Week 07 Laboratory Exercises

However, each CPU instruction is represented in memory as some particular bit pattern. The CPU is able to decode the instruction in order to determine what operation to perform, and which operands (whether it be a register or immediate value) are involved.

In this exercise, we will construct the bit pattern for an addi instruction, given operands.

The MIPS documentation guide provides the general bit pattern for an addi instruction, including the opcode and the bits used for each operand.

The bit pattern here can be broken down into 4 components:

• The opcode: in this case, the opcode 001000 is used to distinguish the instruction as an addi instruction.
• The two registers: sssss and ttttt are used to represent the registers being used as operands for the instruction. Each register number is encoded as a 5-bit value - recall that there are 32 general-purpose registers available in MIPS, numbered $0 to $31. This means that 5 bits is sufficient to refer to a register.
• An immediate value: The last 16 bits are used to represent the final operand - a 16-bit immediate value.

The function addi is given each of the operands for a MIPS addi instruction. Add code so that it returns the encoded binary form of the instruction.

Your task is to add code to this function in addi.c:

// return the encoded binary MIPS for addi $t,$s, i
uint32_t addi(int t, int s, int i) {

    return 42; // REPLACE WITH YOUR CODE

}

1521 spim2hex can be used to print the encoded binary form of a particular addi instruction as a hexadecimal, for example:

echo 'addi $9 $27 42' | 1521 spim2hex
2369002a
./addi 9 27 42
addi(9, 27, 42) returned 0x2369002a
echo 'addi $17 $19 -3' | 1521 spim2hex
2271fffd
./addi 17 19 -3
addi(17, 19, -3) returned 0x2271fffd

Use make(1) to build your code:

make    # or 'make addi'

When you think your program is working, you can use autotest to run some simple automated tests:

1521 autotest addi 

When you are finished working on this exercise, you must submit your work by running give:

give cs1521 lab07_addi addi.c

You must run give before Monday 15 July 12:00 (midday) (2024-07-15 12:00:00) to obtain the marks for this lab exercise. Note that this is an individual exercise, the work you submit with give must be entirely your own.

In the files for this lab, you have been given mips_networking.s, a MIPS assembler program that reads one number and then prints it.

Add code to mips_networking.s to make it equivalent to this C program:

// Reads a 4-byte value and reverses the byte order, then prints it

#include <stdint.h>
#include <stdio.h>

#define BYTE_MASK 0xFF

int main(void) {
    int32_t network_bytes;
    scanf("%d", &network_bytes);

    int32_t computer_bytes = 0;
    uint32_t byte_mask = BYTE_MASK;

    computer_bytes |= (network_bytes & byte_mask) << 24;
    computer_bytes |= (network_bytes & (byte_mask << 8)) << 8;
    computer_bytes |= (network_bytes & (byte_mask << 16)) >> 8;
    computer_bytes |= (network_bytes & (byte_mask << 24)) >> 24;

    printf("%d\n", computer_bytes);

    return 0;
}

In other words, it should read one number, network_bytes, reverse the byte order inside it, and print the result.

The MIPS documentation guide provides a set of bitwise operations that will be useful for this.

For example:

1521 mipsy mips_networking.s
5
83886080
1521 mipsy mips_networking.s
255
-16777216
1521 mipsy mips_networking.s
-1
-1
1521 mipsy mips_networking.s
2147483647
-129

When you think your program is working, you can use autotest to run some simple automated tests:

1521 autotest mips_networking 

When you are finished working on this exercise, you must submit your work by running give:

give cs1521 lab07_mips_networking mips_networking.s

You must run give before Monday 15 July 12:00 (midday) (2024-07-15 12:00:00) to obtain the marks for this lab exercise. Note that this is an individual exercise, the work you submit with give must be entirely your own.

Floating-point Representation

Single-precision floating-point numbers that follow the IEEE 754 standard have an internal structure that looks like this:

The value is determined as: \[ {-1}^{\text{sign}} \times \left(1 + \text{frac}\right) \times 2^{\text{exp} - 127} \] The 32 bits in each float are used as follows:

sign

is a single bit that indicates the number's sign. If set to 0, the number is positive; if set to 1, the number is negative.

exp

is an unsigned 8-bit value (giving a range of \([0\cdots255]\)) which is interpreted as a value in the range \([-127\cdots128]\) by subtracting 127 (the bias) from the stored 8-bit value. It gives a multiplier for the fraction part (i.e., \( 2^{\text{exp}-127} \)).

frac

is a value in the range \( [0\cdots1] \), determined using positional notation: \[ \frac{\text{bit}_{22}}{2^{1}} + \frac{\text{bit}_{21}}{2^{2}} + \frac{\text{bit}_{20}}{2^{3}} + \cdots + \frac{\text{bit}_{2}}{2^{21}} + \frac{\text{bit}_{1}}{2^{22}} + \frac{\text{bit}_{0}}{2^{23}} \] The overall value of the floating-point value is determined by adding 1 to the fraction: we assume that the "fraction" part is actually a value in the range \( [1\cdots2] \), but save bits by not explicitly storing the leading 1 bit.

