Question

Write a MIPS Assembly language program to request a file name from the user, open the file, read the contents, and write it out to the console.

Answer 1

.data  

msg1: .asciiz "Enter the file name: "

msg2: .asciiz "Output:\n"

filename: .ascii ""   

buffer: .space 1024

.text

#prints the message

li $v0, 4 #system call for print string

la $a0, msg1 #msg to display

syscall #Display the message

#take the filename from user

li $v0, 8 #8 for take the input from user

la $a0, filename #store the file name

li $a1, 30 #Size of file name in bytes

syscall   

# Open file for reading

li $v0, 13 # 13 for open file

la $a0, filename

li $a1, 0 # assign zero for read operation

li $a2, 0 # permissions

syscall # open a file

move $s0, $v0 # move the file descriptor to $s0

# reading from file

li $v0, 14 # 14 for reading from file

move $a0, $s0

la $a1, buffer # address of buffer from which to read

li $a2, 1024 # No of bytes to read

syscall # read from file

#prints the message

li $v0, 4 #4 for print string

la $a0, msg2 #msg to display

syscall #Display the message

# Printing File Content

li $v0, 4 # 4 to print string

la $a0, buffer # buffer contains the string

syscall   

#Closing the file

li $v0, 16 # 16 close the file

add $a0, $s0, $0 # $s0 contains fd

syscall

#Exit from program

li $v0, 10

syscall