0A) in your arduino code?
QByteArray ba;
ba.resize(3);
ba[0] = 0x5c; //'R'
ba[1] = 0x0d;
ba[2] = 0x0a;
Or append it to your string with
QByteArray data("R\r\n");
Or
QByteArray data("R\n");
I think I have found a partial solution but it is still incomplete.
When I press debug the first time, qt does not send any signal to Arduino, but when I press debug for the second time it behaves as expected.
So, is'nt it so weird that one has to run it twice to get it working???
Let me know if the problem exists somewhere else,
any help...
Related
int i, a[] = {0, 1, 2};
void foo(int x) {
i++;
x++;
cout << a[0] << " " << a[1] << " " << a[2];
}
void main() {
i = 0;
foo(a[i]);
}
So, the printing output will be:
By value-result: 0 - 1 - 2
By reference: 1 - 1 - 2
By name: 0 - 2 - 2
By constant reference: 0 - 1 - 2
Right ?
Beware, the cout stream and << operators are pure C++ primitives, you are NOT in C !
You also have to understand that side-effect inside a sequence of cout << foo << bar << fuzz; will produce a totally impredictable output depending on the choices made by the compiler (and NOT triggered by the language specification because the variables are supposed to stay constant along the evaluation of the expression).
To illustrate what I am saying, try to compile the following small program (example.cpp):
#include <iostream>
using namespace std;
int main ()
{
int i = 0;
cout << "i++: " << i++ << i++ << i++ << i++ << endl;
i = 0;
cout << "++i: " << ++i << ++i << ++i << ++i << endl;
cout << "The address of i is: " << &i << endl;
return 0;
}
When compiled with g++ -o example example.cpp, you should get something like:
i++: 3210
++i: 4321
The address of i is: 0xbfb1c44c
Then, try to compile it with g++ -O2 -o example example.cpp, you should get something like:
i++: 3210
++i: 4444
The address of i is: 0xbf9af0cc
In fact, the difference of these two execution comes from the fact that once you trigger the optimization in g++, the compiler assumes that you conform to the specification of C++ and that there will be no side-effect inside the cout << ... << endl; expression. So, it will use the last value of i all the time.
I am trying to build a "run length encoder" which produces a report of occurrences of runs within a file using CUDA Thrust. I will use this "report" to perform the run length encoding step later.
e.g.
Input sequence:
inputSequence = [a, a, b, c, a, a, a];
Output sequences:
runChar = [a, a];
runCount = [2, 3];
runPosition = [0, 4];
The output desribes a run of 2 a's starting at position 0 and a run of 3 a's starting at the position 4.
The Thrust run length encoder example described below outputs two arrays - one for the output char and one for its length.
I would like to modify this so runs of less than 2 are excluded and it also outputs the position each run occurs.
// input data on the host
const char data[] = "aaabbbbbcddeeeeeeeeeff";
const size_t N = (sizeof(data) / sizeof(char)) - 1;
// copy input data to the device
thrust::device_vector<char> input(data, data + N);
// allocate storage for output data and run lengths
thrust::device_vector<char> output(N);
thrust::device_vector<int> lengths(N);
// print the initial data
std::cout << "input data:" << std::endl;
thrust::copy(input.begin(), input.end(), std::ostream_iterator<char>(std::cout, ""));
std::cout << std::endl << std::endl;
// compute run lengths
size_t num_runs = thrust::reduce_by_key
(input.begin(), input.end(), // input key sequence
thrust::constant_iterator<int>(1), // input value sequence
output.begin(), // output key sequence
lengths.begin() // output value sequence
).first - output.begin(); // compute the output size
// print the output
std::cout << "run-length encoded output:" << std::endl;
for(size_t i = 0; i < num_runs; i++)
std::cout << "(" << output[i] << "," << lengths[i] << ")";
std::cout << std::endl;
return 0;
One possible approach, building on what you have shown already:
Take your output lengths, and do an exclusive_scan on them. This creates a corresponding vector of the starting indexes of each run.
