$29
Project 2 Dot-Matrix Printer (Vertical) Solution
The purpose of this project is for you to practice writing assembly language to interact with output hardware. The hardware for this project is a primitive dot-matrix printer. The Dot-Matrix Printer (Mars Tool) that we are going to use for this project is shown below. This tool can be found in DotMatrixPrinterVerticalRegister.zip located in the CourseWeb under this project. Extract all les to your [..]/mars4 5/mars/tools directory. If you extract all les to the right directory, when you run the MARS program, you should see "Dot-Matrix Printer - Vertical (Register) V0.1" under the "Tools" menu.
Introduction to the Dot-Matrix Printer (Mars Tool)
The Dot-Matrix Printer - Vertical (Mars Tool) is a very primitive printer. The print head consists of only 8 pins arranged as a vertical line. Therefore, it can only print at most eight vertical dots at a time. This print head only prints when it moves from left to right. Once the print head moves all the way to the right, it will stop printing, move the print head back all the way to the left, feed the paper up just a little bit (equal to the height of eight dots), and then continue printing. For this printer, each row consists of 480 dots.
To print to this printer, we have to write an 8-bit data to the register $t8 and then set the register $t9 to 1 to tell the printer that the data is available to be printed. When the printer nishes printing the given 8-bit data, it will set the register $t9 to 0 to let you know that it is ready to receive the next data. An 8-bit data will be mapped to 8 dots on the paper. A 1 represents a black dot, and a 0 represents no dots. For example, consider the following program:
.text
add
$t9, $zero, $zero
# Clear $t9 to 0
ori
$s0, $zero, 0xff
# $s0 contains 8-bit data to be printed
add
$s1, $zero, $zero
# Counter
addi
$s2, $zero,
8
# $s2 = 8
loop:
beq
$s1, $s2, done
# Check whether counter
== 8
add
$t8, $zero, $s0
# Set $t8
to 8-bit data
to be printed
addi
$t9, $zero, 1
# Set $t9
to 1 to print
wait: bne
$t9, $zero, wait
# Wait until $t9 is back to 0
srl
$s0, $s0, 1
# Change the 8-bit data
by shifting it to the right
by 1
addi
$s1, $s1, 1
# Increase the counter by 1
j
loop
# Go back
to loop
done:
addi
$v0, $zero, 10
# Syscall
10: Terminate
program
syscall
# Terminate the program
The result of the above program is shown below:
As you may notice, the MSB of an 8-bit data corresponds to the dot at the top and the LSB corresponds to the dot at the bottom. Let’s look at the data that we set to the register $t8 iteration-by-iteration:
10000000
11000000
11100000
11110000
11111000
11111100
11111110
11111111
^ ^
| |
+- $t8 (last iteration) is 0x01 +- $t8 (first iteration) is 0xff
So, every time you set the register $t8 to something and set $t9 to 1, the printer will print one vertical line consisting of eight dots. The next time you set $t9 to 1 again, it will print another vertical line next to the previous line. However, if the printer already print the right-most vertical line, it will move the print head back all the way to the left, feed the paper up just a little bit (equal to the height of eight dots).
Horizontally, there are total of 480 dots. To print the whole line, you need to send the total of 480 8-bit data to the printer. From the above code example, instead of terminate the program, if you set $t8 to 0 and send it to the printer (set $t9 to 1 and wait until it is back to 0) for
8 = 472 times, the print head will be right under the rst vertical line (all the way to the
left).
The \Clear" button is used to reset the printer. Note that you should stop your program before click the \Clear" button. Otherwise, the printer may keep printing.
