Oct 24, 2013 · The equivalent half-open range is [0, n), and rand() % n == rand() % (n-0) + 0. So the lesson is: don't confuse half-open ranges for closed ranges. A second lesson is that this shows another way in which <random> is easier to use than rand() and manually computing your own distributions.

Sep 27, 2011 · The c++ rand() function gives you a number from 0 to RAND_MAX (a constant defined in <cstdlib>), which is at least 32767. (from the c++ documentation) The modulus (%) operator gives the remainder after dividing. When you use it with rand() you are using it to set an upper limit (n) on what the random number can be.

Oct 28, 2015 · Like rand(), rand_r() returns a pseudo-random integer in the range [0, RAND_MAX]. The seedp argument is a pointer to an unsigned int that is used to store state between calls. If rand_r() is called with the same initial value for the integer pointed to by seedp, and that value is not modified between calls, then the same pseudo-random sequence ...

Jul 30, 2009 · Related: How to generate a random int in C?. Here is my answer there, which contains the definition for my int utils_rand(int min, int max) func, which returns a random number using rand() which is in the specific range from min to …

rand and srand are usually implemented as a simple LCG, you can easily write your own (it's few lines of code) without looking for the sources of rand and srand. Notice that, if you need random numbers for "serious" purposes (e.g. cryptography), there are much better RNGs than LCG.

You get the same sequence because rand() is automatically seeded with the a value of 1 if you do not call srand(). Edit. Due to comments. rand() will return a number between 0 and RAND_MAX (defined in the standard library). Using the modulo operator (%) gives the remainder of the division rand() / 100. This will force the random number to be ...

Now, assuming you obtained integers in the range from 0 to RAND_MAX - 1 (inclusively) by using std::rand() % RAND_MAX, the chance of getting a 0 would now be doubled, since it will be the result when std::rand() returns either 0 or RAND_MAX. Any other number will only be obtained when directly returned by std::rand().

No, because RAND_MAX is typically expanded to MAX_INT. So adding one (apparently) puts it at MIN_INT (although it should be undefined behavior as I'm told), hence the reversal of sign. To get what you want you will need to move the +1 outside the computation: r = ((double) rand() / (RAND_MAX)) + 1;

Nov 12, 2017 · np.random.rand is for Uniform distribution (in the half-open interval [0.0, 1.0)) np.random.randn is for Standard Normal (aka. Gaussian) distribution (mean 0 and variance 1) You can visually explore the differences between these two very easily:

Oct 18, 2018 · None of the answers here explains the real reason of being rand() bad. rand() is a pseudo-random number generator (PRNG), but this doesn't mean it must be bad. Actually, there are very good PRNGs, which are statistically hard or …