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Home Backend Development PHP Tutorial 较长数字 字符串 的比较问题

较长数字 字符串 的比较问题

Jun 23, 2016 pm 02:39 PM

本帖最后由 anyilaoliu 于 2013-12-13 15:04:43 编辑

如题 
遇到这样的问题

$a = "222111565652521142";		$b = "222111565652521139";		$c = 222111565652521142;		$d = 222111565652521139;		var_dump($a);		echo "<br/>";		var_dump($b);		echo "<br/>";		var_dump($c);		echo "<br/>";		var_dump($d);		echo "<br/>";		if ($a == $b) {			echo 3;		}else{			echo 31;		}		echo "<br/>";		if ($c == $d) {			echo 4;		}else{			echo 41;		}		echo "<br/>";
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结果是


string '222111565652521142' (length=18)string '222111565652521139' (length=18)float 2.2211156565252E+17float 2.2211156565252E+17314
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问题来了,第二种情况 如何进行两者的比较

问题二: 从数据库中读出来的这两组数字,var_dump后显示是string(18)  但是比较结果和上边的例子恰好相反,两个string的 比较相等 也通过了.

求解

回复讨论(解决方案)

长度超过11位的整数请按字符串处理
浮点数自身的特征使其无法在精度要求较高的场合中使用

长度超过11位的整数请按字符串处理
浮点数自身的特征使其无法在精度要求较高的场合中使用

这一点学习了

另外,今天又再次尝试了下发现我的环境下,两个string格式的18位长度的阿拉伯数字,  ==运算符比较依然会忽略后两位, 应该是内部进行了数字转换

而在另一个环境下  则没有出现这个问题,两个字符串可以正确的按照字符串来使用 == 运算符

求教大约是什么设置或者扩展导致这个现象的发生.

我们使用的php环境大多是32位的,而在32位环境中一个长整型数是用4个字节表示的
即 -pow(2,31) 到 pow(2,31) 之间
pow(2,31) = 2147483648
pow(2,32) = 4294967296
这显然不能满足一般的需要,所以 php 在 logn 类型的基础上做了一些扩充。用了6个字节存储1个整型数
也就是可表示 -99999999999999 到 99999999999999 之间的正数,即 14 位整数

而在64位的php中就可轻易的突破这一限制了,因为64位的环境中,一个logn类型本身就是8个字节


用===不会转换

我们使用的php环境大多是32位的,而在32位环境中一个长整型数是用4个字节表示的
即 -pow(2,31) 到 pow(2,31) 之间
pow(2,31) = 2147483648
pow(2,32) = 4294967296
这显然不能满足一般的需要,所以 php 在 logn 类型的基础上做了一些扩充。用了6个字节存储1个整型数
也就是可表示 -99999999999999 到 99999999999999 之间的正数,即 14 位整数

而在64位的php中就可轻易的突破这一限制了,因为64位的环境中,一个logn类型本身就是8个字节
非常感谢xu大   解惑很到位,之前我为什么没有想到64位和32位的问题呢... 当然,这也跟我没有细看过PHP本身的代码和文档有关系

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