std::stof() rounds numbers, how to avoid - c++

Im trying to get a float value from a file.txt into a string. When I output that value with std::stof(str) it gets rounded. Example, in the text file there's "101471.71", whet i use the std::stof(str) it returns "101472", how to I avoid this?
Here's a part of that code (some parts are in spanish, sorry :p):
double CaptureLine(std::string filepath, int fileline, int linesize)
{
std::fstream file;
std::string n_str, num_n;
int current_line = 0, n_size, filesize = FileSize(filepath);
char ch_n;
double n_float = 0.0;
int n_line = filesize - fileline;
file.open("registros.txt");
if (file.is_open()) {
while (!file.eof()) {
current_line++;
std::getline(file, n_str);
if (current_line == n_line) break;
}
if (current_line < n_line) {
std::cout << "Error" << std::endl;
return 1;
}
file.close();
}
n_size = n_str.length();
for (int i = linesize; i < n_size; i++) {
ch_n = n_str.at(i);
num_n.push_back(ch_n);
}
std::cout << ">>" << num_n << "<<\n";
n_float = std::stof(num_n); //Here's the error
return n_float;
}

The issue probably isn't with std::stof, but is probably with the default precision of 6 in std::cout. You can use std::setprecision to increase that precision and capture all of your digits.
Here's a program that demonstrates:
#include <iostream>
#include <iomanip>
#include <string>
int main() {
std::cout << 101471.71f << "\n";
std::cout << std::stof("101471.71") << "\n";
std::cout << std::setprecision(8) << 101471.71f << "\n";
std::cout << std::stof("101471.71") << "\n";
return 0;
}
Outputs:
101472
101472
101471.71
101471.71
Be aware that std::setprecision sticks to the std::cout stream after it's called. Notice how the above example calls it exactly once but its effect sticks around.

Related

How to convert hex representation from URL (%) to std::string (chinese text)?

Intro
I have some input that I need to convert to the correct Chinese characters but I think I'm stuck at the final number to string conversion. I have checked using this hex to text converter online tool that e6b9af corresponds to the text 湯.
MWE
Here is a minimal example that I made to illustrate the problem. The input is "%e6%b9%af" (obtained from an URL somewhere else).
#include <iostream>
#include <string>
std::string attempt(std::string path)
{
std::size_t i = path.find("%");
while (i != std::string::npos)
{
std::string sub = path.substr(i, 9);
sub.erase(i + 6, 1);
sub.erase(i + 3, 1);
sub.erase(i, 1);
std::size_t s = std::stoul(sub, nullptr, 16);
path.replace(i, 9, std::to_string(s));
i = path.find("%");
}
return path;
}
int main()
{
std::string input = "%E6%B9%AF";
std::string goal = "湯";
// convert input to goal
input = attempt(input);
std::cout << goal << " and " << input << (input == goal ? " are the same" : " are not the same") << std::endl;
return 0;
}
Output
湯 and 15120815 are not the same
Expected output
湯 and 湯 are the same
Additional question
Are all characters in foreign languages represented in 3 bytes or is that just for Chinese? Since my attempt assumes blocks of 3 bytes, is that a good assumption?
Based on your suggestions and changing an example from this other post. This is what I came up with.
#include <iostream>
#include <string>
#include <sstream>
std::string decode_url(const std::string& path)
{
std::stringstream decoded;
for (std::size_t i = 0; i < path.size(); i++)
{
if (path[i] != '%')
{
if (path[i] == '+')
decoded << ' ';
else
decoded << path[i];
}
else
{
unsigned int j;
sscanf(path.substr(i + 1, 2).c_str(), "%x", &j);
decoded << static_cast<char>(j);
i += 2;
}
}
return decoded.str();
}
int main()
{
std::string input = "%E6%B9%AF";
std::string goal = "湯";
// convert input to goal
input = decode_url(input);
std::cout << goal << " and " << input << (input == goal ? " are the same" : " are not the same") << std::endl;
return 0;
}
Output
湯 and 湯 are the same

