[assembly] How to Calculate Jump Target Address and Branch Target Address?

Usually you don't have to worry about calculating them as your assembler (or linker) will take of getting the calculations right. Let's say you have a small function:


func:
  slti $t0, $a0, 2
  beq $t0, $zero, cont
  ori $v0, $zero, 1
  jr $ra
cont:
  ...
  jal func
  ... 

When translating the above code into a binary stream of instructions the assembler (or linker if you first assembled into an object file) it will be determined where in memory the function will reside (let's ignore position independent code for now). Where in memory it will reside is usually specified in the ABI or given to you if you're using a simulator (like SPIM which loads the code at 0x400000 - note the link also contains a good explanation of the process).

Assuming we're talking about the SPIM case and our function is first in memory, the slti instruction will reside at 0x400000, the beq at 0x400004 and so on. Now we're almost there! For the beq instruction the branch target address is that of cont (0x400010) looking at a MIPS instruction reference we see that it is encoded as a 16-bit signed immediate relative to the next instruction (divided by 4 as all instructions must reside on a 4-byte aligned address anyway).

That is:

Current address of instruction + 4 = 0x400004 + 4 = 0x400008
Branch target = 0x400010
Difference = 0x400010 - 0x400008 = 0x8
To encode = Difference / 4 = 0x8 / 4 = 0x2 = 0b10

Encoding of beq $t0, $zero, cont

0001 00ss ssst tttt iiii iiii iiii iiii
---------------------------------------
0001 0001 0000 0000 0000 0000 0000 0010

As you can see you can branch to within -0x1fffc .. 0x20000 bytes. If for some reason, you need to jump further you can use a trampoline (an unconditional jump to the real target placed placed within the given limit).

Jump target addresses are, unlike branch target addresses, encoded using the absolute address (again divided by 4). Since the instruction encoding uses 6 bits for the opcode, this only leaves 26 bits for the address (effectively 28 given that the 2 last bits will be 0) therefore the 4 bits most significant bits of the PC register are used when forming the address (won't matter unless you intend to jump across 256 MB boundaries).

Returning to the above example the encoding for jal func is:

Destination address = absolute address of func = 0x400000
Divided by 4 = 0x400000 / 4 = 0x100000
Lower 26 bits = 0x100000 & 0x03ffffff = 0x100000 = 0b100000000000000000000

0000 11ii iiii iiii iiii iiii iiii iiii
---------------------------------------
0000 1100 0001 0000 0000 0000 0000 0000

You can quickly verify this, and play around with different instructions, using this online MIPS assembler i ran across (note it doesn't support all opcodes, for example slti, so I just changed that to slt here):

00400000: <func>    ; <input:0> func:
00400000: 0000002a  ; <input:1> slt $t0, $a0, 2
00400004: 11000002  ; <input:2> beq $t0, $zero, cont
00400008: 34020001  ; <input:3> ori $v0, $zero, 1
0040000c: 03e00008  ; <input:4> jr $ra
00400010: <cont>    ; <input:5> cont:
00400010: 0c100000  ; <input:7> jal func