String not displaying all characters in c++ [duplicate] - c++

Question

This question already has answers here:
Print leading zeros with C++ output operator?
(6 answers)
Closed 2 years ago.
I want cout to output an int with leading zeros, so the value 1 would be printed as 001 and the value 25 printed as 025. How can I do this?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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;
}

Answer 7

I would use the following function. I don't like sprintf; it doesn't do what I want!!
#define hexchar(x) ((((x)&0x0F)>9)?((x)+'A'-10):((x)+'0'))
typedef signed long long Int64;
// Special printf for numbers only
// See formatting information below.
//
// Print the number "n" in the given "base"
// using exactly "numDigits".
// Print +/- if signed flag "isSigned" is TRUE.
// Use the character specified in "padchar" to pad extra characters.
//
// Examples:
// sprintfNum(pszBuffer, 6, 10, 6, TRUE, ' ', 1234); --> " +1234"
// sprintfNum(pszBuffer, 6, 10, 6, FALSE, '0', 1234); --> "001234"
// sprintfNum(pszBuffer, 6, 16, 6, FALSE, '.', 0x5AA5); --> "..5AA5"
void sprintfNum(char *pszBuffer, int size, char base, char numDigits, char isSigned, char padchar, Int64 n)
{
char *ptr = pszBuffer;
if (!pszBuffer)
{
return;
}
char *p, buf[32];
unsigned long long x;
unsigned char count;
// Prepare negative number
if (isSigned && (n < 0))
{
x = -n;
}
else
{
x = n;
}
// Set up small string buffer
count = (numDigits-1) - (isSigned?1:0);
p = buf + sizeof (buf);
*--p = '\0';
// Force calculation of first digit
// (to prevent zero from not printing at all!!!)
*--p = (char)hexchar(x%base);
x = x / base;
// Calculate remaining digits
while(count--)
{
if(x != 0)
{
// Calculate next digit
*--p = (char)hexchar(x%base);
x /= base;
}
else
{
// No more digits left, pad out to desired length
*--p = padchar;
}
}
// Apply signed notation if requested
if (isSigned)
{
if (n < 0)
{
*--p = '-';
}
else if (n > 0)
{
*--p = '+';
}
else
{
*--p = ' ';
}
}
// Print the string right-justified
count = numDigits;
while (count--)
{
*ptr++ = *p++;
}
return;
}