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CSCI2500-Lab 7 Solved
. Checkpoint 1: For the first checkpoint, you will make use of integer division in MIPS to help a cashier make change using only coins, i.e., only quarters, dimes, nickels, and pennies. More specifically, write a procedure called change that accepts as input in register $a0 an unsigned positive integer. This value is the number of cents to make change for. As an example, if this value is 137, then change must be made for $1.37, which would be five quarters, one dime, and two pennies.
To figure this out using MIPS, in your change procedure, use the divu instruction to first determine how many quarters you need. For the 137 example, dividing 137 by 25 cents gives you a quotient of 5 and a remainder of 12. The quotient here indicates that you need five quarters, whereas the remainder is then used to determine the number of dimes, nickels, and pennies.
Note that the divu instruction only has two operands. And it produces both a quotient and a remainder. More specifically, it stores the quotient in a special Lo register and the remainder in a Hi register. You therefore need to use the Move From Hi (mfhi) and Move From Lo
(mflo) instructions accordingly.
As an example, the code snippet below divides 5 by 3, yielding a 1 in $t2 with a remainder of 2 in $t3:
li $t0,5 li $t1,3 divu $t0,$t1 mflo $t2 mfhi $t3
Output your results using the format shown below:
Enter positive integer:
137
For $1.37, you need:
5 quarters
1 dime
2 pennies
Feel free to use a loop to ask the user for more inputs. This will make testing quicker. And carefully match the output format above.
2. Checkpoint 2: For the second checkpoint, download the adder.c code, which partially implements a one-bit adder. After all the MIPS coding we’ve been doing, this might be the first time in your life that you’re happy to see C code!
For this checkpoint, fill in C code for the add_two_bits() function.
Also, to receive full credit for this checkpoint, describe what the following line of code does (and why it “cleans” the input):
d0 = !!d0;
Note that in C, there are a variety of bitwise operations that we can use for this checkpoint. Remember that we can implement a logical bitwise AND via the & operator. How is this different from the better-known && operator?
Further, we can implement a logical bitwise OR via the | operator. Again, how is this different from the better-known || operator?
We can also implement a logical bitwise XOR via the ^ operator. And to implement a logical bitwise NOT, we use the familiar ! operator (the only unary operator).
3. Checkpoint 3: For the third checkpoint, begin by sketching a two-bit adder, showing the proper carry in and carry out lines connected up correctly to handle two-bit addition. Once you understand how the one-bit adders combine together, implement a four-bit adder in C by only making use of your add_two_bits() code from the first checkpoint.
More specifically, in main(), fill in the following code:
int main()
{ int i; int d0[4]; int d1[4]; int ci[4]; int sum[4];
/* Call add_two_bits() multiple times to implement a four-bit adder: */ return
To figure this out using MIPS, in your change procedure, use the divu instruction to first determine how many quarters you need. For the 137 example, dividing 137 by 25 cents gives you a quotient of 5 and a remainder of 12. The quotient here indicates that you need five quarters, whereas the remainder is then used to determine the number of dimes, nickels, and pennies.
Note that the divu instruction only has two operands. And it produces both a quotient and a remainder. More specifically, it stores the quotient in a special Lo register and the remainder in a Hi register. You therefore need to use the Move From Hi (mfhi) and Move From Lo
(mflo) instructions accordingly.
As an example, the code snippet below divides 5 by 3, yielding a 1 in $t2 with a remainder of 2 in $t3:
li $t0,5 li $t1,3 divu $t0,$t1 mflo $t2 mfhi $t3
Output your results using the format shown below:
Enter positive integer:
137
For $1.37, you need:
5 quarters
1 dime
2 pennies
Feel free to use a loop to ask the user for more inputs. This will make testing quicker. And carefully match the output format above.
2. Checkpoint 2: For the second checkpoint, download the adder.c code, which partially implements a one-bit adder. After all the MIPS coding we’ve been doing, this might be the first time in your life that you’re happy to see C code!
For this checkpoint, fill in C code for the add_two_bits() function.
Also, to receive full credit for this checkpoint, describe what the following line of code does (and why it “cleans” the input):
d0 = !!d0;
Note that in C, there are a variety of bitwise operations that we can use for this checkpoint. Remember that we can implement a logical bitwise AND via the & operator. How is this different from the better-known && operator?
Further, we can implement a logical bitwise OR via the | operator. Again, how is this different from the better-known || operator?
We can also implement a logical bitwise XOR via the ^ operator. And to implement a logical bitwise NOT, we use the familiar ! operator (the only unary operator).
3. Checkpoint 3: For the third checkpoint, begin by sketching a two-bit adder, showing the proper carry in and carry out lines connected up correctly to handle two-bit addition. Once you understand how the one-bit adders combine together, implement a four-bit adder in C by only making use of your add_two_bits() code from the first checkpoint.
More specifically, in main(), fill in the following code:
int main()
{ int i; int d0[4]; int d1[4]; int ci[4]; int sum[4];
/* Call add_two_bits() multiple times to implement a four-bit adder: */ return
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