Description

1. If the data segment of a particular MIPS program starts at the address 0x10000020, then what addresses are the following labels associated with, and what value is stored in each 4-byte memory cell?

2. Give MIPS directives to represent the following variables:
a. int u;
b. int v = 42;
c. char w;
d. char x = ‘a’;
e. double y;
f. int z[20];
Assume that we are placing the variables in memory, at an appropriately-aligned address, and with a label which is the same as the C variable name.
3. Consider the following memory state:
Address Data Definition
0x10010000 aa: .word 42
0x10010004 bb: .word 666
0x10010008 cc: .word 1
0x1001000C .word 3
0x10010010 .word 5
0x10010014 .word 7
What address will be calculated, and what value will be loaded into register $t0, after each of the following statements (or pairs of statements)?

4. What is a breakpoint?
When is it useful in debugging?
5. Translate this C program to MIPS assembler

6. Translate this C program to MIPS assembler

7. Translate this C program to MIPS assembler

8. Translate this C program to MIPS assembler

9. The following loop determines the length of a string, a ”-terminated character array:

Write MIPS assembly to implement this loop.
Assume that the variable string is implemented like:
Assume that the variable s is implemented as register $t0, and variable length is implemented as register $t1. And, assume that the character ” can be represented by a value of zero.
10. Conceptually, the MIPS pseudo-instruction to load an address could be encoded as something like the following:

Since addresses in MIPS are 32-bits long, how can this instruction load an address that references the data area, such as 0x10010020?
11. Implement the following C code in MIPS assembly instructions, assuming that the variables x and y are defined as global variables (within the .data region of memory):

Assume that the product might require more than 32 bits to store.
12. Write MIPS assembly to evaluate the following C expression, leaving the result in register $v0.
((x*x + y*y) – x*y) * z
Write one version that minimises the number of instructions, and another version that minimises the number of registers used (without using temporary memory locations).
Assume that: all variables are in labelled locations in the .data segment; the labels are the same as the C variable names; all results fit in a 32-bit register (i.e., no need to explicitly use Hi and Lo).
For all enquiries, please email the class account at cs1521@cse.unsw.edu.au
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