Mips Labs Using SPIM - Lab 1

Opening SPIM
      SPIM - A Whirlwind tour
The Program
      Analysis of the Program
      Assembling OpCodes of Instructions
Loading the Program
      Executing the Program
Questions
Conclusion

Opening SPIM

Start PCSpim on your workstation. Detailed instructions on where to get it and how to run it can be found here.

• Top Window - Registers
• II Window - Code
• III Window - Data, Stack and Kernel Data
• Bottom Window - Blurb

Registers

Code.

Data, Stack and Kernel Data

Blurb

Memory Organization

#Comments begin with a '#' and continue till
#the end of the line
#
#Lab 1: Register Swapping & SPIM orientation
#
#
# Assembler Directive to load the
# data segment, text segment
	.data
	.word 7
	.word 3

	.text
	.globl	main
main:
	lui $s0, 0x1001
	#load upper part of register s0(16) with 0x1001  s0 = 0x10010000

	lw $s1, 0($s0)  
	#load s1 with the contents of memory address 0x10010000 = 7, 
	#since we loaded the data there.

	lw $s2, 4($s0)  
	#load s2 with the contents of memory address 0x10010004 = 3, 
	#since we loaded the data there.

	sw $s2, 0($s0)  
	#store contents of s2 into memory address 0x10010000 

	sw $s1, 4($s0)   
	#store contents of s1 into memory address 0x10010004

Written together, it will look like 001111 00000 10000 0001 0000 0000 0001

To express this binary string in hexadecimal, we need to group them into groups of four each, thus:
0011 1100 0001 0000 0001 0000 0000 0001

This binary string, expressed in hexadecimal, is 0x3c101001, which is the machine code for the instruction

An example of the machine code of a load instruction is:

Note: note the way that the opcode is constructed. 0x23 does not have the usual representation of 0010 0011 because 8 bits are not available. It is represented as 10 0011 in 6 bits.

The machine code, therefore, is 1000 1110 0001 0001 0000 0000 0000 0000 or, 0x8e11000.

1. The code is loaded into addresses starting from 0x00400024
2. The data 7 and 3 is loaded into addresses 0x10010000 and 0x10010004
3. The machine code for the mnemonics in the program is as calculated

The stack is still empty as we have not started executing the code.

Once the program has been loaded, we will step through the execution of the program.
To set a breakpoint, select [Simulator->Breakpoints].
Set a breakpoint at the memory address of the first line of our code, 0x00400024.
The display of the first line of the code shows "break $1"

Select [Simulator->Go].
When it reaches the breakpoint, it will ask us whether to continue. Press no.
Note that the stack has changed because code is executing. It grows upwards, towards lower addresses.

Step through the code using < F10 >. After the first step, observe Register 16 (s0) in the topmost window. It has become 0x1001000.

Press < F10 >. Register 17 (s1) now contains 7, the contents of 0x1001000.
Press < F10 >. Register 18 (s2) now contains 3, the contents of 0x1001004.
Press < F10 >. Observe the third window. Memory address 0x1001000 now contains 3.
Press < F10 >. Memory address 0x1001004 now contains 7.

The contents in the memory have been swapped.

The following questions should be answered and submitted to the TA, Yuly, via email: ysuvorov@iastate.edu. Make sure your subject line is "Comp Sci 321 Lab1"

1. After the values have been swapped in memory, what value is in the memory location immediately following the new location of the 7 (0x10010008)?

2. What is the very first instruction listed in the instruction window after the program has been loaded? (Hint: it includes the comment "# argc")

3. What is the full text of the very first instruction in the kernel code (Hint: it is an 'addu' instruction)

In this lab, we have seen:

• The spim environment
• How to load, and track the execution of a piece of code in SPIM