User
User

Reputation: 66071

Format number with commas in C++

I want to write a method that will take an integer and return a std::string of that integer formatted with commas.

Example declaration:

std::string FormatWithCommas(long value);

Example usage:

std::string result1 = FormatWithCommas(7800);
std::string result2 = FormatWithCommas(5100100);
std::string result3 = FormatWithCommas(201234567890);
// result1 = "7,800"
// result2 = "5,100,100"
// result3 = "201,234,567,890"

What is the C++ way of formatting a number as a string with commas?

(Bonus would be to handle doubles as well.)

Upvotes: 88

Views: 108410

Answers (14)

fsmoke
fsmoke

Reputation: 185

Right solution is

struct comma_out : std::numpunct<char>
{
    char do_thousands_sep()   const { return ','; }  // separate with spaces
    std::string do_grouping() const { return "\3"; } // groups of 3 digit
};
 
int main()
{
    std::cout << "default locale: " << 12345678 << '\n';
    std::cout.imbue(std::locale(std::cout.getloc(), new comma_out));
    std::cout << "locale with modified numpunct: " << 12345678 << '\n';
}

YOU MUST NOT USE std::fixed - it's only for floating point

Upvotes: 2

Phillip
Phillip

Reputation: 11

Here is a very simple way to get what you are asking for. You will need a long long to handle result3 201+ billion. The function formatWithCommas() takes in a long long and returns a formatted string with commas at every third decimal place. Use string method to_string() from the standard string library to convert your number to a string. Then use string.length() to get the length of your numeric string. Setup a for loop that starts at the length of your string minus three. Use another function from the string library insert() to place a comma at every third position as you decrement by three, i -= 3. Notice that I use a string "," instead of a character ','. If you don't change datatypes you only need two arguments, the index and in this case a sub-string. If you change datatypes you will need to use a third argument to state how many characters you want to place starting at the given index. Here is an example, str.insert(i, ",") or str.insert(i, ',', 1). The conditional section of your loop i > 0 will prevent the loop from running on a string with less than four characters as str.length() -3 must be at least four to start with making i equal to one for i > 0 to be a true statement.

If you need to work with possibly negative numbers you can set a bool flag and manipulate index zero before the for loop and insert the negative symbol - back after the commas have been correctly placed before returning the formatted numeric string.

/*
Program:Format Number With Comma
Author:Phillip McCullough
Date:1/29/2023
C/Cpp:11+ ANSI Standard

This program uses the function formatWithCommas() to add one or more commas at
every third decimal place as conventional written formatting includes.
-------10--------20--------30--------40--------50--------60--------70--------80
*/
#include <iostream> 
#include <string> 

std::string formatWithCommas(long long number);

int main()
{ 
    std::cout << formatWithCommas(7800) << '\n'// result1 = 7,800
        << formatWithCommas(5100100) << '\n'// result2 = 5,100,100
        << formatWithCommas(201234567890) << '\n';// result3 = 201,234,567,890

    return 0;
}

// ---------------------------------------------------------------------------- 

std::string formatWithCommas(long long number)
{
    std::string number_str = std::to_string(number);

    for (int i = number_str.length() - 3; i > 0; i -= 3)
        number_str.insert(i, ",");

    return number_str;
}

// ----------------------------------------------------------------------------

Upvotes: 1

yano
yano

Reputation: 5285

Inspired by @carljalal's answer, I found a similar-but-different answer using reverse iterators. Probably less efficient, but a bit more compact:

#include <string>
#include <iterator>
#include <cctype>  // for isdigit
...

void addCommas(std::string& num)
{
   // begin at the 3rd digit from the back, advance 3 each time
   for (auto it = str.rbegin()+3; it < str.rend(); it += 3)
   {
      // this handles a negative sign (ignores it)
      if (isdigit(*it))
      {
         // there are currently no string functions dealing with reverse_iterators,
         // so use .base() to get it's corresponding forward_iterator

         // inserting into a string invalidates any iterators, so "reset" `it`, and
         // we have to make the return value into a reverse_iterator
         it = std::make_reverse_iterator(str.insert(it.base(), ','));
      }
   }
}

Note, there is some wariness using .base(), but all seems to be well for the inputs I tried. Above handles any length number string, with or without a preceding negative sign, but does not handle decimals. Also note that std::make_reverse_iterator requires c++14 or later.