Let’s look at another example. Consider the following code:
.text
add
$t9,
$zero,
$zero
# Clear
$t9 to 0
add
$s0,
$zero,
$zero
# $s0 is a counter
addi
$s1,
$zero,
480
# $s1 =
480
ori
$t8,
$zero,
0xaa
# $s0 =
0...010101010
loop1: beq
$s0,
$s1, done1
# Check
whether $s0
== $s1
addi
$t9,
$zero,
1
# Set $t9 to 1 (to print)
wait1: bne
$t9,
$zero,
wait1
# Wait until $t9 is
0
addi
$s0,
$s0, 1
# Increase counter by 1
j
loop1
# Go back to loop 1
done1: add
$s0,
$zero,
$zero
# $s0 =
0 (counter)
ori
$t8,
$zero,
0xf3
# $s0 =
0...011110011
loop2: beq
$s0,
$s1, done2
# Check
whether $s0
== $s1
addi
$t9,
$zero,
1
# Set $t9 to 1 (to print)
wait2: bne
$t9,
$zero,
wait2
# Wait until $t9 is
0
addi
$s0,
$s0, 1
# Increase counter by 1
j
loop2
# Go back to loop 1
done2: addi $v0,
$zero,
10
# Syscall 10: Terminate program
syscall
# Terminate the program
The result of the above program is shown below:
IMPORTANT Make sure you are familiar with this printer hardware. Think about a simple pattern and try to write a program to print it to ensure that you understand how to control this printer correctly. Do not forget to press the \Clear" button before you run your program.
What you need to do for this project is to print an image onto this printer hardware. The image le format that we are going to use is a simple BMP (indexed with only black and white) format.
Read Data From a File
This project requires you to read data from bitmap (BMP) les. So, let’s see how to read data from a le in MIPS assembly.
In this section, we are going to learn how to read the le named testFile.bin which contains a 14-byte data as shown below
42 4D 7B 00 00 00 0B 00 16 00 C8 01 00 00
The rst two bytes contains two character ’B’ (0x42 in ASCII) and ’M’ (0x4D in ASCII). The next four bytes is a 4-byte integer. Note that MIPS is a little-endian. The four byte integer shown above is 7B 00 00 00 which is 0x0000007B in little-endian format. This number is 123 in decimal. The next two bytes is a 2-byte integer 0x000B (again in little-endian) which is 11 in decimal. The next two byte is another 2-byte integer 0x0016 which is 22 in decimal. The last four bytes is a 4-byte integer 0x000001C8 which is 456 in decimal.
To read a le, we need to open the le rst. This can be done by system call number 13. To open a le, set $v0 to 13, set $a0 to the address of null-terminated string containing lename, set $a1 to 0, set $a2 to 0, and execute the syscall instruction. For example, to open the le test.bin, simply use the following sequence of instructions:
.data
filename:
.asciiz "testFile.bin"
.text
addi $v0,
$zero, 13
# Syscall 13: Open file
la
$a0,
filename
# $a0
is the address of filename
add
$a1,
$zero, $zero
# $a1
= 0
add
$a2,
$zero, $zero
# $a2
= 0
syscall
# Open file
add
$s0,
$zero, $v0
# Copy the file descriptor to $s0
The result is a positive integer represents a le descriptor in $v0. Note that if the le descriptor in $v0 is a negative value, it means the program cannot open the le. Most likely cause of this problem is the le cannot be found. You have to make sure that the le that you want to open is located in the root directory of your Mars program. So, it is a good idea to always check the le descriptor. Your program should notify user and possibly terminate the program if the le descriptor is negative.
