6. MIPS Instruction Set4

생성일: 2021년 10월 3일 오후 8:25

Registers

32 MIPS registers are partitioned as follows

• Register 0 : $zero always stores the constant 0
• Regs 2-3 : $v0, $v1 return values of a procedure
• Regs 4-7 : $a0-$a3 input arguments to a procedure
• Regs 8-15 : $t0-$t7 temporaries
• Regs 16-23: $s0-$s7 variables
• Regs 24-25: $t8-$t9 more temporaries
• Reg 28 : $gp global pointer
• Reg 29 : $sp stack pointer
• Reg 30 : $fp frame pointer
• Reg 31 : $ra return address

Procedures

Each procedure (function, subroutime) maintains a scratchpad of register values

• When another procedure is called (the callee(수신자)), the new procedure takes over the scratchpad
• Values in the register file may have to be saved so we can safely return them to the caller

1. Parameters (arguments) are placed where the callee can see them
2. Control is transferred to the callee
3. Acquire storage resources for the callee
4. Execute the procedure
5. Place result value where the caller can access it
6. Return control to caller

Jump-and-Link (jal)

• A special register (storage not part of the register file) maintains the address of the instruction currently being executed – this is the program counter (PC)
• The procedure (function) call is executed by invoking the jump-and-link (jal) instruction – the current PC (actually, PC+4) is saved in the register $ra and we jump to the procedure’s address (the PC is accordingly set to this address)

💡 jal addr

• To return from the procedure, we can simply execute “jr $31” (or jal addr stroes PC+4 in register $31, and then jumps to location addr)

Supporting procedure calls

• Problem
  • What if procB calls another procedure?
    • Content of $ra will be overwritten
  • What if we have more than 4 parameters?
• Solution
  • Use stack in main memory for storing
  • Stack은 동적으로 증가하고 축소되는 메모리 영역

Stack

A procedure’s values are therefore backed up in memory on a stack

• Example