reveal.js

# Pause for Breath

---

CS 130 // 2021-10-25

## Administrivia
- Assignment 2 returned on Gradescope
- 
 Minor updates to schedule
- 
 Assignment 3 due tonight
    + Last 20-minutes of class today will be dedicated to working on the assignment
- 
 The rest we will do exam practice problems

# Exam 2
- Wednesday, October 27th

## Exam 2
- **Format of the exam**:
    + Written, in-class, 75-minute exam
    + Closed textbook/notes/internet
    + Can bring a 8.5 x 11, single-sided "cheat sheet"
- 
 **Topics**:
    + Basic MIPS, C to MIPS translation, binary, two's complement representation, hexadecimal, R-type and I-type machine code

# Questions
## ...about anything?

# Exam 2 Practice

## Exercise 1
- Is the following MIPS machine-code instruction an R-type or an I-type instruction? Why?
    + $00000010101100001000000000100101$

## Exercise 2
- Convert the following two (unsigned) binary integers into decimal (base 10)
    + 10101
    + 10000

## Exercise 3
- Convert the following (unsigned) binary integer into hexadecimal (base 16)
    + 100101

## Exercise 4
- Using your MIPS sheet
    + What instruction has opcode 0 and funct $25_\text{hex}$?
    + What registers correspond to 21 and 16?
- 
 Using this, write down the equivalent MIPS instruction the following machine code represents
    + $000000\\;10101\\;10000\\;10000\\;00000\\;100101$

## Exercise 5
- Consider the following 16-bit number:
    + $1111111111001010$
- Convert it to hexadecimal
- Convert it to decimal, assuming it is a 16-bit **signed** two's complement representation

## Exercise 6
- Convert the following C code to MIPS
    ```c
    int log(int n)
    {
        int count = 0;
        int value = 1;
        while (value < n) {
            count++;
            value = 2 * value;
        }
        return count;
    }
    ```

## Exercise 7
- Convert the following C code to MIPS
    ```c
    int foo(int x)
    {
        return x + 5;
    }

int bar(int y)
    {
        return 4 * foo(2 * y);
    }
    ```

## Exercise 8
- Convert the following C code to MIPS where `arr` is an array of 4-byte integers with a memory address stored in `$s0`

```c
arr[0] = arr[1] + arr[2];
```

# Questions
## ...about anything?

# Assignment 3 Time