Question

Possible Duplicate:
How do promotion rules work when the signedness on either side of a binary operator differ?

When casting from an unsigned integer to a signed integer, I know the represention of the variable’s bits changes. For instance, 255 may become -1, when converting from uint8 to int8. However, I was never sure what a ‘cast’ or ‘conversion’ entailed for the underlying bits themselves.

My question is, is the raw bit pattern of an integer variable guaranteed to remain the same after a static_cast between signed and unsigned types, or is it possible that it be transformed by the cast in some way?

Out of curiosity too, does a static_cast between integer signage types generate assembly, or is it used only so the compiler knows what asm instructions to generate?

edit:

Here’s an example of the kind of scenario I would want to know about:

unsigned int uintvar = random();
unsigned int control = uintvar;
assert(control == static_cast<unsigned int>(static_cast<signed int>(uintvar)));

Ignoring the fact the double cast would get optimized away, would this example be guarenteed to always hold true?

Answer 1

The bit pattern doesn’t change at all (on most architectures you’re likely to encounter in practice). The difference is in the instructions generated by the compiler to manipulate the values.