storing structure data in circular buffer

I need to store a string along with an integer in a circular buffer and then have to iterate through it, searching (with a substr) for existence of a string which then gets copied along with the integer into result buffer.
I've written the below proof of concept code but it terminates pre-maturely after the first time &res gets copied into out and I'm not exactly sure why, can anyone help me here?
This is what I have so far:
#include <iostream>
#include <boost/circular_buffer.hpp>
#include <boost/algorithm/string/predicate.hpp>
#define CB_SZ 4
#define ARR 7
struct cb_dat_t{
std::string lp;
int cnf;
};
int buffer_check(cb_dat_t &in, boost::circular_buffer<cb_dat_t> &buff, cb_dat_t *out);
int main(void)
{
int i = 0;
cb_dat_t in[ARR];
cb_dat_t out;
boost::circular_buffer<cb_dat_t> cb(CB_SZ);
in[0].lp = "ABC";
in[0].cnf = 78;
in[1].lp = "ABCDE";
in[1].cnf = 63;
in[2].lp = "AB";
in[2].cnf = 92;
in[3].lp = "1234";
in[3].cnf = 85;
in[4].lp = "23";
in[4].cnf = 71;
in[5].lp = "ABC";
in[5].cnf = 63;
in[6].lp = "BC";
in[6].cnf = 71;
for (i=0; i<ARR; i++) {
buffer_check(in[i], cb, &out);
std::cout << "result[" << i << "] " << out.lp << " " << out.cnf << std::endl;
}
std::cout << "all done!" <<std::endl;
return 0;
}
int buffer_check(cb_dat_t &in, boost::circular_buffer<cb_dat_t> &buff, cb_dat_t *out)
{
cb_dat_t res;
if (!buff.size()){
std::cout << "buff.size() " << buff.size() << std::endl;
buff.push_back(in);
memcpy(out,&in,sizeof(cb_dat_t));
return 0;
}
boost::circular_buffer<cb_dat_t>::iterator itr = buff.begin();
while (itr!=buff.end()) {
if (boost::contains(itr->lp,in.lp)) {
std::cout << itr->lp << " contains " << in.lp << std::endl;
memcpy(&res,&itr,sizeof(cb_dat_t));
} else {
std::cout << itr->lp << " does not contain " << in.lp <<std::endl;
memcpy(&res,&in,sizeof(cb_dat_t));
}
itr++;
}
buff.push_back(in);
memcpy(out,&res,sizeof(cb_dat_t));
std::cout << "buff.size() " << buff.size() << std::endl;
return 0;
}
Where the output is:
./circular
buff.size() 0
result[0] ABC 78
ABC does not contain ABCDE
buff.size() 2
Command terminated
I'm not sure why g++ needs me to do memcpy(&res,&itr,sizeof(cb_dat_t));, itr is a pointer already, isn't it? It complains when I do memcpy(&res,itr,sizeof(cb_dat_t)); instead.
You shouldn't be using memcpy at all. Since cb_dat_t is not a POD type (in this case because it contains a member with a constructor), you should be using the assignment operation to copy cb_dat_t objects. The four memcpy calls in buffer_check can be replaced with
*out = in;
res = *itr;
res = in;
*out = res;
memcpy will not handle std::string properly. The compiler generated default assignment operator for cb_dat_t will properly copy all members of the structure.