For example:

raw 32 bits:  01000000001000000000000000000000
partitioned:  0 10000000 01000000000000000000000
sign:         0, so positive
exp:          10000000 = 128, so multiplier is 2128-127 = 21
frac:         0×2-1 + 1×2-2 + ... = 0.25, but we need to add 1

final value:  1.25 × 21 = 2.5

Exercise Description

Your task is to add code to this function in float_bits.c:

// separate out the 3 components of a float
float_components_t float_bits(uint32_t f) {
    // PUT YOUR CODE HERE
}

The function float_bits is given the bits of a float as type uint32_t. Add code so that it returns a struct float_components, containing the sign, exponent and fraction fields of the struct.

You also should add appropriate code to the functions is_nan, is_positive_infinity, is_negative_infinity, and is_zero. All four functions take a struct float_components as their argument, and return 0 or 1 depending on some property of the float it represents:

• is_nan returns 1 iff the struct float_components corresponds to the not-a-number value, NaN, and 0 otherwise;

• is_positive_infinity returns 1 iff the struct float_components corresponds to the positive infinity, inf, and 0 otherwise;

• is_negative_infinity returns 1 iff the struct float_components corresponds to the negative infinity, -inf, and 0 otherwise; and

• is_zero returns 1 iff the struct float_components corresponds to either positive or negative zero.

These function must be implemented only using bit operations and integer comparisons.

Once these functions are completed, you should get output like:

./float_bits -42
float_bits(-42) returned
sign=0x1
exponent=0x84
fraction=0x280000
is_nan returned 0
is_positive_infinity returned 0
is_negative_infinity returned 0
is_zero returned 0
./float_bits 3.14159
float_bits(3.14159012) returned
sign=0x0
exponent=0x80
fraction=0x490fd0
is_nan returned 0
is_positive_infinity returned 0
is_negative_infinity returned 0
is_zero returned 0
./float_bits -inf
float_bits(-inf) returned
sign=0x1
exponent=0xff
fraction=0x000000
is_nan returned 0
is_positive_infinity returned 0
is_negative_infinity returned 1
is_zero returned 0

Use make(1) to build your code:

make    # or 'make float_bits'

When you think your program is working, you can use autotest to run some simple automated tests:

1521 autotest float_bits 

When you are finished working on this exercise, you must submit your work by running give:

give cs1521 lab07_float_bits float_bits.c

You must run give before Monday 15 July 12:00 (midday) (2024-07-15 12:00:00) to obtain the marks for this lab exercise. Note that this is an individual exercise, the work you submit with give must be entirely your own.

Your task is to add code to this function in float_2048.c:

// float_2048 is given the bits of a float f as a uint32_t
// it uses bit operations and + to calculate f * 2048
// and returns the bits of this value as a uint32_t
//
// if the result is too large to be represented as a float +inf or -inf is returned
//
// if f is +0, -0, +inf or -inf, or Nan it is returned unchanged
//
// float_2048 assumes f is not a denormal number
//
uint32_t float_2048(uint32_t f) {
    // PUT YOUR CODE HERE

    return 42;
}

Add code to the function float_2048 so that, given a float as a uint32_t, it multiplies that value by 2048 using only bit operations and addition (+), and returns the result. If, after multiplication, the result is too large to be represented as a float, return inf if the original float was positive, and -inf if it was negative.

Once your program is working, you should see something like:

./float_2048 1
2048
./float_2048 3.14159265
6433.98193
./float_2048 -2.718281828e-20
-5.56704133e-17
./float_2048 1e38
inf
./float_2048 -1e37
-inf
./float_2048 inf
inf
./float_2048 -inf
-inf
./float_2048 nan
nan

Use make(1) to build your code:

make    # or 'make float_2048'

When you think your program is working, you can use autotest to run some simple automated tests:

1521 autotest float_2048 

When you are finished working on this exercise, you must submit your work by running give:

give cs1521 lab07_float_2048 float_2048.c

You must run give before Monday 15 July 12:00 (midday) (2024-07-15 12:00:00) to obtain the marks for this lab exercise. Note that this is an individual exercise, the work you submit with give must be entirely your own.