如何在Objective-C中创建随机浮点数？

I'm trying to create a random float between 0.15 and 0.3 in Objective-C. The following code always returns 1:

我正在尝试在Objective-C中创建0.15到0.3之间的随机浮点数。以下代码始终返回1：

int randn = (random() % 15)+15;
float pscale = (float)randn / 100;

What am I doing wrong?

我究竟做错了什么？

8 个解决方案

#1

Try this:

尝试这个：

 (float)rand() / RAND_MAX

Or to get one between 0 and 5:

或者在0到5之间得到一个：

 float randomNum = ((float)rand() / RAND_MAX) * 5;

Several ways to do the same thing.

有几种方法可以做同样的事情。

#2

Here is a function

这是一个功能

- (float)randomFloatBetween:(float)smallNumber and:(float)bigNumber {
float diff = bigNumber - smallNumber;
return (((float) (arc4random() % ((unsigned)RAND_MAX + 1)) / RAND_MAX) * diff) + smallNumber;
}

#3

use arc4random() or seed your random values
使用arc4random（）或种子随机值

try

尝试

float pscale = ((float)randn) / 100.0f;

#4

Your code works for me, it produces a random number between 0.15 and 0.3 (provided I seed with srandom()). Have you called srandom() before the first call to random()? You will need to provide srandom() with some entropic value (a lot of people just use srandom(time(NULL))).

你的代码适用于我，它产生一个0.15到0.3之间的随机数（假设我用srandom（）种子）。你是否在第一次调用random（）之前调用了srandom（）？你需要为srandom（）提供一些熵值（很多人只使用srandom（time（NULL）））。

For more serious random number generation, have a look into arc4random, which is used for cryptographic purposes. This random number function also returns an integer type, so you will still need to cast the result to a floating point type.

对于更严重的随机数生成，请查看arc4random，它用于加密目的。此随机数函数也返回整数类型，因此您仍需要将结果转换为浮点类型。

#5

Easiest.

最容易的。

+ (float)randomNumberBetween:(float)min maxNumber:(float)max
{
    return min + arc4random_uniform(max - min + 1);
}

#6

Using srandom() and rand() is unsafe when you need true randomizing with some float salt.

当你需要使用一些浮点盐进行真正的随机化时，使用srandom（）和rand（）是不安全的。

On MAC_10_7, IPHONE_4_3 and higher you can use arc4random_uniform(upper_bound)*. It allows to generate true random integer from zero to *upper_bound*.

在MAC_10_7，IPHONE_4_3及更高版本上，您可以使用arc4random_uniform（upper_bound）*。它允许生成从零到* upper_bound *的真随机整数。

So you can try the following

所以你可以尝试以下方法

u_int32_t upper_bound = <some big enough integer>;

float r = 0.3 * (0.5 + arc4random_uniform(upper_bound)*1.0/upper_bound/2);

#7

To add to @Caladain's answer, if you want the solution to be as easy to use as rand(), you can define these:

要添加到@ Caladain的答案，如果您希望解决方案与rand（）一样易于使用，您可以定义以下内容：

#define randf() ((CGFloat)rand() / RAND_MAX)
#define randf_scaled(scale) (((CGFloat)rand() / RAND_MAX) * scale)

Feel free to replace CGFloat with double if you don't have access to CoreGraphics.

如果您无法访问CoreGraphics，请随意用双倍替换CGFloat。

#8

-5

I ended up generating to integers one for the actual integer and then an integer for the decimal. Then I join them in a string then I parse it to a floatvalue with the "floatValue" function... I couldn't find a better way and this works for my intentions, hope it helps :)

我最终生成一个整数，用于实际整数，然后生成小数的整数。然后我将它们连接成一个字符串，然后我用“floatValue”函数将它解析为floatvalue ...我找不到更好的方法，这对我的意图有效，希望它有帮助:)

int integervalue = arc4random() % 2;
int decimalvalue = arc4random() % 9;   
NSString *floatString = [NSString stringWithFormat:@"%d.%d",integervalue,decimalvalue];
float randomFloat = [floatString floatValue];

#1