Java中的双重算术和等式 [英] double arithmetic and equality in Java

问题描述

这是一个奇怪的（至少对我来说）。此例程打印真实：

  double x = 11.0; 
 double y = 10.0; 
 if（xy == 1.0）{
 // print true 
} else {
 // print false 
} 
  

但是这个例程打印出错误：

  double x = 1.1; 
 double y = 1.0; 
 if（xy == 0.1）{
 // print true 
} else {
 // print false 
} 
  

任何人都在意解释这里发生了什么？我猜这与 int 的整数运算有关，因为$ code> float s。此外，还有其他基础（除了 10 ），具有此属性？

解决方案

1.0有一个确切的二进制表示。 0.1不是。

也许你在问为什么0.1没有作为尾数存储为1，而指数为-10？但这不是它的工作原理。它不是十进制数加上指数，而是二进制数。所以10次不是一件自然的事情。

抱歉，也许最后一部分不清楚。最好把这个指数看作是一个位的移位。没有位的移位会将无限序列如0.1（十进制）转换成有限的一个。

Here's an oddity (to me, at least). This routine prints true:

double x = 11.0;
double y = 10.0;
if (x-y == 1.0) {
    // print true
} else {
    // print false
}

But this routine prints false:

double x = 1.1;
double y = 1.0;
if (x-y == 0.1) {
    // print true
} else {
    // print false
}

Anyone care to explain what's going on here? I'm guessing it has something to do with integer arithmetic for ints posing as floats. Also, are there other bases (other than 10) that have this property?

解决方案

1.0 has an exact binary representation. 0.1 does not.

perhaps you are asking why 0.1 is not stored as a mantissa of 1 and an exponent of -10? but that's not how it works. it's not a decimal number plus an exponent, but a binary number. so "times 10" is not a natural thing.

sorry, maybe the last part is unclear. it's better to think of the exponent as a shift of bits. no shift of bits will convert an infinite sequence like 0.1 (decimal) into a finite one.

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