Some sample inputs and outputs:

1 ==>                  1
12 ==>                 12
123 ==>                123
-12 ==>               -12
-123 ==>              -123
1234 ==>               1,234
-1234 ==>             -1,234
12134 ==>              12,134
-12134 ==>            -12,134
328947328949893 ==>    328,947,328,949,893
-328947328949893 ==>  -328,947,328,949,893
9328947328949893 ==>   9,328,947,328,949,893
-9328947328949893 ==> -9,328,947,328,949,893

Demonstration

Upvotes: 0

riPmAn
riPmAn

Reputation: 31

Based on Jacob's answer above...

This is a Windows solution.

// Compiler : Visual Studio 2022
// I tested on Windows 10 64bit
// DATE: 2022.Sep.15

#include <windows.h>
#include <string>
#include <locale>
#include <sstream>

using namespace std;

template<class T>
std::string FormatWithCommas(T value)
{    
    std::stringstream ss;

    // otherwise, use "en_US.UTF-8" as the locale name
    ss.imbue(std::locale("ko_KR.UTF-8"));
    ss << std::fixed << value;

    return ss.str();
}

int main()
{
    LARGE_INTEGER filesize;

    filesize.QuadPart = 1234591046976109;

    // This only works for ASCII strings
    string aaa = (FormatWithCommas(filesize.QuadPart)).c_str();
    std::wstring widestr = std::wstring(aaa.begin(), aaa.end());
    const wchar_t* widecstr = widestr.c_str();

    wprintf_s(L"\nFile Size =  %s bytes\n", widecstr);

    getchar();
    return 0;
}

The execution result is as follows.

The execution result is as follows.

Upvotes: 0

Richard Day
Richard Day

Reputation: 49

I have seen so many ways of doing this, reversing the string(Twice!), using setlocale(sometimes works sometimes not) This is a template solution, I then add explicit specializations. This works for char*, wchar*, string and wstring. I dont translate from numeric to string format here, I highly recommend to_string and to_wstring they are way faster than the 'C' functions such as _itoa etc...

    template<typename T, typename U>
    T StrFormatNumber(const T Data) {
const size_t Length = Data.length();
assert(Length > 0);
// if( 0 == Length ) I would log this and return 
if (Length < 4) { // nothing to do just return
    return Data;
}
constexpr size_t buf_size{ 256 };
assert(((Length)+(Length / 3)) + 1 < buf_size);
if (((Length)+(Length / 3)) + 1 >= buf_size) {
     throw std::invalid_argument( "Input buffer too large" );
}
std::array<U, buf_size > temp_buf{};
auto p{ 0 };
temp_buf[0] = Data[0];
for (auto y{ 1 }; y < Length; y++) {
    if ((Length - y) % 3 == 0) {
        temp_buf[y + p] = ',';
        p++;
    }
    temp_buf[(y + p)] = Data[y];
}
return temp_buf.data();
}
    template<typename T = const char*>
    std::string StrFormatNum(const char* Data) {
        return StrFormatNumber<std::string, char>(std::string(Data));
    }
    template<typename T= std::string>
    std::string StrFormatNum(const std::string Data) {
         return StrFormatNumber<std::string, char>(Data);
     }
    template<typename T = std::wstring>
    std::wstring StrFormatNum( const std::wstring Data) {
        return StrFormatNumber<std::wstring, wchar_t>(Data);
    }
    template<typename T = const wchar_t*>
    std::wstring StrFormatNum( const wchar_t* Data) {
        return StrFormatNumber<std::wstring, wchar_t>(std::wstring(Data));
    }

    void TestStrFormatNumber() {
        constexpr auto Iterations{ 180 };
        for (auto l{ 0 }; l < Iterations; l++)
        {
            {  // std::string
                std::string mystr{ "10" };
                for (int y{ 0 }; y < Iterations; y++) {

