How does this expression work: Num*rand() / RAND_MAX?

Question

How does this expression work?

 Num*rand() / RAND_MAX

If, for example, I replace the number with 7.7, it gives me values from 0. to 7. (double).

Have a look at this: http://stackoverflow.com/questions/10219355/does-n-rand-rand-max-make-a-skewed-random-number-distribution — NathanOliver, Mar 31 '15 at 16:55

score 3 · Accepted Answer · answered Mar 31 '15 at 17:06

3

The rand() function generates an integer value between 0 and RAND_MAX. so rand()/RAND_MAX generates real value between 0 and 1. So in the following code we have
0 <= val <= n where n can be an integer or a real number.

double val = n * rand()/RAND_MAX;

answered Mar 31 '15 at 17:06

MrGreen

479
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`n` cannot be an integer. Otherwise you'll get 0 or 1 due to integer division. – erip May 20 '15 at 23:09

score 1 · Answer 2 · answered Mar 31 '15 at 19:46

1

Tagging onto MrGreen's answer, I have found that when n is an integer, I will get an output of 0 when using double val = n * rand()/RAND_MAX;. However, this can be addressed by a slight addition:

double val = (double) n * rand()/RAND_MAX;

answered Mar 31 '15 at 19:46

coder-don

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How does this expression work: Num*rand() / RAND_MAX?

2 Answers2