Use stream compaction (remove_if) to remove elements from all arrays (output, lengths, and indexes) whose corresponding length is 1. We do this in two steps, the first remove_if operation to clean up output and indexes, using lengths as the stencil, and the second operating directly on lengths. This can probably be significantly improved by operating on all 3 at once, which will make the output length calculation a bit more complicated. How you handle this exactly will depend on which sets of data you intend to retain.
Here is a fully worked example, extending your code:
$ cat t601.cu
#include <iostream>
#include <thrust/device_vector.h>
#include <thrust/copy.h>
#include <thrust/reduce.h>
#include <thrust/scan.h>
#include <thrust/iterator/constant_iterator.h>
#include <thrust/iterator/zip_iterator.h>
struct is_not_one{
template <typename T>
__host__ __device__
bool operator()(T data){
return data != 1;
}
};
int main(){
// input data on the host
const char data[] = "aaabbbbbcddeeeeeeeeeff";
const size_t N = (sizeof(data) / sizeof(char)) - 1;
// copy input data to the device
thrust::device_vector<char> input(data, data + N);
// allocate storage for output data and run lengths
thrust::device_vector<char> output(N);
thrust::device_vector<int> lengths(N);
// print the initial data
std::cout << "input data:" << std::endl;
thrust::copy(input.begin(), input.end(), std::ostream_iterator<char>(std::cout, ""));
std::cout << std::endl << std::endl;
// compute run lengths
size_t num_runs = thrust::reduce_by_key
(input.begin(), input.end(), // input key sequence
thrust::constant_iterator<int>(1), // input value sequence
output.begin(), // output key sequence
lengths.begin() // output value sequence
).first - output.begin(); // compute the output size
// print the output
std::cout << "run-length encoded output:" << std::endl;
for(size_t i = 0; i < num_runs; i++)
std::cout << "(" << output[i] << "," << lengths[i] << ")";
std::cout << std::endl;
thrust::device_vector<int> indexes(num_runs);
thrust::exclusive_scan(lengths.begin(), lengths.begin()+num_runs, indexes.begin());
thrust::device_vector<char> foutput(num_runs);
thrust::device_vector<int> findexes(num_runs);
thrust::device_vector<int> flengths(num_runs);
thrust::copy_if(thrust::make_zip_iterator(thrust::make_tuple(output.begin(), indexes.begin())), thrust::make_zip_iterator(thrust::make_tuple(output.begin()+num_runs, indexes.begin()+num_runs)), lengths.begin(), thrust::make_zip_iterator(thrust::make_tuple(foutput.begin(), findexes.begin())), is_not_one());
size_t fnum_runs = thrust::copy_if(lengths.begin(), lengths.begin()+num_runs, flengths.begin(), is_not_one()) - flengths.begin();
std::cout << "output: " << std::endl;
thrust::copy_n(foutput.begin(), fnum_runs, std::ostream_iterator<char>(std::cout, ","));
std::cout << std::endl << "lengths: " << std::endl;
thrust::copy_n(flengths.begin(), fnum_runs, std::ostream_iterator<int>(std::cout, ","));
std::cout << std::endl << "indexes: " << std::endl;
thrust::copy_n(findexes.begin(), fnum_runs, std::ostream_iterator<int>(std::cout, ","));
std::cout << std::endl;
return 0;
}
$ nvcc -arch=sm_20 -o t601 t601.cu
$ ./t601
input data:
aaabbbbbcddeeeeeeeeeff
run-length encoded output:
(a,3)(b,5)(c,1)(d,2)(e,9)(f,2)
output:
a,b,d,e,f,
lengths:
3,5,2,9,2,
indexes:
0,3,9,11,20,
$
I'm certain that this code can be improved upon, but my purpose is to show you one possible general approach.
In my opinion, for future reference, it's not very helpful for you to strip off the include headers from your sample code. I think it's better to provide a complete, compilable code. Not a big deal in this case.
Also note that there are thrust example codes for run length encoding and decoding.
Here is a brief discription of the books programming project I am trying to do...
