如何在另一种语言的随机函数中复制C ++ rand？

Question

I am using an internal language in my company and they only have Random() that returns a float between 0 and 1.

I need to port a piece of C++ code that use rand() :

int b = rand() % (i+1);

I looked at the docs, but not sure how I can use my Random() to generate a number between 0 and i+1 which is what the above code does, right?

I tried multiplying i+1 with Random() but didn't get the same results, that's why I am not sure if what I am doing is correct.

I expect difference between the results due to different random functions, but still I want to be sure I am translating it correctly.

Answer 1

You need to multiply Random() with i and not i+1 .

C++ rand() returns an integer between 0 and RAND_MAX but your Random() returns a float between 0 and 1 , so multiplying the output of Random() with i and taking the integer portion of the result will give you an integer in [0,i] which is what rand() %(i+1) gives.

Answer 2

Rand() give you a number between 0 and RAND_MAX, which after applying the mod operator you end up with a number between 0 and i (including i).

To do the same with Random() you'll need to multiply by ( i+1 ), then take the floor of that (round down):

b = floor( Random() * (i+1) ) 

This will give you a number from 0 to i (including the fence posts) as required.