Question

Assume my system as 32 bit machine. Considering this if I use long int for n>63 I will get my value as 0. How to solve it?

Answer 1

double is perfectly capable of storing powers of two up to 1023 exactly. Don’t let someone tell you that floating point numbers are somehow always inexact. This is a special case where they aren’t!

double x = 1.0;
for (int n = 0; n <= 200; ++n)
{
    printf("2^%d = %.0f\n", n, x);
    x *= 2.0;
}

Some output of the program:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
...
2^196 = 100433627766186892221372630771322662657637687111424552206336
2^197 = 200867255532373784442745261542645325315275374222849104412672
2^198 = 401734511064747568885490523085290650630550748445698208825344
2^199 = 803469022129495137770981046170581301261101496891396417650688
2^200 = 1606938044258990275541962092341162602522202993782792835301376