I'm currently using
1 + (int)(rand() * 999.0 / RAND_MAX)
Your observation that one digit numbers don't occur as often as two and three digit numbers is not surprising.
There are only 9 one digit numbers (not including zero), but there are 90 two-digit ones, and 900 three digit ones. So a uniform random number generator will draw numbers in that frequency.
To generate random numbers in the range [1, 999] such that the probability of their having 1, 2, and 3 digits is equal, use your favourite generator to generate a random number
p, say, in the range [0, 1) (see the new random library functions in C++ to do that), and transform it using
You should note that the resulting distribution will not be piecewise-uniform: that is to say that the probability of drawing a number with a certain number of digits is not the same as the probability of drawing any other number with that number of digits. But it does have a continuous and differentiable cumulative density function which can be important mathematically.
(For the mathematically-inclined, the transformation I'm applying is the quantile function of the distribution that the OP needs).