Write a program that reads in a C++ source file and converts all ‘<’ symbols to “<” and all ‘>’ symbols to “>” . Also add the tag <PRE> to the beginning of the file and </PRE> to the end of the file. This tag preserves whitespace and formatting in the HTML document. Your program should create a new file with the converted output. To implement this, you should write a function ‘convert’ that takes the input and output streams as parameters.
I am having issues trying to get the program to work correctly. What's happening is the program will create a new file with .html but it is not converting anything in the file. (i.e. adding <PRE> to the beginning and </PRE> to the end and converting all '<' symbols to < and '>' to >).
I've been messing with it for a while now and I'm honestly not sure where I am going wrong. I'm super new to programming in general and even more new to c++ so please be nice haha.
Here is my code, any help is greatly appreciated!!
Thanks!
Scott
#include <iostream>
#include <fstream>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;
// main function
int main() {
// Input file to convert
string filename;
// Output file with .html on the end
string outputname;
char c;
int i;
ifstream inStream;
ofstream outStream;
cout << "Enter filename you woudl like to convert: " << endl;
cin >> filename;
// Open the input file
inStream.open(filename.c_str());
if (inStream.fail()) {
cout << "I/O failure opening file." << endl;
exit(1);
}
// Create the output file
outputname = filename + ".html";
outStream.open(outputname.c_str());
// First, output the <PRE> tag
outStream << "<PRE>" << endl;
// Loop through the input file intil nothing else to get
while (!inStream.eof()) {
inStream.get(c); // Get one character
// Output < or > or original char
if (c == '<') {
outStream << "<";
}
else if (c=='>') {
outStream << ">";
}
else outStream << c;
}
// Output end /PRE tag
outStream << "</PRE>" << endl;
inStream.close();
outStream.close();
cout << "Conversion done. Results are in file " << outputname << endl;
}
The new thrust::tabulate function works for me on the host but not on the device. The device is a K20x with compute capability 3.5. The host is an Ubuntu machine with 128GB of memory. Help?
I think that the unified addressing is not the problem since I can sort a unifiedly addressed array on the device.
#include <iostream>
#include <thrust/device_vector.h>
#include <thrust/execution_policy.h>
#include <thrust/tabulate.h>
#include <thrust/version.h>
using namespace std;
// Print an expression's name then its value, possible followed by a
// comma or endl. Ex: cout << PRINTC(x) << PRINTN(y);
#define PRINT(arg) #arg "=" << (arg)
#define PRINTC(arg) #arg "=" << (arg) << ", "
#define PRINTN(arg) #arg "=" << (arg) << endl
// Execute an expression and check for CUDA errors.
#define CE(exp) { \
cudaError_t e; e = (exp); \
if (e != cudaSuccess) { \
cerr << #exp << " failed at line " << __LINE__ << " with error " << cudaGetErrorString(e) << endl; \
exit(1); \
} \
}
const int N(10);
int main(void) {
int major = THRUST_MAJOR_VERSION;
int minor = THRUST_MINOR_VERSION;
cout << "Thrust v" << major << "." << minor
<< ", CUDA_VERSION: " << CUDA_VERSION << ", CUDA_ARCH: " << __CUDA_ARCH__
<< endl;
cout << PRINTN(N);
cudaDeviceProp prop;
cudaGetDeviceProperties(&prop, 0);
if (!prop.unifiedAddressing) {
cerr << "Unified addressing not available." << endl;
exit(1);
}
cudaGetDeviceProperties(&prop, 0);
if (!prop.canMapHostMemory) {
cerr << "Can't map host memory." << endl;
exit(1);
}
cudaSetDeviceFlags(cudaDeviceMapHost);
int *p, *q;
CE(cudaHostAlloc(&p, N*sizeof(int), cudaHostAllocMapped));
CE(cudaHostAlloc(&q, N*sizeof(int), cudaHostAllocMapped));
thrust::tabulate(thrust::host, p, p+N, thrust::negate<int>());
thrust::tabulate(thrust::device, q, q+N, thrust::negate<int>());
for (int i=0; i<N; i++)
cout << PRINTC(i) << PRINTC(p[i]) << PRINTN(q[i]);
}
Output:
Thrust v1.7, CUDA_VERSION: 6000, CUDA_ARCH: 0
N=10
i=0, p[i]=0, q[i]=0
i=1, p[i]=-1, q[i]=0
i=2, p[i]=-2, q[i]=0
i=3, p[i]=-3, q[i]=0
i=4, p[i]=-4, q[i]=0
i=5, p[i]=-5, q[i]=0
i=6, p[i]=-6, q[i]=0
i=7, p[i]=-7, q[i]=0
i=8, p[i]=-8, q[i]=0
i=9, p[i]=-9, q[i]=0
The following does not add any info content to my post but is required before stackoverflow will accept it: Much of the program is error checking and version checking.
