How to cast the size_t to double or int C++

A cast, as Blaz Bratanic suggested:

size_t data = 99999999;
int convertdata = static_cast<int>(data);

is likely to silence the warning (though in principle a compiler can warn about anything it likes, even if there's a cast).

But it doesn't solve the problem that the warning was telling you about, namely that a conversion from size_t to int really could overflow.

If at all possible, design your program so you don't need to convert a size_t value to int. Just store it in a size_t variable (as you've already done) and use that.

Converting to double will not cause an overflow, but it could result in a loss of precision for a very large size_t value. Again, it doesn't make a lot of sense to convert a size_t to a double; you're still better off keeping the value in a size_t variable.

(R Sahu's answer has some suggestions if you can't avoid the cast, such as throwing an exception on overflow.)

If your code is prepared to deal with overflow errors, you can throw an exception if data is too large.

size_t data = 99999999;
if ( data > INT_MAX )
{
   throw std::overflow_error("data is larger than INT_MAX");
}
int convertData = static_cast<int>(data);