Profil: hazal7291

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hazal7291 (göksu emin)
Java altına konu açtı.

9 yıl

ARKADAŞLAR wifi sense ile ilgili bi network proje ödevim var...ücret karşılığı yaptırmak istiyorum...ilgilenen olursa özelden yazsın lütfen...

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hazal7291 (göksu emin)
PHP / CGI altına konu açtı.

9 yıl

PHP PROJESİ

ARKADAŞLAR ÖĞRENCİ yurtları için bir php proje ödevim var...ücret karşılığı yaptırmak istiyorum...ilgilenen olursa özelden yazsın lütfen...

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hazal7291 (göksu emin)
Projeler altına konu açtı.

11 yıl

PROJEME YARDIM EDEBİLECEKLER VAR MI?

arkadaşlar merhaba,
bi projem var çarşamba gününe kadar teslimi var acil yardım edin yoksa dersten kalıcam:((

C kodu üzerinde gidilerek lex ve yacc da yapılacak
ilgilenen olursa özelden projeyi atayım?

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hazal7291 (göksu emin)
Projeler altına konu açtı.

11 yıl

assembly PROJESİ yapan var mı

assembly dersi alıyorum 1 hafta sonra proje teslimi var.

proje motorola 6800 ve shell script ile ilgili...

ücreti ile projemi yaptırmak istiyorum...
PROJE:

Part 1
Your parser should be able to read a M6800 assembly language code given in a file, and parse it into its tokens. After parsing the
tokens, you should print the type of each token: label, operation (opcode), operand or reference. You should also print the
addressing mode of the instruction. An example input-output is given below:
Input:
LDAA 0H
LDX #10H
LOOP: STAA 10H,X
INX
BRA LOOP
Output:
LDAA – operation
0H – operand
* Direct Mode *
LDX – operation
10H operand
* Immediate Mode *
LOOP – label
STAA – operation
10H – operand
X – denotes indexed mode
* Indexed Mode *
INX – operation
* Inherent Mode *
BRA – operation
LOOP – reference
* Relative Mode *
Assume labels consist of only one word, and have an adjacent semicolon at the end. Other than that, your program should be able
to deal with whitespaces. Also, don’t forget that ‘#’ denotes immediate values, and all numerical values without an H are in
decimal form.
Your parser should read the input from the file character by character until a separation character (such as “:” , “,” , whitespace
character, end of line character, etc.). Determine the type of this token and print it. Then repeat these steps until the end of file is
reached.
While implementing this part, keep in mind that you will need the parser in Part 2. You won’t need the exact output produced in
Part 1, but you will need to parse the instructions into operations, operands, labels and references; and you will need the
addressing modes.
Part 2
In Part 2 you will implement the interpreter. Your interpreter must follow the steps below:
1. Initially read the code from the input file line-by-line and store it in a text buffer array. During this process, prepare SDT. Use
the following structures for text buffer and SDT.Text Buffer
IP 0 L1: LDAA 0H
1 LDX #10H
2 LOOP: STAA 10H,X
3 INX
4 BRA LOOP
struct txt {
char line[20];
}
struct txt textbuffer[100];
SDT
Text buffer index
L1 0
LOOP 2
struct st {
char label[5];
int textindex;
}
struct st sdt[20];
Then, set instruction pointer (IP) to 0 and start interpretation.
2. Using your parser in Part 1, parse the instruction pointed by IP.
3. Check instruction type and call the related execute routine to process it. (You will write an execute routine for each
instruction.)
4. Each execute routine must check the addressing mode of the instruction. Then, it must process the instruction and update the
contents of the following units: memory, accumulators A and B, index register, CCR, etc. It must also display the contents of
these units.
5. Set IP as follows and go to Step 2.
IP  IP+1 (for sequential) or
IP  SDT entry for the label (for branch)
Use the following structures for the memory, accumulators A and B, index register, CCR.
int A; // Accumlator A
int B; // Accumlator B
int IX; // Index register
int IP; // Instruction Pointer
char CCR; // Condition Code Register
int Memory[4096] // Memory
Implement the following instructions and addressing modes:
Symbol Addressing modes
LDAA, LDAB Immediate, direct
ASLA, ASLB Inherent
ASRA, ASRB Inherent
STAA, STAB Direct
ADDA, ADDB Immediate, Direct
SUBA, SUBB Immediate, Direct
INCA, INCB Inherent
DECA, DECB Inherent
INC Direct
CMPA, CMPB Immediate, Direct
BRA Relative
BEQ Relative
BPL Relative
Upon starting, your program must assume a blank memory, setting all memory locations to 0; accumulators will also contain 0.
During the interpretation, your program must parse each instruction, process the instruction, and display the contents of the
accumulators, CCR and memory location in the following format:
A= … B= … CCR= … M[..]= …
Ex: LDAA 100H
A= -4 B= 12 CCR= 1101000 M[100]= -4
There must be at least one loop structure in your input M6800 code

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hazal7291 (göksu emin)
Projeler altına konu açtı.

12 yıl

data projesi ücretli acil

arkadaşlar aranızda data projesi yapmak isteyen var mı?
konu şu : kullanıcı big integer giricek...big integer mesela 30 basamaklı bir sayı olsun...
bu sayıları toplayıp çıkarabileceğimiz bir proje olucak...bir de hoca liked list kullanmamızı istiyor...
yardımcı olabilirmisiniz? yoksa dersten kalıcam:(
belli bi ücret de anlaşabilirsek lütfen bana yazın pazar günü teslim etmem gerekli!