C语言中long long的实现

Question

Recently I was doing some research about types in c++ and I have couple questions about long double . For instance I have some number long double x = -48.12e4 , then how accurate should I translate 0.12 to hex or bin (I prefer to use hex , easer to read). In the scheme of implementation I have 1 extra bit and I have no idea about its role -

//scheme 
1bit    15bit   1bit   63bit 
sign    e        1       m

For example lets take number that I have written before.

1)translate decimal to hex 
-48.12 = -3.1E (are 2 digits after decimal enough?)

2)normalization 
0011.0001 1110  * 10^0 =  001.1 0001 1110 * 10^1

3)calculation of "e"
16 383 + 4 + 1 = 16 388 = 4004(hex) = 0100 0000 0000 0100

4)collecting everything together
1    0100 0000 0000 0100   1    1000 1111 0000 0..0
sign|        e           |bit|      mantisa 
What is that 1 bit for?
5)reverse order
0..000  0011 1100 0110  0000 0010  1010 0000 

Have I done everything right?

Answer 1

From cppreference (emphasis mine) :

long double - extended precision floating point type. Does not necessarily map to types mandated by IEEE-754. Usually 80-bit x87 floating point type on x86 and x86-64 architectures.

You assume a specific layout of long double that is by no means guaranteed by the standard but may vary from compiler to compiler and depending on the target architecture.

Answer 2

The problem you have is that you're doing decimal math. That's logical for humans, but not how computers work.

In particular, -48.12e4 is specified as sign,mantissa,exponent, and that's also how the C++ implementation of long double works, but the implicit base for the exponent differs. e4 signifies a power of 10, but long double uses base-2 (or base-16, or another binary base).

The consequence is that the compiler cannot just convert exponents. 10^3 is approximately 2^10, but not even close enough for float let along long double .

So, the proper solution is a lot harder. Perhaps the easiest solution is to calculate long double(4812) and long double(1e2) , and to multiply those. That way, you only need to implement the conversion of integers to long double and the conversion of powers of 10 to long double .

The conversion of small positive powers of 10 to long double is also easy, reuse the integer code. For large even positive powers, use squaring. For negative powers of 10, use the fact that pow(10,n)=1.0/pow(10,-n) .

To get an idea how real compilers do it (which is a fair bit more efficient), look at open-source implementations of strtod() .