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im new to Regex and C++.
My problem is, that '=' is matching when I search for [a-zA-Z]. But this is only a-z without '='?
Can anyone help me please?
string string1 = "s=s;";
enum states state = s1;
regex statement("[a-zA-Z]+[=][a-zA-Z0-9]+[;]");
regex rg_left_letter("[a-zA-Z]");
regex rg_equal("[=]");
regex rg_right_letter("[a-zA-Z0-9]");
regex rg_semicolon("[;]");
for (const auto &s : string1) {
cout << "Current Value: " << s << endl;
// step(&state, s);
if (regex_search(&s, rg_left_letter)) {
cout << "matching: " << s << endl;
} else {
cout << "not matching: " << s << endl;
}
// cout << "Step Executed with sate: " << state << endl;
}
This outputs:
Current Value: s
matching: s
Current Value: =
matching: =
Current Value: s
matching: s
Current Value: ;
not matching: ;
When you write
regex_search(&s, rg_left_letter)
you basically search the C-String &s for a match character-wise, beginning at the character s. Therefore, your loop will search for a match in the remaining sub-strings
s=s;
=s;
s;
;
Which will always succeed, except in the last case, as there is always one character in the entire string that fits your regex. Note however that this assumes that std::string has some 0-termination added, which is, as far as I can tell, not guaranteed if you do not explicitely use the c_str() method, making your code UB.
What you really want to use is the function regex_match, together with your original regex just as simple as:
#include <iostream>
#include <regex>
int main()
{
std::regex statement("[a-zA-Z]+[=][a-zA-Z0-9]+[;]");
if(std::regex_match("s=s;", statement)) { std::cout << "Hooray!\n"; }
}
This is working for me:
int main(void) {
string string1 = "s=s;";
enum states state = s1;
regex statement("[a-zA-Z]+[=][a-zA-Z0-9]+[;]");
regex rg_left_letter("[a-zA-Z]");
regex rg_equal("[=]");
regex rg_right_letter("[a-zA-Z0-9]");
regex rg_semicolon("[;]");
//for (const auto &s : string1) {
for (int i = 0; i < string1.size(); i++) {
cout << "Current Value: " << string1[i] << endl;
// step(&state, s);
if (regex_match(string1.substr(i, 1), rg_left_letter)) {
cout << "matching: " << string1[i] << endl;
} else {
cout << "not matching: " << string1[i] << endl;
}
// cout << "Step Executed with sate: " << state << endl;
}
cout << endl;
return 0;
}
That's my first question here, so I would be glad to receive some support on the style I used to refer to my problem :). Here is the finished program, its main purpose is to split given words into halves and create words replacing the origin ones. Replaced words are build from its origins by spliting them into halves and taking even ones from the 1st half begining with the first letter of a word. Heres the complete code:
#include <iostream>
#include <string>
#include <cstdio>
#include <math.h>
using namespace std;
void obcinaczSlow(int);
int main(){
int ilosc;
cout << "Prosze o podanie ilosci prob: ";
cin>>ilosc;
cout << endl;
obcinaczSlow(ilosc);
cin.ignore();
cin.get();
return 0;
}
void obcinaczSlow(int ilosc_prob){
int i=0,j=0,dlugosc_slowa=0,dlugosc_polowy=0;
string *tablica_slow,budowane_slowo,aktualne_slowo,dodane;
tablica_slow = new string [ilosc_prob];
cout << "Prosze o podanie " << ilosc_prob << " slow" << endl;
cin.sync();
for(i=0;i<ilosc_prob;i++){
cout << "Prosze o podanie slowa numer: " << i+1 << endl;
cin>>aktualne_slowo;
tablica_slow[i] = aktualne_slowo;
}
for(i=0;i<ilosc_prob;i++){
aktualne_slowo = tablica_slow[i];
cout << "Aktualne slowo do przerobienia: " << aktualne_slowo << endl;
dlugosc_slowa = aktualne_slowo.length();
cout << "Dlugosc slowa do przerobienia: " << dlugosc_slowa << endl;
dlugosc_polowy = floor(dlugosc_slowa/2);
cout << "Dlugosc polowy slowa int: " << dlugosc_polowy << endl;
budowane_slowo.clear();
dodane.clear();
cout << "Budowane slowo to: " << budowane_slowo << endl;
for(j=0;j<=dlugosc_polowy;j=+2){
dodane = aktualne_slowo.at(j);
budowane_slowo.append(dodane);
}
tablica_slow[i] = budowane_slowo;
}
cout << "Slowa po transformacji wygladaja nastepujaco: " << endl;
for(i=0;i<ilosc_prob;i++){
cout << "Slowo o numerze " << i+1 << " : " << tablica_slow[i] << endl;
}
delete [] tablica_slow;
cin.sync();
}
The problem raises when program reaches the loop, that is supposed to append the letter pointed by the j-index using '.at' method from the string class. I can't find a solution even trying to debug it. Could You help me :)?
