Name: Box: Date:

HW9 solution

  1. You are writing a MIPS assembly program. Your program has a beq instructions at line 0x0100 0400. You wish this beq to branch to location 0x0200 0000.

    1. (3 points) How close can the beq get to the goal address? Express your answer as the address nearest to the target address as possible.

      The beq is at 0x0100 0400, so PC+4 is at 0x0100 0404. The biggest positive number that can be put in the immediate value is 0x7fff. So
      0x0100 0404 + SE(0x7fff << 2) = 0x0102 0400

    2. (3 points) What is a possible method to allow long-range conditional branches?

      Since branches can only go as far as the 16-bit immediate allows (\(-2^{15} -- 2^{15}-1\) ), longer ranges can be implemented by branching to a j instruction. For even longer ranges, the target can be loaded into a register and jr can be used to jump.

  2. You are writing a program for a processor with 8-bit instructions. The branch instructions for this processor only allow 3-bit branch immediate.

    1. (3 points) Assuming the immediate is sign extended, what is the branch range of the processor?

      Since there are 3 bits, \(2^3 = 8\) possible locations, for a range of -4 to 3

    2. (3 points) What is the problem with such a branch range?

      It is very limited and makes writing loops and conditions very hard.

  1. Compute the correct hex value for the branch targets below.

    Address Instruction Hex representation
    0x00400028
          j test            
    0x08QQQQQQ
    0x0040002c
    loop: lw   $t1, 0($11)  
    0x8d690000
    0x00400030
          addi $t3, $t3, -4 
    0x216bfffc
    0x00400034
    test: bne  $t1, $0, loop
    0x1520RRRR
    1. (3 points) What is the hex value for QQQQQQ that represents the jump target? Show your work.

      Given the target 0x0040 0034 and PC+4 at 0x0040 002c:
      0x00400034 = top4(0x0040002c) : imm << 2
      Solving results in: imm = 0x001000d

    2. (3 points) What is the hex value for RRRR that represents the branch not equal target? Show your work.

      Given a target of 0x0040 002c and a PC+4 of 0x0040 0038:
      0x0040002c = 0x00400038 + SE(imm << 2)
      Solving results in:
      imm = 0xfffd