Question

8.6 Assume that register $s0 = 0x70000000 and $s1 0x10000000. For the following code, will there be overflow?

Add $s0,$s0,$s1

Add $s0,$s0,$s1

8.7 Assume that register $s0 = 0x40000000 and $s1= 0x20000000. For the code above, will there be overflow?

Answer 1

8.6 :The short answer is no, there will not be an overflow, because

$s0 = 0x70000000 in hex or 0111000000000000000000000000000 in binary

similarly,

$s1 = 0x10000000 or 00010000000000000000000000000000 in binary

on performing first command (add $s0,$s0,$s1), the value in $s0

$s0=0x80000000, which is 10000000000000000000000000000000 in binary

after the second add $s0, $s0, $s1;

$s0=0x90000000, which is 10010000000000000000000000000000 in binary

since the 32 bit is sufficient to provide the correct sum and no bit was carried over, there will be no overflow bit.

8.7 : No carry bit, no over flow.

$s0 = 0x40000000 in hex or 0100000000000000000000000000000 in binary

similarly,

$s1 = 0x20000000 or 00100000000000000000000000000000 in binary

on performing first command (add $s0,$s0,$s1), the value in $s0

$s0=0x60000000, which is 01100000000000000000000000000000 in binary

after the second add $s0, $s0, $s1;

$s0=0x80000000, which is 10000000000000000000000000000000 in binary

since the 32 bit is sufficient to provide the correct sum and no bit was carried over, there will be no overflow bit.