A simple binary tree program -- in assembler

I believe every programmer must learn assembly programming. It brings the programmer and the machine closer and strengthens the bond between them. I learned the term GIGO (Garbage In Garbage Out) when I got first introduced to computer programming. Nothing has changed since. The languages and their environments have become better beyond doubt. Still, a computer and its resources are as good as its programmer.

Assembly language makes you understand how the machine is working beyond all that glittery higher level languages. If you are a system programmer, learning assembly becomes a must because there are things that C can't handle. I will leave more discussion on "Why learning assembly is important" for later post.

Every new architecture that I learn, I learn its assembler first. When I got first introduced to MIPS, I learned its assembly language. Below is a program that I wrote to achieve that.

 

 

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# # This program takes 10 integer values from user, # inserts them into a binary tree and then prints # the tree "inorder" # # # The tree_node is like this: # struct tree_node { # int value; # struct tree_node *left; # struct tree_node *right; # } .text #----------------------------------------------------------- # # This function prints a given node. # The nodes pointer is provided in $a0. # print_node: subu $sp, $sp, 32 # space on stack sw $ra, 28($sp) # store the return address sw $fp, 24($sp) # store the current frame pointer. addiu $fp, $sp, 32 # $fp to start of stack. move $t0, $a0 # save the nodes pointer to $t0 lw $a0, ($t0) # move the integer value to $a0. jal print_int_with_newline # print the integer. nop lw $ra, 28($sp) lw $fp, 24($sp) addiu $sp, $sp, 32 # reclaim the stack. jr $ra #---------------------------------------------------------- # This function prints the given integer and # then prints a new line after it. # $a0 - Integer to be printed. print_int_with_newline: subu $sp, $sp, 32 #minium stack frame sw $ra, 28($sp) #store the return address. sw $fp, 24($sp) #store the frame pointer. sw $a0, 20($sp) #store the argument. addiu $fp, $sp, 32 #base of the frame. #print the integer first. li $v0, 1 syscall #print the new line. la $a0, newline li $v0, 4 syscall #regenerate the previous frame. lw $ra, 28($sp) lw $fp, 24($sp) lw $a0, 20($sp) addiu $sp, $sp, 32 #reclaim the stack frame. jr $ra #return to caller. #------------------------------------------------------------ # This function creates a new tree node, assigns a given # value to it. Initializes it properly and returns the # pointer to it. # $a0 -> Integer to be assigned in value. # $v0 -> on success, returns with node pointer here else zero. # create_tree_node: subu $sp, $sp, 32 sw $ra, 28($sp) sw $fp, 24($sp) addiu $fp, $sp, 32 move $t0, $a0 li $a0, 12 li $v0, 9 syscall beqz $v0, out_of_memory sw $t0, ($v0) sw $zero, 4($v0) sw $zero, 8($v0) out_of_memory: lw $ra, 28($sp) lw $fp, 24($sp) addiu $sp, $sp, 32 jr $ra #--------------------------------------------------------------- # # This function gets number of specified integers in the # given buffer. # # $a0 -> Pointer to the destination buffer. # $a1 -> number of integers to be read out. # get_num_integers: addiu $sp, $sp, -32 # min stack. sw $ra, 28($sp) # store ra sw $fp, 24($sp) # store callers frame pointer. addiu $fp, $sp, 32 # point to our frame. move $t0, $a0 # destination buffer move $t1, $a1 # number of integers. read_loop: beqz $t1, read_loop_exit addiu $t1, $t1, -1 # decrement the count li $v0, 5 syscall sw $v0, ($t0) addiu $t0, $t0, 4 # point to next number(int are 4 bytes) b read_loop nop read_loop_exit: lw $ra, 28($sp) lw $fp, 24($sp) addiu $sp, $sp, 32 jr $ra #--------------------------------------------------------------- # This function adds a node to the given node. If it cannot be # added to the current node, it calls itself recursively until # it finds a node where it can add the given number. # # $a0 -> The starting node # $a1 -> The node to add. .globl add_node_to_tree add_node_to_tree: addiu $sp, $sp, -32 sw $ra, 28($sp) sw $fp, 24($sp) sw $a0, 20($sp) sw $a1, 16($sp) addiu $fp, $sp, 32 lw $t0, ($a0) # value of the starting current node. lw $t1, ($a1) # value of the node to add. bgt $t1, $t0, is_greater lw $t2, 4($a0) #load the left pointer. beqz $t2, null_left move $a0, $t2 # take this left pointer start loc now. jal add_node_to_tree nop b __return_add_node_to_tree nop is_greater: lw $t2, 8($a0) #load the right pointer. beqz $t2, null_right move $a0, $t2 # take this right pointer start loc now. jal add_node_to_tree nop b __return_add_node_to_tree nop null_left: # null left means that we can add the pointer here. sw $a1, 4($a0) b __return_add_node_to_tree nop null_right: sw $a1, 8($a0) b __return_add_node_to_tree nop __return_add_node_to_tree: lw $ra, 28($sp) lw $fp, 24($sp) lw $a0, 20($sp) lw $a1, 16($sp) addiu $sp, $sp, 32 jr $ra #---------------------------------------------------------------- # # This function prints the tree in order. # # $a0 -> root node. .globl print_inorder print_inorder: addiu $sp, $sp, -32 sw $ra, 28($sp) sw $fp, 24($sp) sw $a0, 20($sp) addiu $fp, $sp, 32 lw $t1, 4($a0) beqz $t1, print_curr move $a0, $t1 jal print_inorder nop print_curr: lw $a0, 20($sp) lw $t0, ($a0) move $a0, $t0 jal print_int_with_newline nop lw $a0, 20($sp) lw $t1, 8($a0) beqz $t1, __return_print_inorder move $a0, $t1 jal print_inorder nop __return_print_inorder: lw $ra, 28($sp) lw $fp, 24($sp) lw $a0, 20($sp) addiu $sp, $sp, 32 jr $ra #------------------------------------------------------------------ main: # readout 10 integers in buffer. la $a0, buffer li $a1, 10 jal get_num_integers nop li $t6, 10 la $t8, buffer li $t7, 0 tree_create_loop: beqz $t6, tree_create_done addiu $t6, $t6, -1 lw $a0, ($t8) jal create_tree_node addiu $t8, $t8, 4 #point to next number beqz $v0, alloc_failed nop beqz $t7, init_root move $a1, $v0 move $a0, $t7 jal add_node_to_tree nop b tree_create_loop nop init_root: move $t7, $v0 b tree_create_loop nop tree_create_done: move $a0, $t7 jal print_inorder nop b exit nop alloc_failed: la $a0, OOM_Error li $v0, 4 syscall exit: li $v0, 10 syscall .data buffer: .space 1024 OOM_Error: .asciiz "Out of memory!\n" newline: .asciiz "\n"