Defining '999e999' value without using char type in C++

Is it possible to define 999e999 value without using the char type?

No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long type in c++.

A long double type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.

What can be achieved with your current CPU architecture can be explored using the std::numeric_limits facilities like this:

#include <iostream>
#include <limits>

int main() {   
   std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10  << std::endl;
}

Output:

max_exponent10: 4932

See the online demo

You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.

In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long or long double (Though some may just have it be floating point infinity).

You may instead be interested in std::numeric_limits<T>::infinity() (for float, double or long double) or std::numeric_limits<T>::max() instead.

I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.

Of course it does. A long long is typically 64 bits long, which gives you log(2^64) ≅ 19 decimal digits of precision. 999e999 ≅ (10^3)^1000, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999 isn't just too big for a long long, it's too big by an enormous margin.

Is it possible to define 999e999 value without using the char type?

Sure. You could define an integer-like type based on an array of some sort of integers, like long long. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float and double are useful.