3 ms·

Part of the problem is that 80 bit operations were never designed to be a visible data type, you were expected to round to a 64 bit type at the end of a series

by mark-r 5y ago

Part of the problem is that 80 bit operations were never designed to be a visible data type, you were expected to round to a 64 bit type at the end of a series of operations. There was never a good way of indicating to your compiler where and when those transitions would happen.

Sesse__ 5y ago
Not so strange, as they were never intended to be used for compilers to begin with (you'd write assembler code by hand). The basic idea with having 80-bit internal precision is that you could load your 64-bit values from RAM, compute on them with so much extra precision you could basically ignore precision issues within your expressions, and then store the result back correctly rounded.