bcadd

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

bcadd — Añade dos números de precisión arbitrária

Descripción

bcadd(string $left_operand, string $right_operand, int $scale = 0): string

Suma operando_izquierda y operando_derecha.

Parámetros

left_operand: El left_operand, como una cadena
right_operand: El right_operand, como una cadena
scale: Este parámetro opcional es usado para establecer el número de dígitos después del punto decimal en el resultado. Si se omite, su valor será el de la escala establecida globalmente con la función bcscale(), o se recurrirá a 0 si este valor no ha sido establecido.

Valores devueltos

La suma de dos operandos, como una cadena.

Ejemplos

Ejemplo #1 Ejemplo bcadd()

<?php

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

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

?>

Ver también

bcsub() - Resta un número de precisión arbitraria de otro

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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

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