C++ 使用 cout << 运算符时，如何用前导零填充 int？
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How can I pad an int with leading zeros when using cout << operator?
提问by jamieQ
I want coutto output an int with leading zeros, so the value 1would be printed as 001and the value 25printed as 025. How can I do this?
我想cout，以输出带前导零一个int，因此该值1将被打印成001和值25打印为025。我怎样才能做到这一点？
回答by AraK
With the following,
有了以下，
#include <iomanip>
#include <iostream>
int main()
{
std::cout << std::setfill('0') << std::setw(5) << 25;
}
the output will be
输出将是
00025
setfillis set to the space character (' ') by default. setwsets the width of the field to be printed, and that's it.
setfill' '默认设置为空格字符 ( )。setw设置要打印的字段的宽度，就是这样。
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.
如果您有兴趣了解一般如何格式化输出流，我为另一个问题写了一个答案，希望它有用： 格式化 C++ 控制台输出。
回答by shashwat
Another way to achieve this is using old printf()function of C language
实现此目的的另一种方法是使用printf()C 语言的旧功能
You can use this like
你可以像这样使用
int dd = 1, mm = 9, yy = 1;
printf("%02d - %02d - %04d", mm, dd, yy);
This will print 09 - 01 - 0001on the console.
这将打印09 - 01 - 0001在控制台上。
You can also use another function sprintf()to write formatted output to a string like below:
您还可以使用另一个函数sprintf()将格式化输出写入如下所示的字符串：
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.hheader file in your program for both of these functions
不要忘记stdio.h在程序中包含这两个函数的头文件
Thing to be noted:
需要注意的事项：
You can fill blank space either by 0 or by another char (not number).
If you do write something like %24dformat specifier than this will not fill 2in blank spaces. This will set pad to 24and will fill blank spaces.
您可以用 0 或另一个字符（不是数字）填充空格。
如果您确实编写了%24d格式说明符之类的内容，则不会填充2空格。这将设置 pad24并填充空格。
回答by quest333
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:
这会产生输出：
-12345
****-12345
-12345****
****-12345
-****12345
回答by lyricat
cout.fill( '0' );
cout.width( 3 );
cout << value;
回答by Thomas J Younsi
I would use the following function. I don't like sprintf; it doesn't do what I want!!
我会使用以下功能。我不喜欢sprintf；它没有做我想要的！！
#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 = '#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;
}
';
// 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by user8163172
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
在单个数字值的实例上使用零作为填充字符输出日期和时间的另一个示例：2017-06-04 18:13:02
##代码##