随笔 - 1685  文章 - 623 评论 - 449 trackbacks - 0

故事背景

《搜神记》:

干将莫邪的故事--java比较操作注意要点

 

楚干将、莫邪为楚王作剑,三年乃成。王怒,欲杀之。剑有雌雄。其妻重身当产。夫语妻曰:“吾为王作剑,三年乃成。王怒,往必杀我。汝若生子是男,大,告之曰:‘出户望南山,松生石上,剑在其背。’”于是即将雌剑往见楚王。王大怒,使相之:“剑有二,一雄一雌,雌来雄不来。”王怒,即杀之。

莫邪子名赤,比后壮,乃问其母曰:“吾父所在?”母曰:“汝父为楚王作剑,三年乃成。王怒杀之。去时嘱我:‘语汝子,出户望南山,松生石上,剑在其背。’”于是子出户南望,不见有山,但睹(dû)堂前松柱下石低之上。即以斧破其背,得剑,日夜思欲报楚王。

王梦见一儿,眉间广尺,言欲报仇。王即购之千金。儿闻之,亡去,入山行歌。客有逢者,谓:“子年少,何哭之甚悲耶(yé)?”曰:“吾干将、莫邪子也,楚王杀吾父,吾欲报之!”客曰:“闻王购子头千金,将子头与剑来,为子报之。”儿曰:“幸甚!”即自刎(wên),两手捧头及剑奉之,立僵。客曰:“不负子也。”于是尸乃仆。

客持头往见楚王,王大喜。客曰:“此乃勇士头也,当于汤镬(huò)煮之。”王如其言。煮头三日三夕,不烂,头踔出汤中,瞋目大怒。客曰:“此儿头不烂,愿王自往临视之,是必烂也。”王即临之。客以剑拟王,王头随坠汤中,客亦自拟己头,头复坠汤中。三首俱烂,不可识辨。乃分其汤肉葬之,故通名“三王墓”,今在汝南北宜春县界。

java中的雌雄双剑0与-0

干将莫邪的故事--java比较操作注意要点

 

0与-0按理说都是一样的,但在程序中不一定就相等哦。请看下面的程序:

    public static void main(String[] args) {
        System.out.println(Float.compare(-0.0f, 0.0f));
        System.out.println(Math.max(-0.0f, 0.0f));
        System.out.println(Math.min(-0.0f, 0.0));
        System.out.println(-0.0f<0.0f);
        System.out.println(-0.0f<=0.0f);
    }

上面的结果可能让你大吃一惊。

-1
0.0
-0.0
false
true

让我们看看JSL上怎么说吧

15.20.1. Numerical Comparison Operators <, <=, >, and >=
The type of each of the operands of a numerical comparison operator must be a type that is convertible (§5.1.8) to a primitive numeric type, or a compile-time error occurs.
Binary numeric promotion is performed on the operands (§5.6.2).
Note that binary numeric promotion performs value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).
If the promoted type of the operands is int or long, then signed integer comparison is performed.
If the promoted type is float or double, then floating-point comparison is performed.
Comparison is carried out accurately on floating-point values, no matter what value sets their representing values were drawn from.
The result of a floating-point comparison, as determined by the specification of the IEEE 754 standard, is:
If either operand is NaN, then the result is false.
All values other than NaN are ordered, with negative infinity less than all finite values, and positive infinity greater than all finite values.
Positive zero and negative zero are considered equal.
For example, -0.0<0.0 is false, but -0.0<=0.0 is true.
Note, however, that the methods Math.min and Math.max treat negative zero as being strictly smaller than positive zero.
Subject to these considerations for floating-point numbers, the following rules then hold for integer operands or for floating-point operands other than NaN:
The value produced by the < operator is true if the value of the left-hand operand is less than the value of the right-hand operand, and otherwise is false.
The value produced by the <= operator is true if the value of the left-hand operand is less than or equal to the value of the right-hand operand, and otherwise is false.
The value produced by the > operator is true if the value of the left-hand operand is greater than the value of the right-hand operand, and otherwise is false.
The value produced by the >= operator is true if the value of the left-hand operand is greater than or equal to the value of the right-hand operand, and otherwise is false.
干将莫邪的故事--java比较操作注意要点

 

Positive zero and negative zero are considered equal.(浮点数中0.0和-0.0进行关系比较是相等的)

For example, -0.0<0.0 is false, but -0.0<=0.0 is true.(例如:-0.0<0.0是false,但-0.0<=0.0 是true)

Note, however, that the methods Math.min and Math.max treat negative zero as being strictly smaller than positive zero.(主意:然而,使用Math.min和Math.max进行浮点数比较时,-0.0时小于0.0的)

java中浮点数在循环中使用要慎重

    public static void main(String[] args) {
        final int START = 2000000000;
        int count = 0;
        for (float f = START; f < START + 50; f++)
        count++;
        System.out.println(count);
        }

f的初始值接近于Integer.MAX_VALUE,因此它需要用31 位来精确表示,而float 类型只能提供24 位的精度。对如此巨大的一个float 数值进行增量操作将不会改变其值。因此,这个程序看起来应该无限地循环下去,因为f 永远也不可能解决其终止值。但是,如果你运行该程序,就会发现它并没有无限循环下去,事实上,它立即就终止了,并打印出0。怎么回事呢?

问题在于终止条件测试失败了,其方式与增量操作失败的方式非常相似。这个循环只有在循环索引f 比(float)(START + 50)小的情况下才运行。在将一个int与一个float 进行比较时,会自动执行从int 到float 的提升[JLS 15.20.1]。遗憾的是,这种提升是会导致精度丢失的三种拓宽原始类型转换的一种[JLS5.1.2]。(另外两个是从long 到float 和从long 到double。)

f 的初始值太大了,以至于在对其加上50,然后将结果转型为float 时,所产生的数值等于直接将f 转换成float 的数值。换句话说,(float)2000000000 ==2000000050,因此表达式f < START + 50 即使是在循环体第一次执行之前就是false,所以,循环体也就永远的不到机会去运行。

总结

数据比较的时候,尽量避开浮点数,使用int,long等。

参考资料

【1】https://baike.baidu.com/item/%E5%B9%B2%E5%B0%86%E8%8E%AB%E9%82%AA/550371?fr=aladdin

【2】https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.20.1

posted on 2019-10-15 16:55  一天不进步,就是退步  阅读(...)  评论(...编辑  收藏