CyberShot - 1 month ago 14x

C++ Question

So the following code makes 0 < r < 1

`r = ((double) rand() / (RAND_MAX))`

Why does having

`r = ((double) rand() / (RAND_MAX + 1))`

Shouldn't adding one to RAND_MAX make 1 < r < 2?

Edit: I was getting a warning: integer overflow in expression

on that line, so that might be the problem. I just did

`cout << r << endl`

Answer

This is entirely **implementation specific**, but it appears that in the C++ environment you're working in, `RAND_MAX`

is equal to `INT_MAX`

.

Because of this, `RAND_MAX + 1`

exhibits undefined (overflow) behavior, and becomes `INT_MIN`

. While your initial statement was dividing (random # between 0 and `INT_MAX`

)/(`INT_MAX`

) and generating a value `0 <= r < 1`

, now it's dividing (random # between 0 and `INT_MAX`

)/(`INT_MIN`

), generating a value `-1 < r <= 0`

In order to generate a random number `1 <= r < 2`

, you would want

```
r = ((double) rand() / (RAND_MAX)) + 1
```

Source (Stackoverflow)

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