Sign extend a nine-bit number in C

Question

I have a short, instr, that looks like this:

1110xxx111111111 

I need to pull out bits 0-9, which I do with (instr & 0x1FF). This quantity is then stored in a new short. The problem is that when this occurs, it becomes 0x0000000111111111, not 0x1111111111111111 like I want. How can I fix this? Thanks!

EDIT

Here's the code:

short instr = state->mem[state->pc]; unsigned int reg = instr >> 9 & 7; // 0b111 state->regs[reg] = state->pc + (instr & 0x1FF); 

This is a simulator that reads in assembly. state is the machine, regs[] are the registers and pc is the address of the current instruction in mem[].

This is fine if the last nine bits represent a positive number, but if they're representing -1, it's stored as all 1's, which is interpreted as a positive value by my code.

Answer 1

Assuming a short is 16 bits:

You can do it manually: (instr & 0x1FF) | ((instr & 0x100) ? 0xFE00 : 0). This tests the sign bit (the uppermost bit you are retaining, 0x100) and sets all the bits above it if the sign bit is set. You can extend this to 5 bits by adapting the masks to 0x1F, 0x10 and 0xFFE0, being the lower 5 bits, the 5th bit itself and all the bits 5-16 respectively.

Or you can find some excuse to assign the bits to the upper part of a signed short and shift them down (getting a sign-extension in the process): short x = (instr & 0x1FF) << 7; x >>= 7; The latter may actually end up being more straightforward in assembly and will not involve a branch. If instr is signed this can be done in a single expression: (instr & 0x1FF) << 7 >> 7. Since that already removes the upper bits it simplifies to instr << 7 >> 7. Replace 7 with 11 for 5 bits (16-5).