CODEWITHSUNDEEP
assemblymips
James Manes
James Manes

Reputation: 526

Simple Addition in MIPS

I'm currently in a MIPS assembly class and the book we use is out of print so I am relying upon the internet for help so that I may understand. This program is taking in three integers. Two of them to add/sub/mult/div and the third is the operator. Here is the code.

    .text
    .globl __start
__start:

    # Prompt for first int and accept first int
    la $a0,firstint
    li $v0,4
    syscall

    li $v0,5
    move $s0, $v0
    syscall

    # Prompt for second int and accept second int
    la $a0,firstint
    li $v0,4
    syscall

    li $v0,5
    move $s1, $v0
    syscall

    # Prompt for operation
    la $a0,operation
    li $v0,4
    syscall

    li $v0,5
    move $s2, $v0
    syscall

    beq $s2,0,__add0

    li $v0,10
    syscall

__add0:
    la $a0,added
    li $v0,4
    syscall

    add $a0, $s0, $s1
    li $a0,1
    syscall


    .data
firstint:   .asciiz "Enter the first integer: "
secondint:  .asciiz "Enter the second integer: "
operation:  .asciiz "Enter operation (add=0, subtract=1, multiply=2, divide=3): "
added:      .asciiz "The added number is: "

My understanding is that beq will jump to add0 if the value in $s2 is equal to 0.. but it does not seem to be happening. Output stops after entering the operation type. Example output:

Enter the first integer: 10
Enter the first integer: 5
Enter operation (add=0, subtract=1, multiply=2, divide=3): 0

-- program is finished running --

Any ideas?

Upvotes: 3

Views: 5438

Answers (1)

Patrik
Patrik

Reputation: 2692

You have to do the syscall before the move:

li $v0,5
syscall
move $s2, $v0

Upvotes: 3

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