The problem appears to be fixed in the thrust master branch at the moment. This master branch currently identifies itself as Thrust v1.8.
I ran your code with CUDA 6RC (appears to be what you are using) and I was able to duplicate your observation.
I then updated to the master branch, and removed the __CUDA_ARCH__ macro from your code, and I got the expected results (host and device tabulations match).
Note that according to the programming guide, the __CUDA_ARCH__ macro is only defined when it's used in code that is being compiled by the device code compiler. It is officially undefined in host code. Therefore it's acceptable to use it as follows in host code:
#ifdef __CUDA_ARCH__
but not as you are using it. Yes, I understand the behavior is different between thrust v1.7 and thrust master in this regard, but that appears to (also) be a thrust issue, that has been fixed in the master branch.
Both of these issues I expect would be fixed whenever the next version of thrust gets incorporated into an official CUDA drop. Since we are very close to CUDA 6.0 official release, I'd be surprised if these issues were fixed in CUDA 6.0.
Further notes about the tabulate issue:
One workaround would be to update thrust to master
Issue doesn't appear to be specific to thrust::tabulate in my testing. Many thrust functions that I tested seem to fail in that when used with thrust::device and raw pointers, they fail to write values correctly (seem to write all zeroes), but they do seem to be able to read values correctly (e.g. thrust::reduce seems to work)
Another possible workaround is to wrap your raw pointers with thrust::device_ptr<> using thrust::device_ptr_cast<>(). That seemed to work for me as well.
I just started to dig into Boost::Spirit, latest version by now -- V2.4.
The essense of my problem is following:
I would like to parse strings like "1a2" or "3b4".
So the rule I use is:
(double_ >> lit('b') >> double_)
| (double_ >> lit('a') >> double_);
The attribute of the rule must be "vector <double>". And I'm reading it into the container.
The complete code:
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cstring>
int main(int argc, char * argv[])
{
using namespace std;
using namespace boost::spirit;
using namespace boost::spirit::qi;
using boost::phoenix::arg_names::arg1;
char const * first = "1a2";
char const * last = first + std::strlen(first);
vector<double> h;
rule<char const *, vector<double>()> or_test;
or_test %= (double_ >> lit('b') >> double_)
| (double_ >> lit('a') >> double_);
if (parse(first, last, or_test,h)) {
cout << "parse success: ";
for_each(h.begin(), h.end(), (cout << arg1 << " "));
cout << "end\n";
} else cout << "parse error\n" << endl;
return 0;
}
I'm compiling it with g++ 4.4.3. And it returns "1 1 2". While I expect "1 2".
As far as I understand this happens because parser:
goes to the first alternative
reads a double_ and stores it in the container
then stops at "a", while expecting lit("b")
goes to the second alternative
reads two more doubles
My question is -- Is this a correct behavior, and if yes -- why?
That's expected behavior. During backtracking Spirit does not 'unmake' changes to attributes. Therefore, you should use the hold[] directive explicitly forcing the parser to hold on to a copy of the attribute (allowing to roll back any attribute change):
or_test =
hold[double_ >> lit('b') >> double_)]
| (double_ >> lit('a') >> double_)
;
This directive needs to be applied to all alternatives modifying the attribute, except the last one.