花车是不精确

本文介绍了花车是不精确的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我百思不得其解。我也没有解释为什么本次测试使用双击数据类型时通过，但使用浮动数据类型时失败。考虑code的片断。

I am puzzled. I have no explanation to why this test passes when using the double data type but fails when using the float data type. Consider the following snippet of code.

float total = 0.00; for ( int i = 0; i < 100; i++ ) total += 0.01;

照理总为1.00，但它是等于0.99。为什么会出现这样的情况？我用GCC和铿锵编译，编译器都具有相同的结果。

One would anticipate total to be 1.00, however it is equal to 0.99. Why is this the case? I compiled with both GCC and clang, both compilers have the same result.

推荐答案

在十进制0.01的值是pssed作为该系列的前$ P $：A1 *（1/2）+ A2 *（1/2）^ 2 + A3 *（1/2）^ 4 +等，其中在是0或1。

The value for 0.01 in decimal is expressed as the series: a1*(1/2) + a2*(1/2)^2 + a3*(1/2)^4 + etc. where aN is a zero or one.

我让你找出A1，A2和多少小数位（在）的要求的具体数值。在某些情况下，小数部分不能被重新用有限系列（1/2）^ n个值的psented $ P $

I leave it to you to figure out the specific values of a1, a2 and how many fractional bits (aN) are required. In some cases a decimal fraction cannot be represented by a finite series of (1/2)^n values.

有关这一系列来总结，以小数0.01需要一个超越（存储在一个浮的位数比特的全字减去比特为一个数签署和指数）。但由于双有更多的位，然后0.01小数可以/也许/也许（你们做计算）是precisely定义。

For this series to sum to 0.01 in decimal requires that aN go beyond the number of bits stored in a float (full word of bits minus the number of bits for a sign and exponent). But since double has more bits then 0.01 decimal can/might/maybe (you do the calculation) be precisely defined.