C++ accessing vector of vector got segmentation fault

I created a vector of vector (10*10000) and try to access this vector through member function. but I got a segmentation fault. I don't know what's wrong here...
Here is Simple.h
class Simple
{
private:
std::vector<double> data_row;
std::vector<std::vector<double> > data;
public:
Simple():data_row(10000), data(10, data_row){};
/*initialize data vector*/
int getSampleCounts(std::istream &File);
/*return number of packet samples in this file*/
Result getModel(std::istream &File);
/*return average and variance of simple delta time*/
void splitData (std::istream &File, const int & sample_in_fold);
};
#endif /* SIMPLE_H */
here is Simple.cpp
void Simple::splitData(std::istream& File, const int & sample_in_fold) {
double value = 0.0;
bool isFailed = true;
int label = 0;
while (File >> value) {
// for each value, generate a label
srand(time(NULL));
label = rand() % 10; // generate label between 0 to 9
while (isFailed) {
// segmentation fault in the next line!
std::cout << "current data size is: " << this->data.size() <<endl;
std::vector<double>::size_type sz = this->data[label].size();
if (sz <= sample_in_fold) {
std::cout << "current size is " << sz << "< samples in fold: " << sample_in_fold << endl;
this->data[label].push_back(value);
std::cout << "push_back succeed!" << endl;
isFailed = false;
} else {
std::cout << "label " << label << "if full. Next label. \n";
srand(time(NULL));
label = rand() % 10;
sz = this->data[label].size();
}
}
}
}
and I'm attaching the main file here.
#include <iostream>
#include <fstream>
#include <string>
#include <cstdlib> // for system())
#include <sys/types.h>
#include <dirent.h>
#include <vector>
#include <limits.h> // for PATH_MAX
#include "Complex.h"
#include "Result.h"
#include "Simple.h"
#include <math.h>
using namespace std;
int main(int argc, char ** argv) {
struct dirent *pDirent;
DIR *pDir;
std::string line;
// check for args
if (argc == 1) {
printf("Usage: ./main + folder name. \n");
return 1;
}
pDir = opendir(argv[1]);
if (pDir == NULL) {
printf("Cannot open directory '%s' \n", argv[1]);
return 1;
}
// readdir returns a pointer to the next direcctory entry dirent structure
while ((pDirent = readdir(pDir)) != NULL) {
// get file name and absolute path
char *name = pDirent->d_name;
char buf[PATH_MAX + 1];
realpath(name, buf);
// std::cout << "Current file is: " << (pDirent->d_name) << endl;
if (has_suffix(pDirent->d_name, ".txt")) {
printf("[%s]\n", pDirent->d_name);
//printf("absolute path is %s. \n", buf);
ifstream infile;
// open file with absolute path
infile.open(buf, ios::in);
if (!infile) {
cerr << "Can't open input file " << buf << endl;
exit(1);
}
//processing for simple pattern
if (has_suffix(name, "testfile.txt")) {
Simple* simple_obj;
int number = simple_obj->getSampleCounts(infile);
Result simplerst = simple_obj->getModel(infile);
std::cout << "Number of delta time is " << number << endl;
infile.clear();
infile.seekg(0);
write_to_file(pDirent->d_name, simplerst);
// divide data into k = 10 folds, get number of data in each fold
int sample_in_fold = floor(number / 10);
std::cout << sample_in_fold << std::endl;
simple_obj->splitData(infile, sample_in_fold);
}
} else {
// printf("This is not a txt file. Continue\n");
}
}
closedir(pDir);
return 0;
}
And here is a sample testfile.txt. I only copied part of the original file, for illustration.
10.145906000
10.151063000
10.131083000
10.143461000
10.131745000
10.151285000
10.147493000
10.123198000
10.144975000
10.144484000
10.138129000
10.131634000
10.144311000
10.157710000
10.138047000
10.122754000
10.137675000
10.204973000
10.140399000
10.142194000
10.138388000
10.141669000
10.138056000
10.138679000
10.141415000
10.154170000
10.139574000
10.140207000
10.149151000
10.164629000
10.106818000
10.142431000
10.137675000
10.204973000
10.140399000
10.142194000
10.138388000
10.141669000
10.138056000
10.138679000
10.141415000
Here is Result.h
#ifndef RESULT_H
#define RESULT_H
typedef struct Result {
double average;
double sigma;
}Result;
and getModel function in Simple.cpp:
Result Simple::getModel(std::istream &File) {
double value = 0.0;
double average = 0.0;
double sum = 0.0;
double counter = 0.0;
double sumsqr = 0.0;
double var = 0.0;
double sigma = 0.0;
while (File >> value) {
++counter;
sum += value;
sumsqr += value * value;
}
average = sum / counter;
var = sumsqr / counter - average * average; //E(x^2) - (E(x))^2
sigma = sqrt(var);
std::cout << "average is " << average << std::endl;
std::cout << "std deviation is " << sigma << std::endl;
File.clear();
File.seekg(0);
Result result = {average, sigma};
return result;
}
One issue right away:
Simple* simple_obj;
int number = simple_obj->getSampleCounts(infile);
simple_obj is an uninitialized pointer, thus your program exhibits undefined behavior at this point.
Why use a pointer anyway? You could have simply done this to avoid the issue:
Simple simple_obj;
simple_obj.getSampleCounts(infile);
Also, this line may not be an issue, but I'll mention it anyway:
Result simplerst = simple_obj->getModel(infile);
We already know that in your original code, simple_obj is bogus, but that's not the issue here. If Result is an object, and that object does not have correct copy semantics, then that assignment will also cause undefined behavior.
You've got a couple of uses of endl without specifying std::endl (they're not the same thing - you always have to type the std:: ). Is endl silently referring to another variable somewhere else?

