C++中double型数值计算结果为0为什么输出的不是0?

2025-03-16 04:44:05
推荐回答(2个)
回答1:

这是由于double类型的精度引起的。
在C/C++中,float和double类型,被称为浮点型。
浮点型是C/C++中用来存储实数的。而存储的值,并不是实际值,而是近似值。
即,double类型和float类型存储的值均不是准确值,而是一个和准确值接近的值。
这样,在计算中,就会有精度缺失的情况出现,运算越多,这个精度缺失出现的概率越大。
当经过大量计算后,就可能出现结果与真实结果偏差比较大的情况,也就是实际获得的结果并不准确。

回答2:

1、因为double类型都是1.xxxxxxxxx(若干个0和1,二进制)乘以2的若干次幂来表示一个数,所以,和十进制的小数势必不能够一一对应,因为位数有限,总要有一个精度(两个数之间的实数是任意多的,但是很明显double不可能表示任意多的数),那么double的0就只是用一个近似0的数来表示的
2、浮点数比较相等的时候,不能用==比较,要采用if(abs(x-y)<=epsilon)的方式,其实epsilon是预定义的很小的数,也就是要求的精度,比如类似1e-6,也就是说,比较两数之间的差是否在可接受的精度范围内。
3、和0比较,就是if(abs(x)<=epsilon),绝对值小于要求精度,就可以认为是0.

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