((float)rand()/(float)(((1<< 31)-1))的含义 [英] Meaning of ((float) rand() / (float)((1 << 31) - 1))

问题描述

我试图理解一个C语言程序，其中包括带有.h文件的一行

I'm trying to understand a C program which includes a .h file with the line

#define random                  ((float) rand() / (float)((1 << 31) - 1))

C程序还包含<math.h>.

我的猜测是，这只是从间隔[0,1]上的均匀分布中产生一个随机数；这是正确的吗?

My guess is that this simply produces a random number from a uniform distribution on the interval [0,1]; is this correct?

推荐答案

表面上是.但这在两个主要方面是错误的:

Ostensibly yes. But it's wrong in two principal ways:

1. 改为使用RAND_MAX.那就是它的目的.它可能比1 << 31 - 1小得多.

1. Use RAND_MAX instead. That's what it's there for. It might be much smaller than 1 << 31 - 1.

1 << 31将在 32位 int或更少的平台上为您提供未定义的行为，这非常普遍.不要那样做！

1 << 31 will give you undefined behaviour on a platform with a 32 bit int or less, which is remarkably common. Don't do that!

请注意，如果您不想恢复值1(通常是这样)，请在分母上使用RAND_MAX + 1.0. 1.0强制以浮点形式求值:如果编写RAND_MAX + 1，则冒着整数类型溢出的风险.

Note that if you don't want to ever recover the value 1, (as is often the case), then use RAND_MAX + 1.0 on the denominator. The 1.0 forces evaluation in floating point: you run the risk of overflowing an integral type if you write RAND_MAX + 1.

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