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Asked: 2026-05-31T23:37:01+00:00

So the following code makes 0 < r < 1 r = ((double) rand()

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So the following code makes 0 < r < 1 

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

Why does having r = ((double) rand() / (RAND_MAX + 1)) make -1 < r < 0?

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 and it definitely gives me values between -1 and 0

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1 Answer

  1. Editorial Team

    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
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