Question

Assume that 32-bit registers $80 and $1 hold values 80000000nex and D0000000mex respectively, (a) What is the value of $to for the following assembly code? add $t0, $80, $51 (b) is the value in $to the desired result, or has there been overflow?

Answer 1

a)  $t0 = 0x50000000
b)  yes, there was an overflow

Very detailed Explanation:
---------------------------
Hexadecimal     Binary
    0           0000
    1           0001
    2           0010
    3           0011
    4           0100
    5           0101
    6           0110
    7           0111
    8           1000
    9           1001
    A           1010
    B           1011
    C           1100
    D           1101
    E           1110
    F           1111
Use this table to convert from hexadecimal to binary
Converting 80000000 to binary
8 => 1000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
So, in binary 80000000 is 10000000000000000000000000000000

Hexadecimal     Binary
    0           0000
    1           0001
    2           0010
    3           0011
    4           0100
    5           0101
    6           0110
    7           0111
    8           1000
    9           1001
    A           1010
    B           1011
    C           1100
    D           1101
    E           1110
    F           1111
Use this table to convert from hexadecimal to binary
Converting D0000000 to binary
D => 1101
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
0 => 0000
So, in binary D0000000 is 11010000000000000000000000000000

Adding 10000000000000000000000000000000 and 11010000000000000000000000000000 in binary
    10000000000000000000000000000000
    11010000000000000000000000000000
-------------------------------------
 (1)01010000000000000000000000000000
-------------------------------------
Sum produces a carry of 1. We can ignore that carry.
So, sum of these numbers in binary is 01010000000000000000000000000000

let's convert 01010000000000000000000000000000 back to hexadecimal
Converting 01010000000000000000000000000000 to hexadecimal
0101 => 5
0000 => 0
0000 => 0
0000 => 0
0000 => 0
0000 => 0
0000 => 0
0000 => 0
So, in hexadecimal 01010000000000000000000000000000 is 0x50000000
so, value of $t0 is 0x50000000

Verification
---------------
first = 10000000000000000000000000000000
since left most bit is 1, this number is negative number.
so, follow these steps below to convert this into a decimal value.
I. first flip all the bits. Flip all 0's to 1 and all 1's to 0.
   10000000000000000000000000000000 is flipped to 01111111111111111111111111111111
II. Add 1 to above result
01111111111111111111111111111111 + 1 = 10000000000000000000000000000000
III. Now convert this result to decimal value
=> 10000000000000000000000000000000
=> 1x2^31+0x2^30+0x2^29+0x2^28+0x2^27+0x2^26+0x2^25+0x2^24+0x2^23+0x2^22+0x2^21+0x2^20+0x2^19+0x2^18+0x2^17+0x2^16+0x2^15+0x2^14+0x2^13+0x2^12+0x2^11+0x2^10+0x2^9+0x2^8+0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 1x2147483648+0x1073741824+0x536870912+0x268435456+0x134217728+0x67108864+0x33554432+0x16777216+0x8388608+0x4194304+0x2097152+0x1048576+0x524288+0x262144+0x131072+0x65536+0x32768+0x16384+0x8192+0x4096+0x2048+0x1024+0x512+0x256+0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1
=> 2147483648+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
=> 2147483648
Answer: -2147483648

second = 11010000000000000000000000000000
since left most bit is 1, this number is negative number.
so, follow these steps below to convert this into a decimal value.
I. first flip all the bits. Flip all 0's to 1 and all 1's to 0.
   11010000000000000000000000000000 is flipped to 00101111111111111111111111111111
II. Add 1 to above result
00101111111111111111111111111111 + 1 = 00110000000000000000000000000000
III. Now convert this result to decimal value
=> 110000000000000000000000000000
=> 1x2^29+1x2^28+0x2^27+0x2^26+0x2^25+0x2^24+0x2^23+0x2^22+0x2^21+0x2^20+0x2^19+0x2^18+0x2^17+0x2^16+0x2^15+0x2^14+0x2^13+0x2^12+0x2^11+0x2^10+0x2^9+0x2^8+0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 1x536870912+1x268435456+0x134217728+0x67108864+0x33554432+0x16777216+0x8388608+0x4194304+0x2097152+0x1048576+0x524288+0x262144+0x131072+0x65536+0x32768+0x16384+0x8192+0x4096+0x2048+0x1024+0x512+0x256+0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1
=> 536870912+268435456+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
=> 805306368
Answer: -805306368

sum = 01010000000000000000000000000000
since left most bit is 0, this number is positive
so, we can directly convert this into a decimal value
=> 1010000000000000000000000000000
=> 1x2^30+0x2^29+1x2^28+0x2^27+0x2^26+0x2^25+0x2^24+0x2^23+0x2^22+0x2^21+0x2^20+0x2^19+0x2^18+0x2^17+0x2^16+0x2^15+0x2^14+0x2^13+0x2^12+0x2^11+0x2^10+0x2^9+0x2^8+0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 1x1073741824+0x536870912+1x268435456+0x134217728+0x67108864+0x33554432+0x16777216+0x8388608+0x4194304+0x2097152+0x1048576+0x524288+0x262144+0x131072+0x65536+0x32768+0x16384+0x8192+0x4096+0x2048+0x1024+0x512+0x256+0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1
=> 1073741824+0+268435456+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0
=> 1342177280
Answer: 1342177280
-2147483648+-805306368 must be -2952790016
This is not correct since we can verify that -2147483648+-805306368 not equals 1342177280
So, there was an overflow.