Once the le is opened, you can simply read the le using the system call 14 with the returned le descriptor. The system call 14 needs three arguments; (1) the le descriptor in $a0, (2) the address of input bu er in $a1, and (3) the maximum number of characters (bytes) to read in $a2. The following sequence of instructions read the rst two bytes of the le testFile.bin, put the data in the bu er named firstTwo assuming that the le is successfully opened and the le descriptor is stored in $s0, and print the rst two bytes as character using syscall number 11:
.data
firstTwo:
.space 2
.text
:
addi $v0,
$zero, 14
# Syscall 14: Read file
add
$a0,
$zero, $s0
# $a0 is the
file descriptor
la
$a1,
firstTwo
# $a1 is the
address of a buffer (firstTwo)
addi $a2,
$zero, 2
# $s2 is the
number of bytes to read
syscall
# Read file
la
$s1,
firstTwo
# Set $s1 to
the address of firstTwo
addi $v0,
$zero, 11
# Syscall 11: Print character
lb
$a0,
0($s1)
# $a0 is the
first byte of firstTwo
syscall
# Print a character
lb
$a0,
1($s1)
# $a0 is the
second byte of firstTwo
syscall
# Print a character
Now, we are going to read the next 12 bytes in one system call. Note that there are two 4-byte integers (words). For simplicity, we need a 12-bytes bu er but it MUST align with word boundary so that we can simply load the whole word at once using the lw instruction. The following sequence of instruction read the last twelve bytes and print them out according to data size:
.data
:
.align 2
lastTwelve: .space
12
.text
:
addi $v0,
$zero, 14
# Syscall 14: Read
file
add
$a0,
$zero, $s0
# $a0 is the
file descriptor
la
$a1,
lastTwelve
# $a1 is the
address of a buffer (lastTwelve)
addi $a2,
$zero, 12
# $s2 is the
number of bytes to read
syscall
# Read file
la
$s1,
lastTwelve
# Set $s1 to
the address of lastTwelve
addi $v0,
$zero, 1
# Syscall 1:
Print
integer
lw
$a0,
0($s1)
# $a0 is the
first
4-byte integer
syscall
# Print an integer
lh
$a0,
4($s1)
# $a0 is the
first
2-byte integer
syscall
# Print an integer
lh
$a0,
6($s1)
# $a0 is the
second 2-byte integer
syscall
# Print an integer
lw
$a0,
8($s1)
# $a0 is the
second 4-byte integer
syscall
# Print an integer
When you are done reading the le, the le must be closed. This can be done by system call 16. Simply set the le descriptor of the le to be closed in $a0, set $v0 to 16, and execute the syscall instruction as shown below:
add
$v0, $zero, 16
# Syscall 16: Close file
add
$a0, $zero, $s0
#
$a0 is the
file descriptor
syscall
#
Close file
The complete code of the above program readFileExample.asm including the le testFile.bin can be found on the CourseWeb under this project.
Introduction to BMP File Format
The rst two bytes of a BMP le should begin with the magic numbers 0x42 followed by 0x4D. These two numbers are simply the character ’B’ followed by the character ’M’. The next 12 bytes is a header that contains the following data:
O set
Size (bytes)
Description
0x02
4
Size of this BMP le in bytes
0x06
2
Reserved
0x08
2
Reserved
0x1A
4
O set of the bitmap data
Note that o set 0x02 means 2 bytes from the beginning of the le. So, if we put the magic numbers together with the head it will look like the following:
0x00
0x02
0x06
0x08
0x0A
0x0E
B
M
Size in Bytes
Reserved
Reserved
Offset (bitmap data)
Note that BMP les use little-endian format which compatible with MIPS. So, to read a 4-byte data as an integer, simply read the whole world as shown in previous section. Be careful with alignment. You may need to use .align to ensure that your bu er aligned with word boundary.
Immediately after the header at the o set 0x0E, it is the beginning of Bitmap Information (DIB) header. A DIB header does not have a speci c size. Therefore, the rst four bytes of a DIB header contains an integer value represents the size of the DIB header (including this 4 bytes). This will allow programmer to correctly allocate memory and read the whole DIB header. Note that you need to allocate memory using system call number 9 using the size speci ed in the rst four bytes of the DIB header but you need to subtracted by 4 since the size includes the rst four bytes. Similar to the header, DIB header contain bitmap information as shown below:
O set
Size (bytes)
Description
0x0E
4
Size of DIB header
0x12
4
Image width (in number of pixels)
0x16
4
Image height (in number of pixels)
0x1A
2
Number of color planes being used
0x1C
2
Number of bits per pixel
0x1E
4
Compression method (0 for no compression)
0x22
4
Size of the raw bitmap data
0x26
4
Horizontal resolution in pixels per meter
0x2A
4
Vertical resolution in pixels per meter
0x2E
4
Number of colors in the color palette
0x32
4
Number of important colors used
6
Immediately after the DIB header is an array of colors. A color consists of four value (one byte each), blue, green, red, and alpha. Since we are working with 1-bit color (black and white), the array of colors should only contains two colors, black (0, 0, 0, 0) and white (255, 255, 255, 0). Note that the order of colors does matter. If black is the rst color, it means index 0 is for black and index 1 is for white.