You have a typo here
for(j=0;j<=dlugosc_polowy;j=+2)
I assume you meant += instead of =+
for(j=0;j<=dlugosc_polowy;j+=2)
Otherwise you are just assigning 2 to j over and over again.
Your error is reversing two characters:
Change:
`j=+2` to `j+=2`
^^ ^^
(The way it is written j is assigned the value of 2, then, for the rest of its life, stays there.)
for(j=0;j<=dlugosc_polowy;j=+2){
dodane = aktualne_slowo.at(j);
budowane_slowo.append(dodane);
}
replace the j=+2 to j+=2
for(j=0;j<=dlugosc_polowy;j+=2){
dodane = aktualne_slowo.at(j);
budowane_slowo.append(dodane);
}
I am trying to extract values from myString1 using std::stringstream like shown below:
// Example program
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
string myString1 = "+50years";
string myString2 = "+50years-4months+3weeks+5minutes";
stringstream ss (myString1);
char mathOperator;
int value;
string timeUnit;
ss >> mathOperator >> value >> timeUnit;
cout << "mathOperator: " << mathOperator << endl;
cout << "value: " << value << endl;
cout << "timeUnit: " << timeUnit << endl;
}
Output:
mathOperator: +
value: 50
timeUnit: years
In the output you can see me successfully extract the values I need, the math operator, the value and the time unit.
Is there a way to do the same with myString2? Perhaps in a loop? I can extract the math operator, the value, but the time unit simply extracts everything else, and I cannot think of a way to get around that. Much appreciated.
The problem is that timeUnit is a string, so >> will extract anything until the first space, which you haven't in your string.
Alternatives:
you could extract parts using getline(), which extracts strings until it finds a separator. Unfortunately, you don't have one potential separator, but 2 (+ and -).
you could opt for using regex directly on the string
you could finally split the strings using find_first_of() and substr().
As an illustration, here the example with regex:
regex rg("([\\+-][0-9]+[A-Za-z]+)", regex::extended);
smatch sm;
while (regex_search(myString2, sm, rg)) {
cout <<"Found:"<<sm[0]<<endl;
myString2 = sm.suffix().str();
//... process sstring sm[0]
}
Here a live demo applying your code to extract ALL the elements.
