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.
Related
I am pretty new to both html and c++ with a´lot of knowledge in c#, I have just writen my first 100% own c++ program a little banking system and need a litte help on how to integrate html.
For example how would i dispaly the funds in html?
IDE Clion.
int press;
int newFunds;
cout << "Available funds ";
cout << funds;
cout << "$ \n";
cout << "Press 1 for Withdraw \n";
cout << "Press 2 for Deposit \n";
cin >> press;
if(press == 1) {
cout << "Amount";
cin >> withdraw;
funds -= withdraw;
newFunds += withdraw;
if(funds < 0) {
funds += newFunds;
cout << "Not enough funds!";
}
else {
cout << "Remaining funds ", cout << funds, cout << "$ \n";
}
}
else if(press == 2) {
cout << "Amount\n";
cin >> deposit;
funds += deposit;
cout << "Funds ";
cout << funds;
cout << "$";
}
if(press > 2) {
cout << "Invalid";
}
return 0;
}
I welcome any suggestion!
I have tried to look up som tut but they wherent so good!
C++ and HTML can't connect just by creating a localhost web server. To make a "connection" between them, you'll need a C++ API to interact with the server and your C++ program. FastCGI protocol is simple and easy for beginners, but maybe not be the best option for more professional projects. Check information about FastCGI++ that's an API for C++. Here is a useful GitHub link that I recommend to start: https://github.com/eddic/fastcgipp.
I'm experiencing odd behavior while using the thrust::reverse function on a zip_iterator constructed with a thrust::make_zip_iterator( thrust::make_tuple( )) type syntax (see the answer from JackOLantern here for a good example of that combination).
I wish to reverse some arbitrarily-indicated section of multiple device vectors as in the example code below. When I do the reversing in one go by tupling and zipping them together, unexpected behavior ensues. The first half of the range is correctly changed to an inversion of the second half of the range, however, the second half of the range is left unchanged.
I've been using other thrust functions in a similar fashion (sort_by_key, uniqe_by_key, adjacent_difference, etc.) without issue. Am I just executing this incorrectly or is there some reason that this will not work on a fundamental level? A thought I had is that perhaps the zip_iterator is not bidirectional as required for reverse. Is this true? I couldn't find documentation indicating as such.
A workaround is just to reverse the vector individually, which works as shown below. However, I suspect this will be less efficient. Note that in my actual use-case I have vectors with sizes of the order of 10,000 and I'm zipping up anywhere from 3-7 vectors for the operations.
#include <iostream>
#include <ostream>
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <thrust/tuple.h>
#include <thrust/iterator/zip_iterator.h>
#include <thrust/sequence.h>
#include <thrust/reverse.h>
int main(){
// initial host vectors
const int N=10;
thrust::host_vector<int> h1(N);
thrust::host_vector<float> h2(N);
// fill them
thrust::sequence( h1.begin(), h1.end(), 0);
thrust::sequence( h2.begin(), h2.end(), 10., 0.5);
// print initial contents
for (size_t i=0; i<N; i++){
std::cout << h1[i] << " " << h2[i] << std::endl;
}
// transfer to device
thrust::device_vector<int> d1 = h1;
thrust::device_vector<float> d2 = h2;
// what chunk to invert
int iStart = 3; int iEnd = 8;
// attempt to reverse middle via zip_iterators
thrust::reverse(
thrust::make_zip_iterator( thrust::make_tuple( d1.begin()+iStart, d2.begin()+iStart)),
thrust::make_zip_iterator( thrust::make_tuple( d1.begin()+iEnd, d2.begin()+iEnd))
);
// pull back and write out unexpected ordering
thrust::host_vector<int> temp1 = d1;
thrust::host_vector<float> temp2 = d2;
std::cout << "<==========>" << std::endl;
for (size_t i=0; i<N; i++){
std::cout << temp1[i] << " " << temp2[i] << std::endl;
}
// reset device variables
d1 = h1;
d2 = h2;
// reverse individually
thrust::reverse( d1.begin()+iStart, d1.begin()+iEnd);
thrust::reverse( d2.begin()+iStart, d2.begin()+iEnd);
// pull back and write out the desired ordering
temp1 = d1;
temp2 = d2;
std::cout << "<==========>" << std::endl;
for (size_t i=0; i<N; i++){
std::cout << temp1[i] << " " << temp2[i] << std::endl;
}
return 0;
}
Output
0 10
1 10.5
2 11
3 11.5
4 12
5 12.5
6 13
7 13.5
8 14
9 14.5
<==========>
0 10
1 10.5
2 11
7 13.5
6 13
5 12.5
6 13
7 13.5
8 14
9 14.5
<==========>
0 10
1 10.5
2 11
7 13.5
6 13
5 12.5
4 12
3 11.5
8 14
9 14.5
The information from Robert Crovella in the comments combined with the initially given workaround in the initial post appears to answer the question - thus, I will combine them here so the question can be marked as "answered." If others wish to post other solutions, I'm more than willing to look at them and move the "official answer" check mark. That being said...