For better understanding of BMP le (indexed and black and white), let’s consider the black and white image below:
The above image is a 16 16 black and white image (testPattern01.bmp). Green lines are used to separate pixels. The content of the le is as follows:
0x00
|
42
4D
C2
00
00
00
00
00 00 00
82
00
00
00
6C
00
0x10
|
00
00
10
00
00
00
10
00 00 00
01
00
01
00
00
00
0x20
|
00
00
40
00
00
00
13
0B 00 00
13
0B
00
00
02
00
0x30
|
00
00
02
00
00
00
42
47 52 73
00
00
00
00
00
00
0x40
|
00
00
00
00
00
00
00
00 00 00
00
00
00
00
00
00
0x50
|
00
00
00
00
00
00
00
00 00 00
00
00
00
00
00
00
0x60
|
00
00
00
00
00
00
00
00 00 00
02
00
00
00
00
00
0x70
|
00
00
00
00
00
00
00
00 00 00
FF FF FF 00
00 00
0x80
|
00
00
AA 00 00
00 AA 00 00 00 AA 3C 00 00
AA 3C
0x90
|
00
00
AA 3C 00
00 AA 3C 00 00 AA 00 00 00
AA 00
0xA0 |
00
00
0F 00
00
00
0F
FF 00 00 0F 00 00 00
0F FF
0xB0 |
00
00
F0 00
00
00
F0
FF 00 00 F0 00 00 00
F0 FF
0xC0 |
00
00
Numbers on the left column are o sets from the beginning of the le. Now, consider the rst 14 bytes (the magic numbers and header)
O set
Size (bytes)
Short Description
Hexadecimal (little endian)
Decimal
0x00
2
Magic Number
42 4D
BM
0x02
4
File Size
C2 00 00
00
194
0x06
2
Reserved
00 00
0
0x08
2
Reserved
00 00
0
0x0A
4
O set (data)
82 00 00
00
130
The above information tells us that this is a BMP le (magic number), the le size is 194 bytes, and the bitmap data is located at 130 byte from the beginning of the le.
The next four bytes at the index 0x0E represents the size of DIB header. In this case, the value (hexadecimal little endian) is 6C 00 00 00 which is 108 in decimal. Note that once you see the size of the DIB header, you should read the whole DIB header at once for simplicity. Note that the number of bytes to read is now 108 - 4 = 104 since we already read the rst four bytes. So, let’s put the rst 40 bytes of the DIB header into a table, we have
7
O set
Size (bytes)
Short Description
Hexadecimal (little endian)
Decimal
0x0E
4
DIB size
6C 00 00 00
108
0x12
4
Width
10 00 00
00
16
0x16
4
Height
10 00 00
00
16
0x1A
2
Color plane
01 00
1
0x1C
2
Bit per pixel
01 00
1
0x1E
4
Compression Method
00 00 00
00
0
0x22
4
Size of bitmap data
40 00 00
00
64
0x26
4
Horizontal resolution
13 0B 00 00
2835
0x2A
4
Vertical resolution
13 0B 00 00
2835
0x2E
4
Number of colors
02 00 00
00
2
0x32
4
Colors used
02 00 00
00
2
Note that the above information stated that the number of colors is 2. So, the next 4 2 = 8 bytes is the array of colors (4 bytes for each color). In this case, the o set of the color array is 2 + 12 + 108 = 122 = 0x7A bytes from the beginning of the le. At 0x7A, we have two colors FF
FF 00 and 00 00 00 00. This states that the color at index 0 is white and the color at index 1 is black. Now we are at the beginning of the bitmap data at the o set 0x82 which is also stated in the header at o set 0x0A.
