bcadd

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

bcadd — 2つの任意精度の数値を加算する

説明

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

num1 を num2 に加算します。

パラメータ

num1: 左オペランドを表す文字列。
num2: 右オペランドを表す文字列。
scale: This parameter is used to set the number of digits after the decimal place in the result. If null, it will default to the default scale set with bcscale(), or fallback to the value of the bcmath.scale INI directive.

戻り値

二つの数の和を文字列で返します。

エラー / 例外

This function throws a ValueError in the following cases:

num1 or num2 is not a well-formed BCMath numeric string.
scale is outside the valid range.

変更履歴

バージョン	説明
8.0.0	`scale` は nullable になりました。

例

例1 bcadd() の例

<?php

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

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

?>

参考

bcsub() - 任意精度数値の減算を行う

Found A Problem?

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