printf()

(PHP 4, PHP 5, PHP 7)

输出格式化字符串

说明

依据$format格式参数产生输出。

参数

$format描述信息，请参见sprintf()。

返回值

返回输出字符串的长度。

参见

• print- 输出字符串
• sprintf()Return a formatted string
• vprintf() 输出格式化字符串
• sscanf()根据指定格式解析输入的字符
• fscanf()从文件中格式化输入
• flush()刷新输出缓冲

Be careful:
printf ("(9.95 * 100) = %d \n", (9.95 * 100));
'994'
First %d converts a float to an int by truncation.
Second floats are notorious for tiny little rounding errors.

[Editor's Note: Or just use vprintf...]
If you want to do something like  (this doesn't work) instead of  you can use this function:

Use it the following way:

and it will print:
There is a difference between good and evil

instead of writing a function to round off a float (let's call it 'x') accurately, it's much easier to add a small number to x and then truncate it...
For example: if you want to round off to the nearest integer, just add 0.5 to x and then truncate it. if x=12.6, then it would calculate 13.1, and truncate it to 13. If x=14.4, it would calculate 14.9 and truncate it to 14.

You can use this function to format the decimal places in a number:
$num = 2.12;
printf("%.1f",$num);
prints:
2.1
see also: number_format()

Be sure that the output channel is available to write on before executing printf()! Some functions in classes available from various sources (in my case, DB_Sql::query() in PHPLIB) assume that printing will work, even after the default output stream has been closed. 
For me the issue was most notable in PHP4 session management when I was creating my own sess_write() handler. Since I was unable to find any function that checks for the output stream that printf() uses, I just had to drop the crazy use of the printf(). (does some function already exist to check for the presence of an output stream?)

Be careful when relying on typecasting with printf(). For example,
  printf("%d", "17,999") 
returns "17".

If anyone is looking for writing a quine using printf(),
this is my example:

This also helps those who are new to printf() see one way of using the 'mixed args' part, rather than just a single argument as in most examples I've seen.

//If you want to make many Hidden fields you can use the function 
//You can pass the values as array value,This will help you, when you are going to post many hiddend fields:-
function MakeHidden($ArrValues)
  {
    global $dearvar;
    echo $dearvar; 
    if(is_array($ArrValues)){
       foreach($ArrValues as $key=>$values)
       {
         echo $MakeHTML = "";
        }
   }  
   
   
   else {
      echo $MakeHTML = "";
     }
  return $MakeHTML;
}
//example:-
MakeHidden(array("value1","value2","value3"));
//OutPut :- 

copypasted from msdn
A format specification, which consists of optional and required fields, has the following form:
%[flags] [width] [.precision] [{h  |  l  |  I64  |  L}]type
Each field of the format specification is a single character or a number signifying a particular format option. The simplest format specification contains only the percent sign and a type character (for example, %s). If a percent sign is followed by a character that has no meaning as a format field, the character is copied to stdout. For example, to print a percent-sign character, use %%.
The optional fields, which appear before the type character, control other aspects of the formatting, as follows: 
type 
Required character that determines whether the associated argument is interpreted as a character, a string, or a number (see the printf Type Field Characters table. 
flags 
Optional character or characters that control justification of output and printing of signs, blanks, decimal points, and octal and hexadecimal prefixes (see the Flag Characters table). More than one flag can appear in a format specification. 
width 
Optional number that specifies the minimum number of characters output (see printf Width Specification). 
precision 
Optional number that specifies the maximum number of characters printed for all or part of the output field, or the minimum number of digits printed for integer values (see the How Precision Values Affect Type table). 
h  |  l  |  I64  |  L 
Optional prefixes to type-that specify the size of argument (see the Size Prefixes for printf and wprintf Format-Type Specifiers table).

Why rounding is not same as for round()?
Try this code:

Result:
1.03
1.04
In my opion it should be:
1.04
1.04
Why is that?

If your missing features such as "-"*100 to print a single character multiple times you can use the slightly longer and less readable PHP equivalent printf("%'-100s",""); and sprint("%'-100s","").

be careful with integers, they cant hold large values.
printf("%d",10023123553.45634663);
will print out: 1433188961 
and
printf("%.0f",10023123553.45634663);
will print out: 10023123553

Several people have commented about problems with aligning numbers. I just wanted to add a bit of clarification.
According to the spec all True Type Fonts (especially porpotional spaced fonts) use a fixed width for numeric digits. All digits have the same width which is equal to the width of "0".
Where things go haywire when trying to align numbers is that the space character does not have the same width as a digit when using a porportional spaced font.
Therefore if you want to line up a column of numbers, you can not use leading spaces to position them unless you use the same quantity of space characters for each row. e.g. your numbers have leading zeros etc.
The simplest solution is to switch to a monospaced font. Alternativly you can use positioning tags.
Summary: Your numbers will always line up properly regardless of the font used, if you get the starting position to be consistant.

I don't know if this is useful to anyone, but here goes! Example for using the printf function to output an object.
class person
{
  var $name = "";
  function name($newname = NULL)
  {
    if(! is_null($newname))
    {
      $this->name=$newname;
    }
    return $this->name;
  }
  var $surname = "";
  function surname($newsurname = NULL)
  {
    if(! is_null($newsurname))
    {
      $this->surname=$newsurname;
    }
    return $this->surname;
  }
  var $age = "";
  function age($newage = NULL)
  {
    if(! is_null($newage))
    {
      $this->age=$newage;
    }
    return $this->age;
  }
}
$bob = new person;
$bob->name('Bob');
$bob->surname('Builder');
$bob->age('50');
printf("Hi %s, your surname is %s and you are %s years old", $bob->name, $bob->surname, $bob->age);
Outputs:
Hi Bob, your surname is Builder and you are 50 years old

In response to chris dot breen at accurate dot com :
This will work, but i didn't try:

Your problem came from the fact that in the string "\$s", the $ is simply escaped by the \, but but you must not do that with single quoted strings. Try to echo your strings to test it, or just go to http://www.php.net/manual/en/language.types.string.php

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