【分享】PHP int 超大溢出整数的 加减运算函数，如果有更好的方法欢迎探讨
分享一个溢出整数加减的运算函数，刚刚写的，对于溢出的整数可以用这个来进行加减运算。
遗憾的几点是：
<br> 一代码太多；<br> 二只有加减运算，乘除取余都没有；<br>
其实还有一个更简便的方式就是用SQL数据库的：SELECT n1+n2;
<br> mysql> SELECT 11234123413241341234123412341234+1;<br> +------------------------------------+<br> | 11234123413241341234123412341234+1 |<br> +------------------------------------+<br> | 11234123413241341234123412341235 |<br> +------------------------------------+<br> 1 row in set (0.00 sec)<br> <br> <br> mysql> SELECT 11234123413241341234123412341234*12341234123;<br> +----------------------------------------------+<br> | 11234123413241341234123412341234*12341234123 |<br> +----------------------------------------------+<br> | 138642947209487270472850788378836360727782 |<br> +----------------------------------------------+<br> 1 row in set (0.00 sec)<br> <br> <br>
如果有更好的方法，请随时回帖或者发个信息给我。欢迎探讨。
<br>
/* big int operate [by fuzb 20130826] */<br>
function bigintO($num1,$op,$num2)<br>
{<br>
$arr = array();<br>
$endop = '';<br>
$num1o = $num1;<br>
$num2o = $num2;<br>
if($num1 < 0)<br>
{<br>
$c1 = -1;<br>
$num1 = preg_replace('/^(-)/','',$num1);<br>
<br>
} else {<br>
$c1 = 1;<br>
}<br>
<br>
if($num2 < 0)<br>
{<br>
$c2 = -1;<br>
$num2 = preg_replace('/^(-)/','',$num2);<br>
} else {<br>
$c2 = 1;<br>
}<br>
<br>
$len1 = strlen($num1);<br>
$len2 = strlen($num2);<br>
$len = max(strlen($num1),strlen($num2));<br>
if($len1 < $len) $num1 = str_pad('0',$len - $len1).$num1;<br>
if($len2 < $len) $num2 = str_pad('0',$len - $len2).$num2;<br>
<br>
<br>
if($op == '+')<br>
{<br>
if($c1 == $c2)<br>
{ <div class="clear">
</div>
