遇到问题在C四舍五入 [英] Having trouble rounding in c

问题描述

在试图找出如何圆一个浮点数像 1.255 到最近的百分之一，我发现了一些有趣的事情。我用gcc 4.4.5在Debian 6。

While trying to figure out how to round a float like 1.255 to the nearest hundredth, I found something interesting. I'm using gcc 4.4.5 on Debian 6.

int   x = (1.255 * 100) + 0.5;   //  gives me back 125 instead of 126. 
float y = (1.255 * 100) + 0.5;   //  gives me back 126.000000. 

为什么是，当我保存到 INT 我回来 125 ，而不是 126 ？在Fedora当我保存上述前pression到 INT 我回来 126 。这是Debian的一个gcc的bug？任何帮助将大大AP preciated。
谢谢你。

Why is is that when I save to an int I get back 125 and not 126 ? In fedora when I save the above expression to an int I get back 126. Is this a gcc bug in debian ? Any help would be greatly appreciated. Thanks.

推荐答案

虽然这看起来像一个典型的浮点问题，它比这更复杂。

Although this looks like a "typical" floating-point question, it's more complicated than that.

这其中涉及的3件事情的组合：

This one involves a combination of 3 things:

• 的通常浮点再presentation的东西。


• 隐式转换。


• 隐式类型推广。

让我们打破这：

浮点文本的类型双击默认的。因此， 1.255 的类型为双击。

Floating-point literals are of type double by default. Hence 1.255 is of type double.

因此​​，前pression：

Thus the expression:

(1.255 * 100) + 0.5

使用类型做双击。

但由于二进制浮点无法重新present 1.255 正好，前pression计算结果为：

But because binary floating-point can't represent 1.255 exactly, the expression evaluates to:

(1.255 * 100) + 0.5 = 125.99999999999999000000

和类型是双击。

由于这小于 126 ，将其存储到一个整数，将导致 125 。
它存储到浮动将其四舍五入到最接近的浮动，这将导致 126 。

Since this is less than 126, storing it to an integer will result in 125. Storing it to float will round it to the nearest float, which results in 126.

int    x = (1.255 * 100.) + 0.5;
float  y = (1.255 * 100.) + 0.5;
double z = (1.255 * 100.) + 0.5;

cout << fixed;
cout << x << endl;
cout << setprecision(20) << y << endl;
cout << setprecision(20) << z << endl;

输出：

125
126.00000000000000000000
125.99999999999999000000

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