                    mystr += "1";
                    auto p = mystr.length();
                    std::cout << "\r\n  mystr      = " 
                              << std::setw(80) << mystr.c_str()
                              << "\r\n  Length     = "
                              << std::setw(10) << p
                              << "\r\n  modulo % 3 = "
                              << std::setw(10)
                              << p % 3 << "     divided by 3 = "
                              << std::setw(10) << p / 3
                              << "\r\n  Formatted  = " << 
                    StrFormatNum((mystr)).c_str() << "\n";
                }
            }
            {  // std::wstring
                std::wstring mystr{ L"10" };
                for (int y{ 0 }; y < Iterations; y++)
                {
                     mystr += L"2";
                     auto p = mystr.length();
                     std::wcout << "\r\n  mystr      = "
                                << std::setw(80) << mystr.c_str()
                                << "\r\n  Length     = "
                                << std::setw(10) << p
                                << "\r\n  modulo % 3 = "
                                << std::setw(10) << p % 3
                                << "     divided by 3 = "
                                << std::setw(10) << p / 3
                                << "\r\n  Formatted  = "
                                << StrFormatNum((mystr)).c_str()
                                << "\n";
                        }
                    }
                    {   // char*
                        std::string mystr{ "10" };
        for (int y{ 0 }; y < Iterations; y++) {
            mystr += "3";
            auto p = mystr.length();
            std::cout << "\r\n  mystr      = " << std::setw(80) << mystr.c_str()
                << "\r\n    Length     = " << std::setw(10) << p
                << "\r\n  modulo % 3 = " << std::setw(10) << p % 3 << "     divided by 3 = " << std::setw(10) << p / 3
                << "\r\n  Formatted  = " << StrFormatNum((mystr.c_str())).c_str() << "\n";
        }
    }
    {  // wchar*
        std::wstring mystr{ L"10" };
        for (int y{ 0 }; y < Iterations; y++) {
            mystr += L"4";
            auto p = mystr.length();
            std::wcout << "\r\n  mystr      = " << std::setw(80) << mystr.c_str()
                << "\r\n  Length     = " << std::setw(10) << p
                << "\r\n  modulo % 3 = " << std::setw(10) << p % 3 << "     divided by 3 = " << std::setw(10) << p / 3
                << "\r\n  Formatted  = " << StrFormatNum((mystr.c_str())).c_str() << "\n";
        }
    } 
}

}

I have tested up to 1,000 spaces(With a larger buffer of course)

Upvotes: 1

carljalal
carljalal

Reputation: 647

I consider the following answer to be easier than the others:

#include <iostream>
int main() {
   int v = 7654321;
   auto s = std::to_string(v);

   int n = s.length() - 3;
   int end = (v >= 0) ? 0 : 1; // Support for negative numbers
   while (n > end) {
      s.insert(n, ",");
      n -= 3;
   }
   std::cout << (s == "7,654,321") << std::endl;
}   

This will quickly and correctly insert commas into your string of digits.

Upvotes: 43

Hammy Rahardja
Hammy Rahardja

Reputation: 11

I found the solution! just copy this to one of your function, this function is written in static function.

// Convert 100000000 to 100,000,000, put commas on the numbers!

std::string AppManager::convertNumberToString(int number) {
    std::string s = std::to_string(number);
    std::string result = "";
    std::string tempResult = "";
    unsigned long n = s.length() - 3;
    int j = 0;
    for (int i=s.size()-1; i>=0; i--) {
        if (j%3 == 0) {
            result.append(",");
        }
        result.append(s, i, 1);
        j++;
    }
    
    result = result.substr(1, result.size()-1);
    
    //now revert back
    for (int i=result.size()-1; i>=0; i--) {
        tempResult.append(result, i, 1);
    }
    
    return tempResult;
}

Here is the result of those code:

click here for seeing the result above!