fixed amount of digits on creating .txt file

I need to create file with specific file name format (on windows). the format is:
Name_nodeNum_frequency.txt
nodeNum is int and frequency is float.
those two variables should be written with fixed digits:
if nodeNum is 8 --> 008
if frequency is 4.88421 --> 4.884
this is the function:
create_file(int nodeNum, double frequency)
{
char buffer [50];
//convert int to string
itoa(nodeNum, buffer, 10);
string sNodeNum = string(buffer);
//not sure about the double
//tried to_string but I got:
// more than instance of overloaded function matches the argument list
string fileName = ("Name_" + sNodeNum + "_" + sfreq + "MHZ");
FILE* pFile = OpenFile(fileName);
}
I tried to use %d, but it seems like I should not do that:
string fileName = ("Delay_" + "%3d" + "_" + sfreq + "MHZ" , sNodeNum);
I will be happy for some guidance.
thanks!
You seem to be mixing C and C++ here. A simple way to do this in C would be:
#include <stdio.h>
int main()
{
int sNodeNum = 8;
double sfreq = 4.88421;
char filename[50];
sprintf(filename, "Delay_%03d_%.3fMHZ.txt", sNodeNum, sfreq);
FILE* pFile = fopen(filename, "w");
return 0;
}
If on the other hand, if you want to use C++, you should make a few changes:
#include <iomanip>
#include <fstream>
#include <sstream>
#include <iostream>
int main()
{
int sNodeNum = 8;
double sfreq = 4.88421;
std::ostringstream ss;
ss << "Delay_" << std::setfill('0') << std::setw(3) << sNodeNum
<< "_" << std::setprecision(4) << sfreq << "MHZ.txt";
std::string filename(ss.str());
std::ofstream fout(filename.c_str());
return 0;
}
Each of these two approaches opens a file for writing, with the name Delay_008_4.884MHZ.txt.
Live demo link
#include <string>
#include <iomanip>
#include <iostream>
int nodeNum = 8;
float frequency = 4.88421f;
std::ostream& out = std::cout; // put your std::ofstream file or std::ostringstream
std::ios::fmtflags flags = out.flags();
out.width(3);
out.fill('0');
out.setf(std::ios::right, std::ios::adjustfield);
out << nodeNum << std::endl;
flags = out.flags(flags);
out.precision(3);
out.setf(std::ios::fixed, std::ios::floatfield);
out << frequency;
out.flags(flags);
Or even simpler:
out << std::setw(3) << std::setfill('0') << std::right << nodeNum << std::endl;
out << std::setprecision(3) << std::fixed << frequency;
Output:
008
4.884

