Uniform int32 Distribution

Question

Is there any way to get uniform int32_t distribution without a warning? I use this uniform_int_distribution<int32_t> in my code but I get a warning:

54988961.cpp: In function ‘int main()’:
54988961.cpp:6:64: warning: overflow in conversion from ‘double’ to ‘int’ changes value from ‘1.0e+10’ to ‘2147483647’ [-Woverflow]
     std::uniform_int_distribution<std::int32_t> unif(1,std::pow(10,10));
                                                        ~~~~~~~~^~~~~~~

This is exactly my code:

#include <cmath>
#include <cstdint>
#include <random>

int main() {
    std::uniform_int_distribution<std::int32_t> unif(1,std::pow(10,10));
}

Answer 1

pow(10, 10)

This is 10000000000 , an int32 can only hold 2147483647 ( 2^31 - 1 ). You should use int64_t if you want to be able to store your pow(10, 10) .

Since your min value is 1 you could also just go for its unsigned counterpart.

Answer 2

The use of a large double value ( pow() ) for an integer argument causes this warning in the constructor of uniform_int_distribution .

Use an int constant instead. If you need a range that doesn't fit in an int32_t then use an int64_t template argument.