MIPS JAL Conversion

0
votes

I'm having difficulty understanding how to translate a JAL (J-Type) instruction in MIPS. Here is the instruction set that I am working on:

0x00400018          add  $a0, $a3, $0 = 00E02020
0x0040001C          jal  L2
0x00400020  L1:     jr   $ra
0x00400024  L2:     sw   $s1, 0($s2)
0x00400028          bne  $a0, $0, ELSE
0x0040002C          j    L1
0x00400030  ELSE:   addi $a0, $a0, A2 
0x00400034          j    L2

The first line did not pose a problem, but for the second line, I am entirely unsure how to translate jal L2. Using the MIPS reference sheet, I see that to get the address, I need to perform R[31]=PC+8;PC=JumpAddr.

J-Type Instructions are opcode/6bit and address 26bit so the first 6bit are 00 0011 but I don't know how I get the remaining 26bit. Any help would be appreciated.

2
assemblymipsmips32
What do you mean by "get the the remaining 26bit"? What is this for (i.e. what are you developing)? A disassembler? An emulator? A MIPS core for an FPGA? What are your inputs and outputs? – Michael
This is just for practicing to translate MIPS instructions into Machine Code (first binary then hex). Each MIPS instruction (for example add $a0, $a3, $0) is 32bit and so for line two, since I have the opcode and opcodes are 6bit, I don't know how to get the remaining 26bit (I don't know how to translate jal L2). – Carlo
If you've read the full 32 bit instruction word then all you have to do is mask off the unwanted bits. Something like lower_26_bits = instruction_word & 0x3FFFFFF;. – Michael
Please explain this step-by step, in relevance to R[31]=PC+8;PC=JumpAddr. How did you get to 0X3FFFFFF. I guess I don't know what you mean with "mask off the unwanted bits." – Carlo
see stackoverflow.com/questions/6950230/… – markgz

2 Answers

5
votes

I know this is completely irrelevant now, but for anyone that stumbles upon this question and needs help:

You would take the address of L2 --> 0x0040 0024 and shave off the first four bits, and last two. This would leave you with:

0000 0100 0000 0000 0000 0010 01 as your remaining 26 bits.

Hope I was able to help someone!

0
votes

MIPS has a couple of different control-flow instructions:
1. (conditional) branches
2. (unconditional) immediate jumps
3. (unconditional) register jumps

branch b contains a 16 bit signed integer number of instructions, relative to the next instruction, to branch to.

jump j /jump and link jal contains a 26 bit immediate, which is shifted right by two (because all MIPS (32) instructions are 4 bytes long). The upper 4 bits come from the instruction sequentially following the jump.

jump register jr /jump and link register jalr do not contain a branch target. The target is in a register, and as such is already 32 bits long.