You could something more robust like <regex> like in the example below:
#include <iostream>
#include <regex>
#include <string>
int main () {
std::regex e ("(\\+|\\-)((\\d)+)(years|months|weeks|minutes|seconds)");
std::string str("+50years-4months+3weeks+5minutes");
std::sregex_iterator next(str.begin(), str.end(), e);
std::sregex_iterator end;
while (next != end) {
std::smatch match = *next;
std::cout << "Expression: " << match.str() << "\n";
std::cout << " mathOperator : " << match[1] << std::endl;
std::cout << " value : " << match[2] << std::endl;
std::cout << " timeUnit : " << match[4] << std::endl;
++next;
}
}
Output:
Expression: +50years
mathOperator : +
value : 50
timeUnit : years
Expression: -4months
mathOperator : -
value : 4
timeUnit : months
Expression: +3weeks
mathOperator : +
value : 3
timeUnit : weeks
Expression: +5minutes
mathOperator : +
value : 5
timeUnit : minutes
LIVE DEMO
I'd use getline for the timeUnit, but since getline can take only one delimiter, I'd search the string separately for mathOperator and use that:
string myString2 = "+50years-4months+3weeks+5minutes";
stringstream ss (myString2);
size_t pos=0;
ss >> mathOperator;
do
{
cout << "mathOperator: " << mathOperator << endl;
ss >> value;
cout << "value: " << value << endl;
pos = myString2.find_first_of("+-", pos+1);
mathOperator = myString2[pos];
getline(ss, timeUnit, mathOperator);
cout << "timeUnit: " << timeUnit << endl;
}
while(pos!=string::npos);
I'm not sure, but I think I remember there being something in Java that can specify how far from the left of a window that a string or digit begins..
How to easily format a table?
I have this (using setw):
Bob Doe 10.96 7.61 14.39 2.11 47.30 14.21 44.58 5.00 60.23
Helen City 10.44 7.78 16.27 1.99 48.92 13.93 53.79 5.00 70.97
Joe Green 10.90 7.33 14.49 2.05 47.91 14.15 44.45 4.70 73.98
and ideally would like:
Bob Doe BLR 10.96 7.61 14.39 2.11 47.30 14.21 44.58 5.00 60.23 4:27.47
Helen City CUB 10.90 7.33 14.49 2.05 47.91 14.15 44.45 4.70 73.98 4:29.17
Joe Green USA 10.44 7.78 16.27 1.99 48.92 13.93 53.79 5.00 70.97 5:06.59
Is the only way calculations? Or is there some magical even more simple way?
In C++, you have three functions to help you do what you want. There are defined in <iomanip>.
- setw() helps you defined the width of the output.
- setfill() Fill the rest with the character you want (in your case ' ').
- left (or right) allow you to define the alignment.
Here is the code to write your first line :
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
const char separator = ' ';
const int nameWidth = 6;
const int numWidth = 8;
cout << left << setw(nameWidth) << setfill(separator) << "Bob";
cout << left << setw(nameWidth) << setfill(separator) << "Doe";
cout << left << setw(numWidth) << setfill(separator) << 10.96;
cout << left << setw(numWidth) << setfill(separator) << 7.61;
cout << left << setw(numWidth) << setfill(separator) << 14.39;
cout << left << setw(numWidth) << setfill(separator) << 2.11;
cout << left << setw(numWidth) << setfill(separator) << 47.30;
cout << left << setw(numWidth) << setfill(separator) << 14.21;
cout << left << setw(numWidth) << setfill(separator) << 44.58;
cout << left << setw(numWidth) << setfill(separator) << 5.00;
cout << left << setw(numWidth) << setfill(separator) << 60.23;
cout << endl;
cin.get();
}
EDIT :
To reduce the code, you can use a template function :
template<typename T> void printElement(T t, const int& width)
{
cout << left << setw(width) << setfill(separator) << t;
}
That you can use like this :
printElement("Bob", nameWidth);
printElement("Doe", nameWidth);
printElement(10.96, numWidth);
printElement(17.61, numWidth);
printElement(14.39, numWidth);
printElement(2.11, numWidth);
printElement(47.30, numWidth);
printElement(14.21, numWidth);
printElement(44.58, numWidth);
printElement(5.00, numWidth);
printElement(60.23, numWidth);
cout << endl;
Here are the various functions I use to display data in an organized, tabular form, along with an example demonstrating a possible use scenario.
Because the functions use stringstreams, they aren't as fast as other solutions, but for me that never matters --- the computing bottlekneck is elsewhere.
One advantage of using stringstreams is that the functions alter the precision of their own (internal scope) stringstreams, instead of changing the static cout precision. So you never have to worry about unintentionally modifying precision in a way that persists to affect other parts of your code.
DISPLAYING ARBITRARY PRECISION
This prd function (short for "print double") simply prints a double value with a specified precision.
