How can I pad an int with leading zeros when using cout << operator? [duplicate]

Question

With the following,

#include <iomanip>
#include <iostream>

int main()
{
    std::cout << std::setfill('0') << std::setw(5) << 25;
}

the output will be

00025

setfill is set to the space character (' ') by default. setw sets the width of the field to be printed, and that's it.

---

If you are interested in knowing how the to format output streams in general, I wrote an answer for another question, hope it is useful: Formatting C++ Console Output.

Another way to achieve this is using old printf() function of C language

You can use this like

int dd = 1, mm = 9, yy = 1;
printf("%02d - %02d - %04d", mm, dd, yy);

This will print 09 - 01 - 0001 on the console.

You can also use another function sprintf() to write formatted output to a string like below:

int dd = 1, mm = 9, yy = 1;
char s[25];
sprintf(s, "%02d - %02d - %04d", mm, dd, yy);
cout << s;

Don't forget to include stdio.h header file in your program for both of these functions

Thing to be noted:

You can fill blank space either by 0 or by another char (not number).
If you do write something like %24d format specifier than this will not fill 2 in blank spaces. This will set pad to 24 and will fill blank spaces.

cout.fill('*');
cout << -12345 << endl; // print default value with no field width
cout << setw(10) << -12345 << endl; // print default with field width
cout << setw(10) << left << -12345 << endl; // print left justified
cout << setw(10) << right << -12345 << endl; // print right justified
cout << setw(10) << internal << -12345 << endl; // print internally justified

This produces the output:

-12345
****-12345
-12345****
****-12345
-****12345

cout.fill( '0' );    
cout.width( 3 );
cout << value;

In C++20 you'll be able to do:

std::cout << std::format("{:03}", 25); // prints 025

In the meantime you can use the {fmt} library, std::format is based on.

Disclaimer: I'm the author of {fmt} and C++20 std::format.

Another example to output date and time using zero as a fill character on instances of single digit values: 2017-06-04 18:13:02

#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <ctime>
using namespace std;

int main()
{
    time_t t = time(0);   // Get time now
    struct tm * now = localtime(&t);
    cout.fill('0');
    cout << (now->tm_year + 1900) << '-'
        << setw(2) << (now->tm_mon + 1) << '-'
        << setw(2) << now->tm_mday << ' '
        << setw(2) << now->tm_hour << ':'
        << setw(2) << now->tm_min << ':'
        << setw(2) << now->tm_sec
        << endl;
    return 0;
}