bcadd

(PHP 4, PHP 5, PHP 7, PHP 8)

bcadd — Add two arbitrary precision numbers

Опис

bcadd(string $num1, string $num2, ?int $scale = null): string

Sums num1 and num2.

Параметри

num1: The left operand, as a string.
num2: The right operand, as a string.
scale: Необов'язковий параметр, яким вказується кількість знаків після крапки в результаті. Якщо опущено, застосовується межа, що встановлена глобально функцією bcscale(), або ж 0, якщо межа ще не встановлена.

Значення, що повертаються

The sum of the two operands, as a string.

Журнал змін

Версія	Опис
8.0.0	`scale` is now nullable.

Приклади

Приклад #1 bcadd() example

<?php

$a = '1.234';
$b = '5';

echo bcadd($a, $b);     // 6
echo bcadd($a, $b, 4);  // 6.2340

?>

Прогляньте також

bcsub() - Subtract one arbitrary precision number from another

Improve This Page

Learn How To Improve This Page • Submit a Pull Request • Report a Bug

＋add a note

User Contributed Notes 2 notes

down

Nitrogen ¶

15 years ago

I made this to add an unlimited size of numbers together..



This could be useful for those without the BCMath extension.



It allows decimals, and optional $Scale parameter.  If $Scale isn't specified, then it'll automatically adjust to show the correct number of decimals.



<?php



function Add($Num1,$Num2,$Scale=null) {

  // check if they're valid positive numbers, extract the whole numbers and decimals

  if(!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num1,$Tmp1)||

     !preg_match("/^\+?(\d+)(\.\d+)?$/",$Num2,$Tmp2)) return('0');



  // this is where the result is stored

  $Output=array();



  // remove ending zeroes from decimals and remove point

  $Dec1=isset($Tmp1[2])?rtrim(substr($Tmp1[2],1),'0'):'';

  $Dec2=isset($Tmp2[2])?rtrim(substr($Tmp2[2],1),'0'):'';



  // calculate the longest length of decimals

  $DLen=max(strlen($Dec1),strlen($Dec2));



  // if $Scale is null, automatically set it to the amount of decimal places for accuracy

  if($Scale==null) $Scale=$DLen;



  // remove leading zeroes and reverse the whole numbers, then append padded decimals on the end

  $Num1=strrev(ltrim($Tmp1[1],'0').str_pad($Dec1,$DLen,'0'));

  $Num2=strrev(ltrim($Tmp2[1],'0').str_pad($Dec2,$DLen,'0'));



  // calculate the longest length we need to process

  $MLen=max(strlen($Num1),strlen($Num2));



  // pad the two numbers so they are of equal length (both equal to $MLen)

  $Num1=str_pad($Num1,$MLen,'0');

  $Num2=str_pad($Num2,$MLen,'0');



  // process each digit, keep the ones, carry the tens (remainders)

  for($i=0;$i<$MLen;$i++) {

    $Sum=((int)$Num1{$i}+(int)$Num2{$i});

    if(isset($Output[$i])) $Sum+=$Output[$i];

    $Output[$i]=$Sum%10;

    if($Sum>9) $Output[$i+1]=1;

  }



  // convert the array to string and reverse it

  $Output=strrev(implode($Output));



  // substring the decimal digits from the result, pad if necessary (if $Scale > amount of actual decimals)

  // next, since actual zero values can cause a problem with the substring values, if so, just simply give '0'

  // next, append the decimal value, if $Scale is defined, and return result

  $Decimal=str_pad(substr($Output,-$DLen,$Scale),$Scale,'0');

  $Output=(($MLen-$DLen<1)?'0':substr($Output,0,-$DLen));

  $Output.=(($Scale>0)?".{$Decimal}":'');

  return($Output);

}



$A="5650175242.508133742";

$B="308437806.831153821478770";



printf("  Add(%s,%s);\r\n// %s\r\n\r\n",$A,$B,  Add($A,$B));

printf("BCAdd(%s,%s);\r\n// %s\r\n\r\n",$A,$B,BCAdd($A,$B));



/*

  This will produce the following..

    Add(5650175242.508133742,308437806.831153821478770);

  // 5958613049.33928756347877



  BCAdd(5650175242.508133742,308437806.831153821478770);

  // 5958613049

*/



?>



It was a fun experience making, and thought I'd share it.

Enjoy,

Nitrogen.

down

Bo Anders Svensson ¶

19 years ago

Be aware:

$exp1 = "1E5";
$exp2 = "2E4";

$ans1 = bcadd((float)$exp1, (float)$exp2, 3);
$ans2 = bcadd((int)$exp1, (int)$exp2, 3);
$ans3 = bcadd($exp1, $exp2, 3);

echo "1: $exp1 + $exp2 = $ans1\r\n";
echo "2: $exp1 + $exp2 = $ans2\r\n";
echo "3: $exp1 + $exp2 = $ans3\r\n";

1: 1E5 + 2E4 = 120000.000
2: 1E5 + 2E4 = 3.000
3: 1E5 + 2E4 = 0.000

＋add a note