/* Convert double to string with specified number of places after the decimal. */
std::string prd(const double x, const int decDigits) {
stringstream ss;
ss << fixed;
ss.precision(decDigits); // set # places after decimal
ss << x;
return ss.str();
}
The following is just a variant that allows you to specify a blank-space padding to the left of the number. This can be helpful in displaying tables.
/* Convert double to string with specified number of places after the decimal
and left padding. */
std::string prd(const double x, const int decDigits, const int width) {
stringstream ss;
ss << fixed << right;
ss.fill(' '); // fill space around displayed #
ss.width(width); // set width around displayed #
ss.precision(decDigits); // set # places after decimal
ss << x;
return ss.str();
}
CENTER-ALIGN FUNCTION
This function simply center-aligns text, padding left and right with blank spaces until the returned string is as large as the specified width.
/*! Center-aligns string within a field of width w. Pads with blank spaces
to enforce alignment. */
std::string center(const string s, const int w) {
stringstream ss, spaces;
int padding = w - s.size(); // count excess room to pad
for(int i=0; i<padding/2; ++i)
spaces << " ";
ss << spaces.str() << s << spaces.str(); // format with padding
if(padding>0 && padding%2!=0) // if odd #, add 1 space
ss << " ";
return ss.str();
}
EXAMPLE OF TABULAR OUTPUT
So, we could use the prd and center functions above to output a table in the following fashion.
The code:
std::cout << center("x",10) << " | "
<< center("x^2",10) << " | "
<< center("(x^2)/8",10) << "\n";
std::cout << std::string(10*3 + 2*3, '-') << "\n";
for(double x=1.5; x<200; x +=x*2) {
std::cout << prd(x,1,10) << " | "
<< prd(x*x,2,10) << " | "
<< prd(x*x/8.0,4,10) << "\n";
}
will print the table:
x | x^2 | (x^2)/8
------------------------------------
1.5 | 2.25 | 0.2812
4.5 | 20.25 | 2.5312
13.5 | 182.25 | 22.7812
40.5 | 1640.25 | 205.0312
121.5 | 14762.25 | 1845.2812
RIGHT- and LEFT-ALIGN FUNCTIONS
And, of course, you can easily construct variants of the center function that right- or left-align and add padding spaces to fill the desired width. Here are such functions:
/* Right-aligns string within a field of width w. Pads with blank spaces
to enforce alignment. */
string right(const string s, const int w) {
stringstream ss, spaces;
int padding = w - s.size(); // count excess room to pad
for(int i=0; i<padding; ++i)
spaces << " ";
ss << spaces.str() << s; // format with padding
return ss.str();
}
/*! Left-aligns string within a field of width w. Pads with blank spaces
to enforce alignment. */
string left(const string s, const int w) {
stringstream ss, spaces;
int padding = w - s.size(); // count excess room to pad
for(int i=0; i<padding; ++i)
spaces << " ";
ss << s << spaces.str(); // format with padding
return ss.str();
}
I'm sure there are plenty of more-elegant ways to do this kind of thing --- certainly there are more concise ways. But this is what I do. Works well for me.
Just use sprintf with format specifiers to format fields. You can also use MFC CString
#include <iostream>
#include "stdio.h"
using namespace std;
int main()
{
char buf[256];
char pattern[] = "%10s %10s %7.2f %7.2f %7.2f %7.2f %7.2f %7.2f %7.2f %7.2f %7.2f";
sprintf(buf, pattern, "Bob", "Doe", 10.96, 7.61, 14.39, 2.11, 47.30, 14.21, 44.58, 5.00, 60.23);
cout << buf << endl;
sprintf(buf, pattern, "Helen", "City", 10.44, 7.78, 16.27, 1.99, 48.92, 13.93, 53.79, 5.00, 70.97);
cout << buf << endl;
sprintf(buf, pattern, "Joe", "Green", 10.90, 7.33, 14.49, 2.05, 47.91, 14.15, 44.45, 4.70, 73.98);
cout << buf << endl;
}
You could do something like this to simplify the process a bit.
