c# - Dividing by Random.next always results in 0? -

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this 1 puzzling me. writing damage algorithm random variation. when calculate variation looks like.

random random = new random(); double variation = random.next(85, 115) / 100; double damage = restofalgorithm * variation;

when that, variation outputs 0. however, if below, output expected result.

random random = new random(); double variation = random.next(85, 115); double damage = restofalgorithm * (variation / 100);

why happen?

divide double:

double variation = random.next(85, 115) / 100.0;

or

double variation = random.next(85, 115) / (double)100;

otherwise you'll doing integer arithmetic (since random.next returns integer , 100 integer).

i consider best practice know types working , cast type desired. more necessary, compiler implicitly convert values. explicit casts intentions visible looking @ code later.

double variation = (double)random.next(85, 115) / 100d;

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