What is the offset for 0xfe1ff0ef ?

Steps

Convert to binary representation
Extract the imm fields
Reorganize and concatenate
Perform 2s complement conversion (if -ve need additional steps like example here, if positive just read out number in binary).

Instruction in Binary (Bits 31 to 0)

First, convert each hexadecimal digit to its 4-bit binary equivalent and write them in order:

Hexadecimal to Binary Conversion:

Hex	Binary
f	1111
e	1110
1	0001
f	1111
f	1111
0	0000
e	1110
f	1111

Combined Binary Instruction:

#### **Binary Representation**

Bit Position	[31]	[30]	[29]	[28]	[27]	[26]	[25]	[24]	[23]	[22]	[21]	[20]	[19]	[18]	[17]	[16]	[15]	[14]	[13]	[12]	[11]	[10]	[9]	[8]	[7]	[6]	[5]	[4]	[3]	[2]	[1]	[0]
Binary Value	1	1	1	1	1	1	1	0	0	0	0	1	1	1	1	1	1	1	1	1	0	0	0	0	1	1	1	0	1	1	1	1

Grouped by Fields

Field	Bits	Binary Value
imm[20]	[31]	1
imm[10:1]	[30‒21]	1 1 1 1 1 1 0 0 0 0
imm[11]	[20]	1
imm[19:12]	[19‒12]	1 1 1 1 1 1 1 1
rd	[11‒7]	0 0 0 0 1
opcode	[6‒0]	1 1 0 1 1 1 1

Field Breakdown

The RISC-V JAL instruction format divides the 32-bit instruction into the following fields:

imm[20]: Bit [31]
imm[10:1]: Bits [30:21]
imm[11]: Bit [20]
imm[19:12]: Bits [19:12]
rd: Bits [11:7]
opcode: Bits [6:0]

Detailed Breakdown:

Opcode (bits [6:0]): 1 1 0 1 1 1 1 (binary) = 0x6F (hexadecimal)
- Instruction: JAL (Jump and Link)
Destination Register (rd) (bits [11:7]): 0 0 0 0 1 (binary) = 1 (decimal)
- Destination Register: x1
Immediate Fields:
- imm[20] (bit [31]): 1
- imm[10:1] (bits [30:21]): 1 1 1 1 1 1 0 0 0 0
- imm[11] (bit [20]): 1
- imm[19:12] (bits [19:12]): 1 1 1 1 1 1 1 1

Immediate Value Reconstruction

To reconstruct the immediate value, we rearrange the bits according to the JAL instruction format:

Immediate[20]    Immediate[19:12] | Immediate[11] |   Immediate[10:1]   |     
     1         | 1 1 1 1 1 1 1 1 |      1.        |  1 1 1 1 1 1 0 0 0 0 |             

Combined Immediate Bits (after shifting):

Immediate[20:0]: 1 1 1 1  1 1 1 1 1  1 1 1 1 1  1 0 0 0 0 0

Note: The immediate value is left-shifted by 1 (since the address is word-aligned), so we append a 0 at the end.

Since the MSB is a 1, the immediate value is negative. It represents -offset.
Thus, we first figure out what +offset is in decimal and then apply the negative sign.
To find out what +offset is we apply the following steps in binary form. -(-offset) = ~offset + 1

Calculating the Offset (Two’s Complement)

Since the most significant bit is 1, the immediate value is negative.

Two’s Complement Calculation:

Invert the Bits:

Inverted: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1

Add 1:

Two's Complement Value: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

Convert to Decimal:

Binary: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
Decimal Value: 32

Apply Negative Sign:
```
Offset = -32 bytes
```

Final Disassembled Instruction

jal x1, -32

This horizontal representation shows each bit and its corresponding field, providing a clear view of the instruction’s structure.