ICS 53编程讲解、辅导C++程序、c/c++程序语言调试讲解留学生Prolog|讲解SPSS

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ICS 53, Winter 2021
Assignment 4: A Memory Allocator
You will write a program which maintains a heap. Your program will allow a user to allocate
memory, free memory, and see the current state of the heap. Your program will accept user
commands and execute them.
I. Assumptions about the heap
The heap is 127 bytes long and memory is byte-addressable. The first address of the heap is
address 0, so the last address of the heap is address 126. When we refer to a “pointer” in this
assignment we mean an address in memory. All pointers should therefore be values between 0
and 126.
The heap will be organized as an implicit free list. The heap is initially completely
unallocated, so it should contain a single free block which is as big as the entire heap. Memory
is initialized so that all addresses (other than the header and the footer of the initial free block)
contain 0. Each block should have a header and a footer which is a single byte, and the header
and footer byte should be contained in memory, just before the payload of the block(header)
and after payload of a block (footer). The most-significant 7 bits of the header and footer
should indicate the size of the block, including the header and footer itself. The least
significant bit of the header and footer should indicate the allocation of the block: 0 for free,
1 for allocated. The header for the first block (the initial single free block) must be placed at
address 0 in memory and the footer of the first block must be placed at address 126.
You cannot assume that the blocks are aligned. This means that the start address of a block can
be any address in the heap.
II. Operations
Your program should provide a prompt to the user (“>”) and accept the following
commands. Your program should repeatedly accept commands until the user enters “quit”. You
only need to support these commands and can safely assume they will be entered and
formatted in the manner shown below.
1. malloc - This operation allows the user to allocate a block of memory from
your heap. This operation should take one argument, the number of bytes which the
user wants in the payload of the allocated block. This operation should print out a
pointer which is the first address of the payload of the allocated block.
Example:
>malloc 10 // Comment: header at 0, payload from 1-10,
footer at 11
1
>malloc 5 // Comment: header at 12, payload from 13-17,
footer at 18
13
>malloc 2 // Comment: header at 19, payload from 20-21,
footer at 22
20
2. free - This operation allows the user to free a block of memory. This
operation takes one argument, the pointer to the start of the payload of a previously
allocated block of memory. You can assume that the argument is a correct pointer to
the payload of an allocated block.
Example:
>malloc 10
1
>malloc 5
13
>free 13
>free 1
3. blocklist - This operation prints out information about all the blocks in your heap. The
information about blocks should be printed in the order that the blocks are contained in
the heap. The following information should be printed about each block: pointer to the
start of the payload, payload size, and the allocation status (allocated of free). All
three items of information about a single block should be printed on a single line and
should be separated by commas.
Example:
>malloc 10
1
>malloc 5
13
>blocklist
1, 10, allocated.
13, 5, allocated.
20, 106, free.
4. writemem , – This operation writes alpha-numeric characters
into memory. The operation takes two arguments. The first argument is a pointer to the
location in memory and the second argument is a sequence of alpha-numeric characters
which will be written into memory, starting at the address indicated by the pointer. The
first character will be written into the address indicated by the pointer, and each
character thereafter will be written into the neighboring addresses sequentially. For
example, the operation “writemem 3 abc” will write an ‘a’ into address 3, a ‘b’ into
address ‘4’, and a ‘c’ into address 5. Additionally, if a block is freed, you must ensure
that whatever was written to the block is reset back to 0. The index can be any location
in the heap. Since C does not do explicit bounds-checking, the writmem command for
your simulator can very well overwrite parts of another blocks even headers and
footers and payloads of other allocated/free blocks. You are not required to do
bounds checking for your heap as well. Operation like these will destroy the heap
structure and are permitted operations. This can then corrupt subsequent operations
as well.
5. printmem , – This operation prints
out a segment of memory in hexadecimal. The operation takes two arguments. The first
argument is a pointer to the first location in memory to print, and the second argument
is an integer indicating how many addresses to print. The contents of all addresses will
be printed on a single line and separated by a single space. The index along with
number of characters can very well exceed block sizes. Like writemem, you are not
