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I'm currently trying to write a Ninja VM, a task which was given to me as an exercise. Opcode and value are saved in a 32-Bit unsigned int, with the first 8 bit being the opcode and the rest being the value. To deal with a negative value, I use a macro given to me with the exercise

#define SIGN_EXTEND(i) ((i) & 0x00800000 ? (i) | 0xFF000000 : (i))

However, when trying to retrieve the op-code of

unsigned int i = (PUSHC << 24) | SIGN_EXTEND(-2); (where PUSHC equals 1)

by using a bit-wise AND as seen here ->

int opcode = (i>>24)&0xff

my result is 255, although this extraction method works when the value is positive and reproduces the correct op-code in these cases.

While researching this, I found "How I get the value from the Immediate part of a 32 Bit sequence in C?", which seemingly deals with what I'm asking, however the answer given in that question is exactly what I'm doing, anyone know more?

SacrilegiousDuckHerder
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  • This macro seem to be converting from *24*-bit negative values to *32*-bit. So `SIGN_EXTEND(-2)` is using it not as intended – Eugene Sh. May 27 '19 at 20:45

1 Answers1

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SIGN_EXTEND looks like it is intended to be used in decoding, to take the 24 bits that have been extracted from the instruction and convert them to a 32-bit signed integer. It looks like it is not intended to be used in encoding. For encoding, IMMEDIATE(x) appears to be intended for that. Given a 32-bit signed integer, it produces a 24-bit signed integer, providing the value fits.

So:

unsigned int i = (PUSHC << 24) | SIGN_EXTEND(-2);

should be:

unsigned int i = (PUSHC << 24) | IMMEDIATE(-2);
Eric Postpischil
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  • Thank you, that did indeed fix my problem. I tried to research the way the macros work beforehand, but it appears I haven't really grasped them yet. I'd be grateful if anyone could explain what exactly happens when you enter a value into them. `#define IMMEDIATE(x) ((x) & 0x00FFFFFF)` & `#define SIGN_EXTEND(i) ((i) & 0x00800000 ? (i) | 0xFF000000 : (i))` – SacrilegiousDuckHerder May 27 '19 at 21:37