CS 61C Midterm – Spring 2000

Question 1 (2 points)

Convert the binary value 110000001111111111101110 into

a) hexadecimal (base 16)

b) octal (base 8)

Question 2 (4 points)

Assuming a five-bit word length, convert the binary value 11100 to decimal, supposing the representation is

a) unsigned

b) sign-magnitude

c) one’s complement

d) two’s complement

Question 3 (3 points)

Decode the following binary numbers as MIPS instructions and give the equivalent MIPS assembly language (MAL) statements. Show memory addresses, if any, in hexadecimal.

Address Value

0x40 00001100101101110000000000100100

0x44 01000110000001000100000110000010

0x48 00000100111000011111111111111110

Question 4 (2 points)

In the MIPS procedure-calling convention, there exist a compromise between a pure “callee-saved” and a pure “caller-saved” convention. That is, some registers are “callee-saved” ($s registers) and some are “caller-saved” ($t registers).

In one English sentence, explain why the MIPS designers chose this mixed strategy rather than either pure callee-saved or pure caller-saved.

Question 5 (4 points)

Consider this C struct definition:

struct foo {

int a[3];

char b[4];

int *c;

} baz;

Suppose that register $16 contains the address of baz.

Here is a fragment of C code:

baz.c = baz.a;

baz.c[1] = baz.b[2];

Below is a buggy translation into MIPS assembly language. Find and fix all the bugs:

lw $8, 0($16)

sw $8, 28($16)

lw $8, 20($16)

lw $9, 28($16)

sw $8, 4($9)

Question 6 (8 points)

a) Translate the following C procedure, which recursively computes the number of times a given character occurs in a given string, to MAL. Use the convention in which arguments are passed in registers. Use as little stack space as possible.

int count(char ch, char *str) {

if (*str == ‘\0’) return 0;

if (*str == ch) return count(ch, str+1)+1;

return count(ch, str+1);

}

b) Now translate to MAL the following iterative computation of the same function. Again, use the convention in which arguments are passed in registers. Use as little stack space as possible.

int count (char ch, char *str) {

int result=0;

while (*str != ‘\0’) {

if (*str == ch) result++;

str++;

}

return result;

}

Question 7 (3 points)

Using only two-input NOR gates (which return the value NOT(A OR B)), implement the CARRY output of a half-adder.

(Hint: What is A NOR A?)