#include <iomanip>
#include <iostream>
struct TableFormat {
int width;
char fill;
TableFormat(): width(14), fill(' ') {}
template<typename T>
TableFormat& operator<<(const T& data) {
std::cout << data << std::setw(width) << std::setfill(fill);
return *this;
}
TableFormat& operator<<(std::ostream&(*out)(std::ostream&)) {
std::cout << out;
return *this;
}
};
int main() {
TableFormat out;
out << "Bob" << "Doe";
out.width = 8;
out << "BLR" << 10.96 << 7.61 << 14.39 << 2.11 << 47.30;
}
Which would print out (horribly in my case, but it's "customisable" to a degree):
Bob Doe BLR 10.96 7.61 14.39 2.11 47.3
The code is pretty self-explanatory, it's just a wrapper around std::cout to allow you to make the tedious calls easier, the second overload for operator<< is to allow you send std::endl..
C++20 includes <format> but it's not supported by libc++ for now.
I suggest to use {fmt} library since it could be obtained easily in Ubuntu20.
According to the doc, you may specify the width as an argument as well.
Format example: {2:<{0}}
`2` -> Use second arg as value.\
`:` -> Use non-default format.\
`<` -> Align to left\
`{0}` -> Use argument 0 as width.
Live Demo
#include <string>
#include <iostream>
#include <fmt/core.h>
#include <tuple>
#include <vector>
int main()
{
using Row = std::tuple<std::string, std::string, double>;
std::vector<Row> table = {
std::make_tuple("Bob", "Doe", 10.96),
std::make_tuple("Helen", "City", 10.44),
std::make_tuple("Joe", "Green", 10.90)
};
size_t nameWidth{12};
size_t valWidth{7};
for(const auto& row: table){
std::cout << fmt::format("{2:<{0}} {3:<{0}} {4:<{1}} \n",
nameWidth, valWidth, std::get<0>(row), std::get<1>(row), std::get<2>(row) );
}
}
Output
Bob Doe 10.96
Helen City 10.44
Joe Green 10.9
Assuming you want to format your output to resemble a table, what you need is I/O manipulators.
You can use setw() manipulator to set the output width and setfill() to set the filling character.
Considering an example:
string firstname = "Bob";
string lastname = "Doe";
string country = "BLR";
float f1 = 10.96f, f2=7.61f, f3=14.39f, f4=2.11f, f5=47.30f, f6=14.21f, f7=44.58f, f8=5.00f, f9=60.23f;
string time = "4:27.47";
cout << setw(12) << firstname << set(12) << lastname;
cout << setw(5) << country << setprecision(2) << f1 << setprecision(2) << f2 << setprecision(2) << f3..
use setw() to set the width while printing a string
use setprecision to set the precision for floating values
read MSDN
I'm not sure what you wrote so I can't see what's wrong, but you can get the results you want with std::setw:
#include <iostream>
#include <iomanip>
int main() {
std::cout << std::left << std::setw(20) << "BoB" << std::setw(20) << 123.456789 << '\n';
std::cout << std::left << std::setw(20) << "Richard" << std::setw(20) << 1.0 << '\n';
}
http://ideone.com/Iz5RXr
I've been trying to format the output to the console for the longest time and nothing is really happening. I've been trying to use as much of iomanip as I can and the ofstream& out functions.
void list::displayByName(ostream& out) const
{
node *current_node = headByName;
// I have these outside the loop so I don't write it every time.
out << "Name\t\t" << "\tLocation" << "\tRating " << "Acre" << endl;
out << "----\t\t" << "\t--------" << "\t------ " << "----" << endl;
while (current_node)
{
out << current_node->item.getName() // Equivalent tabs don't work?
<< current_node->item.getLocation()
<< current_node->item.getAcres()
<< current_node->item.getRating()
<< endl;
current_node = current_node->nextByName;
}
// The equivalent tabs do not work because I am writing names,
// each of different length to the console. That explains why they
// are not all evenly spaced apart.
}
Is their anything that I can use to get it all properly aligned with each other?