The solution to the question has two parts:
If using an older version of CUDA and upgrading is an option: upgrade to the newest CUDA version and the operation should work (tested to work on CUDA 9.2.148 - thanks Robert!)
If unable to upgrade to a newer version of CUDA: apply reverse to the vectors individually to achieve the same result as given in the initial post. The code with only the working solution is copied below for completeness.
#include <iostream>
#include <ostream>
#include <thrust/device_vector.h>
#include <thrust/host_vector.h>
#include <thrust/tuple.h>
#include <thrust/iterator/zip_iterator.h>
#include <thrust/sequence.h>
#include <thrust/reverse.h>
int main(){
// initial host vectors
const int N=10;
thrust::host_vector<int> h1(N);
thrust::host_vector<float> h2(N);
// fill them
thrust::sequence( h1.begin(), h1.end(), 0);
thrust::sequence( h2.begin(), h2.end(), 10., 0.5);
// print initial contents
for (size_t i=0; i<N; i++){
std::cout << h1[i] << " " << h2[i] << std::endl;
}
// transfer to device
thrust::device_vector<int> d1 = h1;
thrust::device_vector<float> d2 = h2;
// what chunk to invert
int iStart = 3; int iEnd = 8;
// reverse individually
thrust::reverse( d1.begin()+iStart, d1.begin()+iEnd);
thrust::reverse( d2.begin()+iStart, d2.begin()+iEnd);
// pull back and write out the desired ordering
temp1 = d1;
temp2 = d2;
std::cout << "<==========>" << std::endl;
for (size_t i=0; i<N; i++){
std::cout << temp1[i] << " " << temp2[i] << std::endl;
}
return 0;
}
I want to train my tesseract for hindi language . I have many 'hindi' written text images with specific font and I would like to train tesseract ocr for that images .
Several times I tried train tesseract using this link https://code.google.com/p/tesseract-ocr/wiki/TrainingTesseract3 . when I run makebox command it extracts box file but it recognises like english character. I dont understand why this happen. Please help me to train tesseract ocr for Hindi language.
You can check sample image on following link.
sample file
I have been wanting to train a few character sets myself, and have been gathering information first. Maybe this info is of use to you too.
Did you read this document:
http://blog.cedric.ws/how-to-train-tesseract-301
If none of the characters are recognized you will have to train all of the characters, I'm afraid. But important steps seem to be:
include the indication of the language ('eng') in the makebox command line (this would probably be 'hin' in your case.
be aware of the version of tesseract. I have the impression that the training procedure has been changing in the last versions.
Sample program of the recognize the Hindi char from the image and store the respective bounding box values and respective Hindi char store into the one file.
/*
* Char_OCR.cpp
*
* Created on: Jun 23, 2016
* Author: pratik
*/
#include <opencv2/opencv.hpp>
#include <tesseract/baseapi.h>
#include <leptonica/allheaders.h>
#include <iostream>
#include <fstream>
using namespace std;
using namespace cv;
void dumpIntoFile(const char *ocrResult , ofstream &myfile1 ,int x1, int y1,
int x2, int y2, int &);
int main(int argc ,char **argv)
{
Pix *image = pixRead(argv[1]);
if (image == 0) {
cout << "Cannot load input file!\n";
}
tesseract::TessBaseAPI tess;
if (tess.Init("/usr/share/tesseract/tessdata", "hin")) {
fprintf(stderr, "Could not initialize tesseract.\n");
exit(1);
}
tess.SetImage(image);
tess.Recognize(0);
tesseract::ResultIterator *ri = tess.GetIterator();
tesseract::PageIteratorLevel level = tesseract::RIL_SYMBOL;
cout << ri << endl;
ofstream myfile1("Word.txt");
myfile1 << "ID" << '\t' << "CORD_X" << '\t' << "CORD_Y" << '\t' <<
"CORD_W" << '\t' << "CORD_H" << '\t' << "STRING" << endl;
int i=1;
if(ri!=0)
{
do {
const char *word = ri->GetUTF8Text(level);
// cout << word << endl;
//float conf = ri->Confidence(level);
int x1, y1, x2, y2;
ri->BoundingBox(level, &x1, &y1, &x2, &y2);
dumpIntoFile(word, myfile1, x1, y1, x2, y2, i);
delete []word;
} while (ri->Next(level));
delete []ri;
}
}
void dumpIntoFile(const char *ocrResult , ofstream &myfile1 ,int x1, int y1,
int x2, int y2,int &i)
{
int length = strlen(ocrResult);
myfile1 << i++ << '\t' << x1 << '\t' << y1 << '\t' <<
x2 << '\t' << y2 << '\t' ;
//cout << "in the string (" << length << ") ::";
for(int j = 0; j < length && ocrResult[j] != '\n'; j++)
{
myfile1 << ocrResult[j];
}
myfile1 << endl;
}
Currently the Tesseract API provides pre-trained language models for most of the popular languages:
https://tesseract-ocr.github.io/tessdoc/Data-Files-in-different-versions.html
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.
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...