Download Assembly Arrays and Binary Conversions

Computer Organization
CS345
David Monismith
Based upon notes by Dr. Bill Siever and notes
from the Patternson and Hennessy Text
Last Time
• Reviewed arrays
• Discussed the lw (load word) and sw (store
word) instructions
• Discussed the li (load immediate) instruction
• We saw that there are different ways to store
arrays
Recall:
• Array names are defined by a label. The name is
followed by a colon.
• Array type is provided after the name using a directive.
Some examples of directives are listed below.
– .asciiz - string (an array of characters or bytes) that is
null terminated
– .word - array of standard size data units (32-bit units for
MIPS)
– .space - array of bytes
• Arrays are declared in the .data section of an assembly
program and follow the format:
• label: .directive initValues/size
Array Examples
• myString: .asciiz "This is my string\n"
• myIntArray: .word 1, 2, 3, 4, 5
• myBytes: .space 200
• We can load a word (integer) into a register if we have the
address of an array.
• la $t1, myIntArray #here we load the
address of myIntArray into $t1
• lw $t2, 0($t1) #here we load the first
data value from myIntArray into $t2
• Notice that $t2 contains the value 1 decimal.
Arrays
• We can store a value at a position in the
integer array.
• li $t3, 432
• sw $t3, 8($t1)
• Here we store the value 432 in place of the
value 3 in location 2 of the myIntegerArray.
Conversions
• This time we will continue with decimal to
binary conversions.
• We've seen how to convert from binary to
decimal (multiply by 2)
• 1001001 = 2^0 + 2^3 + 2^6 = 1 + 8 + 64 = 73
• We've seen how to convert from binary to hex
• But how do we convert from decimal to
binary?
Decimal to binary
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0=0
1=1
2 = 10
3 = 11
4 = 100 . . .
Notice how the division by 2 method works
4%2=0
2%2=0
1
-> 100 binary = 4 decimal
By repeatedly taking our value modulo 2, dividing by 2 and
repeating, we can find the binary value of a decimal number.
Example
• Take 467
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467 % 2 = 1
233 % 2 = 1
116 % 2 = 0
58 % 2 = 0
29 % 2 = 1
14 % 2 = 0
7%2=1
3%2=1
1
Result
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Result:
1 1101 0011
Let's check our result.
1 D 3 hex
• 256 + 13*16 + 3 = 256 + 208 + 3 = 256 + 211 = 467
• Recall that every four binary digits represent a hex
digit.
Questions for next time:
• What is the issue with plain binary numbers?
– Signs must be represented
– We need an extra bit to do that
• How do we account for overflow? That is, what if we
add two very large binary numbers?
– We need to be careful about adding very large numbers
and consider different number representations.
• How do we represent instructions?
– There are three formats that MIPS uses to represent
instructions.
– R format - all registers
– I format - using registers and a 16 bit integer
– J format - using one instruction and a 26 bit address