For the bitmap data, each row of pixels must be a multiple of four bytes. An easy way to nd the number of bytes for each row of pixel is to divide the size of the raw bitmap data (stated at o set 0x22) by the image height (stated at o set 0x16). In this example, the size of bit map data is 64 and the image height is 16. Thus, each row of pixels is 64=16 = 4 bytes. IMPORTANT, the bitmap data is reversed. The rst row of the image (top row) is at the end of the bitmap data and the last row of the image (bottom row) is at the beginning of the bitmap data. Let’s put all 64 bytes of bitmap data in the reverse order (of row) together with its binary representation and the actual image:
Offset
| Hexadecimal
| Binary
0xBE
|F0FF0000
| 1111000011111111 0...0
0xBA
|F0000000
| 1111000000000000 0...0
0xB6
|F0FF0000
| 1111000011111111 0...0
0xB2
|F0000000
| 1111000000000000 0...0
0xAE
|0FFF0000
| 0000111111111111 0...0
0xAA
|0F000000
| 0000111100000000 0...0
0xA6
|0FFF0000
| 0000111111111111 0...0
0xA2
|0F000000
| 0000111100000000 0...0
0x9E
|AA000000
| 1010101000000000 0...0
0x9A
|AA000000
| 1010101000000000 0...0
0x96
|AA3C0000
| 1010101000111100 0...0
0x92
|AA3C0000
| 1010101000111100 0...0
0x8E
|AA3C
00
00
| 1010101000111100 0...0
0x8A
|AA3C
00
00
| 1010101000111100 0...0
0x86
|AA00
00
00
| 1010101000000000 0...0
0x82
|AA00
00
00
| 1010101000000000 0...0
Recall that index 0 is white and index 1 is black. Do you see how the image on the right is represented by the bitmap data on the left?
8
Note that this description about BMP le format is only for indexed black and white. There are other di erent format BMP format. So, test your program with given BMP les only. Your program may not work with other BMP le.
What to do?
For this project, there are two parts:
(50 points) Ask a user to enter a name of a BMP le and display information about the given BMP le on MARS console screen. The following is an example of the output when a user enter an image from previous section testPattern01.bmp:
Enter a filename: testPattern01.bmp
The first two characters: BM
The size of the BMP file (bytes): 194
The starting address of image data: 130
Image width (pixels): 16
Image height (pixels): 16
The number of color planes: 1
The number of bits per pixel: 1
The compression method: 0
The size of raw bitmap data (bytes): 64
The horizontal resolution (pixels/meter): 2835
The vertical resolution (pixels/meter): 2835
The number of colors in the color palette: 2
The number of important colors used: 2
The color at index 0 (B G R): 255 255 255
The color at index 1 (B G R): 0 0 0
Note that your program must work with all given BMP les.
(50 points) Print the image of all given BMP les to the printer hardware. For example, the output of the le testPattern01.bmp should be as shown below:
Requirements
The following are requirements that you must follow:
Your program should be named printer.asm
Since the Dot-Matrix Printer - Vertical (Mars Tool) uses registers $t8 and $t9, do not use these registers for any other purpose.
9
HINTS
For simplicity, you should separate your program into various functions based on its func-tionality. Focus on register sharing since you may need quite a number of registers. Make sure you follow all calling conventions.
DO NOT FORGET that the bitmap data of a BMP le is in a reverse order. The last row comes rst and the rst row comes last.
Do not forget about the color indexes. 0 may be black or white depending on the index of color. Check the color index rst.
START EARLY
Submission
The due date of this project is stated on the CourseWeb. Late submissions will not be accepted.
You should submit the le printer.asm via CourseWeb.
10
