为什么我的64位计算机不能在C ++中存储双精度数2 ^ 1024？

Question

My computer has an Intel® Core™2 Duo Processor, which I believe is a 64-bit processor. So I had thought that in C++ (or any other programming language), any double-precision number (which requires 64 bits) could be stored.

I read the following link: http://en.wikipedia.org/wiki/Double-precision_floating-point_format

Based on that, I know that a double-precision number has 1 bit for the sign, 11 bits for the exponent, and the remaining 52 bits for the fraction.

So, I thought I'd experiment with a double that has the following binary code:

0 11111111111 0000000000000000000000000000000000000000000000000000

Again, based on what I read in the above link, I think this number translates to a decimal number of 2^1024, the exponenent being equal to the following:

2^10 + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 - 1023

I then tried to print 2^1024 in C++, using the following code:

double x = pow(2.0, 1024);
cout << x << endl;

However, when I run the program, it prints '1.#INF'

Does anyone know why?

Answer 1

Some bit patters are reserved for special use like not-a-number, +infinity and -infinity. You have encountered the one used for +infinity.