Formatting an integer in C++

I have an 8 digit integer which I would like to print formatted like this:
XXX-XX-XXX
I would like to use a function that takes an int and returns a string.
What's a good way to do this?
This is how I'd do it, personally. Might not be the fastest way of solving the problem, and definitely not as reusable as egrunin's function, but it strikes me as both clean and easy to understand. I'll throw it in the ring as an alternative to the mathier and loopier solutions.
#include <sstream>
#include <string>
#include <iomanip>
std::string format(long num) {
std::ostringstream oss;
oss << std::setfill('0') << std::setw(8) << num;
return oss.str().insert(3, "-").insert(6, "-");
};
Tested this, it works.
The format parameter here is "XXX-XX-XXX", but it only looks at (and skips over) the dashes.
std::string foo(char *format, long num)
{
std::string s(format);
if (num < 0) { return "Input must be positive"; }
for (int nPos = s.length() - 1; nPos >= 0; --nPos)
{
if (s.at(nPos) == '-') continue;
s.at(nPos) = '0' + (num % 10);
num = num / 10;
}
if (num > 0) { return "Input too large for format string"; }
return s;
}
Usage:
int main()
{
printf(foo("###-##-###", 12345678).c_str());
return 0;
}
Here's a bit different way that tries to work with the standard library and get it to do most of the real work:
#include <locale>
template <class T>
struct formatter : std::numpunct<T> {
protected:
T do_thousands_sep() const { return T('-'); }
std::basic_string<T> do_grouping() const {
return std::basic_string<T>("\3\2\3");
}
};
#ifdef TEST
#include <iostream>
int main() {
std::locale fmt(std::locale::classic(), new formatter<char>);
std::cout.imbue(fmt);
std::cout << 12345678 << std::endl;
return 0;
}
#endif
To return a string, just write to a stringstream, and return its .str().
This may be overkill if you only want to print out one number that way, but if you want to do this sort of thing in more than one place (or, especially, if you want to format all numbers going to a particular stream that way) it becomes more reasonable.
Here's a complete program that shows how I'd do it:
#include <iostream>
#include <iomanip>
#include <sstream>
std::string formatInt (unsigned int i) {
std::stringstream s;
s << std::setfill('0') << std::setw(3) << ((i % 100000000) / 100000) << '-'
<< std::setfill('0') << std::setw(2) << ((i % 100000) / 1000) << '-'
<< std::setfill('0') << std::setw(3) << (i % 1000);
return s.str();
}
int main (int argc, char *argv[]) {
if (argc > 1)
std::cout << formatInt (atoi (argv[1])) << std::endl;
else
std::cout << "Provide an argument, ya goose!" << std::endl;
return 0;
}
Running this with certain inputs gives:
Input Output
-------- ----------
12345678 123-45-678
0 000-00-000
7012 000-07-012
10101010 101-01-010
123456789 234-56-789
-7 949-67-289
Those last two show the importance of testing. If you want different behaviour, you'll need to modify the code. I generally opt for silent enforcement of rules if the caller can't be bothered (or is too stupid) to follow them but apparently some people like to use the principle of least astonishment and raise an exception :-)
You can use the std::ostringstream class to convert the number to a string. Then you can use the string of digits and print them using whatever formatting you want, as in the following code:
std::ostringstream oss;
oss << std::setfill('0') << std::setw(8) << number;
std::string str = oss.str();
if ( str.length() != 8 ){
// some form of handling
}else{
// print digits formatted as desired
}
int your_number = 12345678;
std::cout << (your_number/10000000) % 10 << (your_number/1000000) % 10 << (your_number/100000) %10 << "-" << (your_number/10000) %10 << (your_number/1000) %10 << "-" << (your_number/100) %10 << (your_number/10) %10 << (your_number) %10;
http://www.ideone.com/17eRv
Its not a function, but its a general method for parsing an int number by number.
#include <iostream>
#include <string>
using namespace std;
template<class Int, class Bi>
void format(Int n, Bi first, Bi last)
{
if( first == last ) return;
while( n != 0 ) {
Int t(n % 10);
n /= 10;
while( *--last != 'X' && last != first);
*last = t + '0';
}
}
int main(int argc, char* argv[])
{
int i = 23462345;
string s("XXX-XX-XXX");
format(i, s.begin(), s.end());
cout << s << endl;
return 0;
}
How's this?
std::string format(int x)
{
std::stringstream ss
ss.fill('0');
ss.width(3);
ss << (x / 10000);
ss.width(1);
ss << "-";
ss.width(2);
ss << (x / 1000) % 100;
ss.width(1);
ss << "-";
ss.width(3);
ss << x % 1000;
return ss.str();
}
Edit 1: I see strstream is deprecated and replaced with stringstream.
Edit 2: Fixed issue of missing leading 0's. I know, it's ugly.
Obviously a char * and not a string, but you get the idea. You'll need to free the output once you're done, and you should probably add error checking, but this should do it:
char * formatter(int i)
{
char *buf = malloc(11*sizeof(char));
sprintf(buf, "%03d-%02d-%03d", i/100000, (i/1000)%100, i%1000);
return buf;
}
You don't require malloc or new, just define buf as char buff[11];