required to do bounds checking.
Example:
>writemem 5 ABC
>printmem 5 3
41 42 43
Notice that the values 41, 42, and 43 are the hexadecimal representations of the ASCII
values of the characters ‘A’, ‘B’, and ‘C’.
6. quit – This quits your program.
III. Requirements about allocation and freeing of memory.
When a block is requested which is smaller than any existing block in the heap, then your
code must perform splitting to create a new block of the appropriate size.
When a block is freed, it must be coalesced with the next block if the next block is free
and the previous block if the previous block is free. (Forward and Backward Coalescing needs
to be supported by your code)
When searching for a block to allocate, be sure to use the first-fit allocation strategy.
IV. Developing Testcases
Testing is an important part of the programming process, so you are required to develop
your own test cases. It is up to you to consider all the possibilities that can occur, within the
limits of the specification, and make your test cases based on that. We will grade your test
cases based on coverage, i.e., you should come up with the test cases such that the whole
suite of test cases is capable of 100% code coverage, if you have a question about how the
program is expected to perform, come to office hours and ask the professor, or go to lab and
ask a TA. We have created a Python script called autocov.py which you should download with
the assignment and use before you submit your assignment in order to ensure that you have
achieved 100%-line coverage. The autocov.py script will compile and execute your code with a
set of test cases which you define, and it will report the line coverage achieved by the test case.
Do not modify the autocov.py script which we provide.
Test Runs
Testing is performed by executing your program several times with different test data.
Each test execution of your code is referred to as a test run (“run” for short) and the test
data associated with each run must be stored in a .run file. A .run file will contain all the
test data which will be supplied to your program during a single test execution. A .run
file is a text file describing all the inputs supplied to your program for a single execution.
The name of each .run file must have the suffix “.run”, but the prefix does not matter.
Each line of a .run file can be no longer than 80 characters.
Input to autocov.py.
The autocov.py script requires your access to the source code of your program and the
set of .run files to be used during testing. Your source code file must be named “hw.c”.
This is important; it is the only file which will be compiled by the script. Any number of
.run files may be provided. All these files, hw.c and all of your .run files, must be
“zipped” together into a single zip file named “hw.zip”. You can create the zip file using
the “zip” command on linux. For example, if you have two .run files called “t1.run” and
“t2.run” then you could create the zip file with the following command: “zip hw.zip hw.c
t1.run t2.run”. The hw.zip file is the only input to the autocov.py script.
Running autocov.py
In order to execute the script, the autocov.py file must be placed in the same directory
as the hw.zip file and another directory called “hw”. The hw directory will be used by
the script to compile and run your program, so it should be empty before running the
script. The script needs to be run using a Python interpreter version 3.6.8 or later. On
the openlab systems the command “python3” invokes the Python 3.x interpreter, so you
can run the autocov.py script by entering the command “python3 autocov.py”. The
script will print several lines to the screen, including program outputs, but the secondto-last
line should look like this, “Lines executed: X% of Y”, where X is the line coverage
and Y is the number of lines in your program.
V. Submission Instructions:
Your source code must be a single c file named ‘hw.c’ containing your solution and any number
of run files containing your suite of test cases zipped into 1 file named ‘hw.zip’. Be sure that
your program must compile on openlab.ics.uci.edu using gcc version 4.8.5 with no compiler
switches other than gcov switches.
Submissions will be done through Gradescope. If you are working with a partner, you must
indicate that through Gradescope. The first line of your submitted file should be a comment
which includes the name and ID number of you and your partner (if you are working with a
partner).
VI. Sample Execution
>malloc 10
1
>malloc 5
13
>blocklist
1, 10, allocated.
13, 5, allocated.
20, 106, free.
>free 1
>blocklist
1, 10, free.
13, 5, allocated.
20, 106, free.
>malloc 5
1
>blocklist
1, 5, allocated.
8, 3, free.
13, 5, allocated.
20, 106, free.
>writemem 1 HELLO
>printmem 1 5
48 45 4C 4C 4F
>free 13
>blocklist // Comment: You can see here that there was a
block allocated between 2 free blocks, on freeing that allocated
block, both the free blocks were coalesced into one free block.
1, 5, allocated.
8, 118, free.
>quit
$ <- back to bash prompt.

ICS 53编程讲解、辅导C++程序、c/c++程序语言调试 讲解留学生Prolog|讲解SPSS

ICS 53编程讲解、辅导C++程序、c/c++程序语言调试讲解留学生Prolog|讲解SPSS