Upvotes: 1

ravin.wang
ravin.wang

Reputation: 1252

Make another solution:

#include <stdio.h>
#include <string>
#include <stdint.h>
#include <inttypes.h>

std::string GetReadableNum(uint64_t n)
{
    std::string strRet;
    char szTmp[256] = { 0 };
    int ccWritten = sprintf(szTmp, "%" PRIu64 "", n);
    if (ccWritten > 0)
    {
        int nGroup = (ccWritten + 2) / 3;
        int nReminder = ccWritten % 3;
        
        strRet.reserve(ccWritten + (nGroup -1) * 3 + 1);
        const char* p = szTmp;
        for (int i = 0; i < nGroup; i++)
        {
            if (nGroup > 1 && i > 0)
                strRet.append(1, ',');

            for (int c = 0; c < (i > 0 || nReminder == 0 ? 3 : nReminder); c++)
                strRet.append(1, *p++);
        }
    }

    return strRet;
}

int main(int argc, const char* argv[])
{
    uint64_t a = 123456789123ULL;

    std::string s = GetReadableNum(a);

    printf("%s\n", s.c_str());

    return 0;
}

Upvotes: 0

NG_
NG_

Reputation: 7181

If you are using Qt, you can use this code:

const QLocale& cLocale = QLocale::c();
QString resultString = cLocale.toString(number);

Also, do not forget to add #include <QLocale>.

Upvotes: 2

Tom Serink
Tom Serink

Reputation: 29

This is pretty old school, I use it in large loops to avoid instantiating another string buffer.

void tocout(long a)
{
    long c = 1;

    if(a<0) {a*=-1;cout<<"-";}
    while((c*=1000)<a);
    while(c>1)
    {
       int t = (a%c)/(c/1000);
       cout << (((c>a)||(t>99))?"":((t>9)?"0":"00")) << t;
       cout << (((c/=1000)==1)?"":",");
    }
}

Upvotes: 2

Marcel
Marcel

Reputation: 1

To make it more flexible, you could construct the facet with a custom thousands sep and grouping string. This way you can set it at runtime.

#include <locale>
#include <iostream>
#include <iomanip>
#include <string>

class comma_numpunct : public std::numpunct<char>
{
public:
   comma_numpunct(char thousands_sep, const char* grouping)
      :m_thousands_sep(thousands_sep),
       m_grouping(grouping){}
protected:
   char do_thousands_sep() const{return m_thousands_sep;}
   std::string do_grouping() const {return m_grouping;}
private:
   char m_thousands_sep;
   std::string m_grouping;
};

int main()
{

    std::locale comma_locale(std::locale(), new comma_numpunct(',', "\03"));

    std::cout.imbue(comma_locale);
    std::cout << std::setprecision(2) << std::fixed << 1000000.1234;
}

Upvotes: -1

Radif Sharafullin
Radif Sharafullin

Reputation: 689

based on the answers above, I ended up with this code:

#include <iomanip>
#include <locale> 

template<class T>
std::string numberFormatWithCommas(T value){
    struct Numpunct: public std::numpunct<char>{
    protected:
        virtual char do_thousands_sep() const{return ',';}
        virtual std::string do_grouping() const{return "\03";}
    };
    std::stringstream ss;
    ss.imbue({std::locale(), new Numpunct});
    ss << std::setprecision(2) << std::fixed << value;
    return ss.str();
}

Upvotes: 1

Node
Node

Reputation: 3509

You can do as Jacob suggested, and imbue with the "" locale - but this will use the system default, which does not guarantee that you get the comma. If you want to force the comma (regardless of the systems default locale settings) you can do so by providing your own numpunct facet. For example:

#include <locale>
#include <iostream>
#include <iomanip>

class comma_numpunct : public std::numpunct<char>
{
  protected:
    virtual char do_thousands_sep() const
    {
        return ',';
    }

    virtual std::string do_grouping() const
    {
        return "\03";
    }
};

int main()
{
    // this creates a new locale based on the current application default
    // (which is either the one given on startup, but can be overriden with
    // std::locale::global) - then extends it with an extra facet that 
    // controls numeric output.
    std::locale comma_locale(std::locale(), new comma_numpunct());

    // tell cout to use our new locale.
    std::cout.imbue(comma_locale);

    std::cout << std::setprecision(2) << std::fixed << 1000000.1234;
}

Upvotes: 65

Jacob
Jacob

Reputation: 34621

Use std::locale with std::stringstream

#include <iomanip>
#include <locale>

template<class T>
std::string FormatWithCommas(T value)
{
    std::stringstream ss;
    ss.imbue(std::locale(""));
    ss << std::fixed << value;
    return ss.str();
}

Disclaimer: Portability might be an issue and you should probably look at which locale is used when "" is passed

Upvotes: 65

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