The functions that I'm calling are self-explanatory and all of different lengths, so that don't align very well with each other.
I've tried just about everything in iomanip.
Think of it like using Microsoft Excel :)
You think of your stream as fields. So you set the width of the field first then you insert your text in that field. For example:
#include <iostream>
#include <iomanip>
#include <string>
int main()
{
using namespace std;
string firstName = "firstName",
secondName = "SecondName",
n = "Just stupid Text";
size_t fieldWidth = n.size(); // length of longest text
cout << setw(fieldWidth) << left << firstName << endl // left padding
<< setw(fieldWidth) << left << secondName << endl
<< setw(fieldWidth) << left << n << endl;
cout << setw(fieldWidth) << right << firstName << endl // right padding
<< setw(fieldWidth) << right << secondName << endl
<< setw(fieldWidth) << right << n << endl;
}
......
......
The field width means nothing but the width of the text + spaces. You could fill anything other than spaces:
string name = "My first name";
cout << setfill('_') << setw(name.size() + 10) << left << name;
.....
output::
My first name__________
......
I think the best way is to figure out your format then, write a new formatter that does all what you want:
#include <iostream>
#include <iomanip>
#include <string>
std::ostream& field(std::ostream& o)
{
// usually the console is 80-character wide.
// divide the line into four fields.
return o << std::setw(20) << std::right;
}
int main()
{
using namespace std;
string firstName = "firstName",
secondName = "SecondName",
n = "Just stupid Text";
size_t fieldWidth = n.size();
cout << field << firstName << endl
<< field << secondName << endl
<< field << n << endl;
}
If you started thinking about parametrized manipulators, only that accept one int or long parameter are easy to implement, other types are really obscure if you are not familiar with streams in C++.
Boost has a format library that allows you to easily format the ourput like the old C printf() but with type safety of C++.
Remember that the old C printf() allowed you to specify a field width. This space fills the field if the output is undersized (note it does not cope with over-sized fields).
#include <iostream>
#include <iomanip>
#include <boost/format.hpp>
struct X
{ // this structure reverse engineered from
// example provided by 'Mikael Jansson' in order to make this a running example
char* name;
double mean;
int sample_count;
};
int main()
{
X stats[] = {{"Plop",5.6,2}};
// nonsense output, just to exemplify
// stdio version
fprintf(stderr, "at %p/%s: mean value %.3f of %4d samples\n",
stats, stats->name, stats->mean, stats->sample_count);
// iostream
std::cerr << "at " << (void*)stats << "/" << stats->name
<< ": mean value " << std::fixed << std::setprecision(3) << stats->mean
<< " of " << std::setw(4) << std::setfill(' ') << stats->sample_count
<< " samples\n";
// iostream with boost::format
std::cerr << boost::format("at %p/%s: mean value %.3f of %4d samples\n")
% stats % stats->name % stats->mean % stats->sample_count;
}
Give up on the tabs. You should be able to use io manipulators to set the field width, the fill character, and the format flag (to get left or right justification). Use the same values for the headings as you do for the data, and everything should come out nicely.
Also beware that you've switched Rating and Acres in your example.
You can write a procedure that always print the same number of characters to standard output.
Something like:
string StringPadding(string original, size_t charCount)
{
original.resize(charCount, ' ');
return original;
}
And then use like this in your program:
void list::displayByName(ostream& out) const
{
node *current_node = headByName;
out << StringPadding("Name", 30)
<< StringPadding("Location", 10)
<< StringPadding("Rating", 10)
<< StringPadding("Acre", 10) << endl;
out << StringPadding("----", 30)
<< StringPadding("--------", 10)
<< StringPadding("------", 10)
<< StringPadding("----", 10) << endl;
while ( current_node)
{
out << StringPadding(current_node->item.getName(), 30)
<< StringPadding(current_node->item.getLocation(), 10)
<< StringPadding(current_node->item.getRating(), 10)
<< StringPadding(current_node->item.getAcres(), 10)
<< endl;
current_node = current_node->nextByName;
}
}