How can I pad an int with leading zeros when using

问题:

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?

回答1:

First include <iomanip>, then:

cout << setfill(\'0\') << setw(5) << 25;

output:
00025

setfill is set to space \' \' 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.

回答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.

回答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

回答4:

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

回答5:

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;
}

回答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;
}