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overflowmipssubtraction

MIPS overflow logic

Good evening. I am trying to figure out how to determine if an integer qualifies 16 bit integer in MIPS.

I understand that 2^15-1 =32767 or 2^(16-1)-1=32767 and that we want 16 bit values for binary number. Anyway, I am trying to determine if an integer passes the test. I wrote this:

addi $s3, $zero, 32767
bgt  $t2, $s3, else  #branch to else if t2>s3 
move $v0, $t2        #if no overflow; place t2 in v0
addi $v1, $zero, 0   #if no overflow; place zero in v1
   else:
      addi $v0, $zero, 0  #if overflow; place 0 in v0
      addi $v1, $zero, -1 #if overflow; place -1 in v1

Anyway, There's a problem with my logic when I try and evaluate negative numbers. I have assignment due tomorrow. I am learning MIPS programming. I am not a programming snob so any helpful advice is appreciate. Thank you for your time.

Solution

  • This is too late for you assignment1.

    It's a bit unclear whenever you want to test a) if a number N encoded as a 32-bit two's complement number can be also encoded as a 16-bit two's complement number or if you want to test b) if a 32-bit number can be encoded as a 16-bit number.

    In the case of b) you just need to test if any bit higher than the 16th is set:

    #Assume $t0 is the number to test
lui $t1, 0xffff                      #$t1 = 0xffff0000
and $t1, $t1, $t0                    #$t1 is zero if all higher bits of $t0 are zero
beq $t1, $0 fits16bits               #Jump to label if fits

#Here the number doesn't fit 16 bits

    For the case a) the key is to understand that just like the number 0x00f1 and the number 0x0000000f1 are the same, the leading zeros are not significant, the number 0xffff and the number 0xffffffff are the same number in two's complement (the number -1).
    To extend a 16-bit two's complement number to 32-bit we need to perform a sign extension, i.e. replicate the most significant bit (the sign bit) of the original number in the upper 16 bits.

    So 0x7fff becomes 0x00007fff, 0xc000 becomes 0xffffc0000.

    A simple way to test that all the upper 17 bits are equal is to shift them right arithmetically so that if they actually are equal we end up with 0x00000000 or 0xffffffff.

    sra $t1, $t0, 15                     #Shift right 16 bits duplicanting the MSb 
beqz $t1, $0, fits16bits             #Jump to label if fits (All zero)

addiu $t1, 1                         #Add 1
beqz $t1, $0, fits16bits             #Jump to label if fits (Before +1 was all ones)

#Here the numbers doesn't fit

    1 Maybe it's better this way.