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0 pointswilliamcotton12y ago0 comments

C expects addressable registers. From what I can tell this doesn't have those!

It really does seem like a Lisp would map much better, with the whole caller/callee and those private data belts that looked like hardware level closures!

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4 comments · 2 top-level

Symmetry12y ago· 2 in thread

To elaborate on what knz42 said, C expects addressable memory but doesn't say anything about registers at all. In fact, the existence of registers is sort of vaguely awkward from the perspective of C's model of computing.

Mindless211212y ago

> doesn't say anything about registers at all

Considering that C has a "register" storage-class specifier keyword, I don't see how you can make that claim. Granted, modern compilers ignore it.

Certainly, the C specification states that "register" storage is non-addressable, so it does say something.

Someone12y ago

That "register" nowadays is just a sequence of characters that, to those familiar with CPU architecture, rings a bell. Modern C standards do not convey any links between the keyword and hardware registers. http://www.open-std.org/JTC1/SC22/wg14/www/docs/n1124.pdf, section 6.7.1:

"A declaration of an identifier for an object with storage-class specifier register suggests that access to the object be as fast as possible. The extent to which such suggestions are effective is implementation-defined."

And 'non-addressable' only applies to the C 'runtime'. The standard does not say anything about the hardware.

For example, a C compiler for the 6502 could attempt to store variables of storage class 'register' in the zero page. Such storage locations have an address, but you are not allowed to take it from standard C.

For an even better example, look at http://en.wikipedia.org/wiki/TMS9900. That CPU stored all its general purpose registers in RAM.

knz4212y ago

C does not expect addressable registers. This architecture seems quite appropriate for C!

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