bcadd()
(PHP 4, PHP 5, PHP 7)
2个任意精度数字的加法计算
说明
bcadd(string $left_operand,string $right_operand[,int $scale]): string
左操作数
和右操作数
求和
参数
- $left_operand
左操作数,字符串类型
- $right_operand
右操作数,字符串类型
- $scale
此可选参数用于设置结果中小数点后的小数位数。也可通过使用bcscale()来设置全局默认的小数位数,用于所有函数。
返回值
2个操作数求和之后的结果以字符串返回
范例
bcadd()示例
<?php $a = '1.234'; $b = '5'; echo bcadd($a, $b); // 6 echo bcadd($a, $b, 4); // 6.2340 ?>
参见
bcsub()
2个任意精度数字的减法
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.
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