# Illustration for the leaf procedure example (similar to example on p. 99
# Sec 2.8. Here we assume ony $s0 needs to be saved). 
# I have added additional code to ilustrate how the leaf procedure works.
# - Yashwant
#You can place breakpoints at [BPA] and [BPB] to observe registers before 
#    and after the procedure.
#
# C code:
# int leaf_example (int g, h, i, j)
# { int f;
#	f = (g + h) - (i + j);
#	return f;
# }
# Arguments g, …, j in $a0, …, $a3
# f in $s0 (hence, need to save $s0 on stack)
# Result in $v0


 .text                           # text section
 .globl main                     # call main by SPIM

 main:  li $a0, 16		#load immediate pseudoinstruction
	li $a1, 4
	li $a2, 9
	li $a3, 1
	li $s0, 7		#the procedure should not change it
        jal leaf_example	# [BPA]

        add $a0, $v0, $zero	#place in $a0 for printing [BPB]

	li $v0, 1		#syscall for printing
        syscall                 # print the content of value in $a0

	la $a0,str		# put string address into a0
	li $v0,4		# system call to print
	syscall			# print string
	
        li $v0, 10              # 10 is the exit syscall.
        syscall                 # do the syscall.


leaf_example:
	addi $sp, $sp, -4	#Save 
	sw   $s0, 0($sp)	#     $s0 on stack
	add  $t0, $a0, $a1
	add  $t1, $a2, $a3
	sub  $s0, $t0, $t1
	add  $v0, $s0, $zero
	lw   $s0, 0($sp)	#restore $s0
	addi $sp, $sp, 4	#	from stack
	jr   $ra

.data
str:	.asciiz ": that is the Result\n"

#that's it.