Difference between revisions of "Assembly Language Programming"

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Programs written in high-level languages such as Java and C++ are usually converted by compilers into assembly language (which in turn is translated into machine language programs - sequences of 1’s and 0’s – by an assembler). Even today, with very good quality compilers available, there is the need for programmers to understand assembly language.  First, it provides programmers with a better understanding of the compiler and its constraints.  Second, on occasion, programmers find themselves needing to program directly in assembly language in order to meet constraints in execution speed or space (for example, writing games for micros).  
Programs written in high-level languages are traditionally converted by compilers into assembly language, which is turned into machine language programs sequences of 1’s and 0’s – by an assembler. Even today, with very good quality compilers available, there is the need for programmers to understand assembly language.  First, it provides programmers with a better understanding of the compiler and its constraints.  Second, on occasion, programmers find themselves needing to program directly in assembly language in order to meet constraints in execution speed or space.  


ACSL chose to define its own assembly language rather than use a “real” one in order to eliminate the many sticky details associated with real languages.  The basic concepts of our ACSL topic description are common to all assembly languages.   
ACSL chose to define its own assembly language rather than use a “real” one in order to eliminate the many sticky details associated with real languages.  The basic concepts of our ACSL topic description are common to all assembly languages.   


== Reference Manual==
== Reference Manual ==


Execution starts at the first line of the program and continues sequentially, except for “branch” instructions, until the “end” instruction is encountered.  The result of each operation is stored in a special word of memory, called the “accumulator” (ACC).  Each line of an assembly language program has the following fields (lower-case indicates optional components):
Execution starts at the first line of the program and continues sequentially, except for branch instructions (BG, BE, BL, BU), until the end instruction (END) is encountered.  The result of each operation is stored in a special word of memory, called the “accumulator” (ACC).  The initial value of the ACC is 0. Each line of an assembly language program has the following fields:


  label OPCODE LOC comments
  LABEL OPCODE LOC


The ''label'' is a character string beginning in the first column.  Valid OPCODE’s are listed in the chart below. The LOC field is either a reference to a label or ''immediate data''.  For example, “LOAD A” would put the contents referenced by the label “A” into the ACC; “LOAD =123” would store the value 123 in the ACC.  Only those instructions that do not modify the LOC field can use the “immediate data” format.  In the following chart, they are indicated by an asterisk in the first column.
The LABEL field, if present, is an alphanumeric character string beginning in the first column. A label must begin with an alphabetic character(A through Z, or a through z), and labels are case-sensitive.  Valid OPCODEs are listed in the chart below; they are also case-sensitive and uppercase. Opcodes are reserved words of the language and (the uppercase version) many not be used a label. The LOC field is either a reference to a label or ''immediate data''.  For example, “LOAD A” would put the contents referenced by the label “A” into the ACC; “LOAD =123” would store the value 123 in the ACC.  Only those instructions that do not modify the LOC field can use the “immediate data” format.  In the following chart, they are indicated by an asterisk in the first column.


{| class="wikitable" style="text-align: left"|
{| class="wikitable" style="text-align: left"|
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!OP CODE
!OP CODE
!DESCRIPTION
!DESCRIPTION
|-
|-
!*LOAD
!*LOAD
|Contents of LOC are placed in the ACC. LOC is unchanged.
|The contents of LOC are placed in the ACC. LOC is unchanged.
 
|-
|-
!*STORE
!STORE
|Contents of LOC are placed in the LOC. ACC is unchanged.
|The contents of the ACC are placed in the LOC. ACC is unchanged.
 
|-
|-
!*ADD
!*ADD
|Contents of LOC are added to the contents of the ACC.  The sum is stored in the ACC. LOC is unchanged. Addition is modulo 1,000,000.
|The contents of LOC are added to the contents of the ACC.  The sum is stored in the ACC. LOC is unchanged. Addition is modulo 1,000,000.
 
|-
|-
!*SUB
!*SUB
|
|The contents of LOC are subtracted from the contents of the ACC.  The difference is stored in the ACC.  LOC is unchanged.  Subtraction is modulo 1,000,000.
Contents of LOC are subtracted from the contents of the ACC.  The difference is stored in the ACC.  LOC is unchanged.  Subtraction is modulo 1,000,000.


|-
|-
!*MULT
!*MULT
|
|The contents of LOC are multiplied by the contents of the ACC.  The product is stored in the ACC.  LOC is unchanged.  Multiplication is modulo 1,000,000.
The contents of LOC are multiplied by the contents of the ACC.  The product is stored in the ACC.  LOC is unchanged.  Multiplication is modulo 1,000,000.


|-
|-
!*DIV
!*DIV
|
|The contents of LOC are divided into the contents of the ACC.  The signed integer part of the quotient is stored in the ACC.  LOC is unchanged.
Contents of LOC are divided into the contents of the ACC.  The signed integer part of the quotient is stored in the ACC.  LOC is unchanged.
.
.
|-
|-
!BG
!BG
|
|
Branch to instruction labeled with LOC if ACC>0.
Branch to the instruction labeled with LOC if ACC>0.


|-
|-
!BE
!BE
|
|
Branch to instruction labeled with LOC if ACC=0.
Branch to the instruction labeled with LOC if ACC=0.
 
|-
|-
!BL
!BL
|
|
Branch to instruction labeled with LOC if ACC<0.
Branch to the instruction labeled with LOC if ACC<0.
 
|-
|-
!BU
!BU
|Branch to instruction labeled with LOC  
|Branch to the instruction labeled with LOC.
 
|-
|-
!READ
!READ
Line 64: Line 68:
|
|
Print the contents of LOC.
Print the contents of LOC.
|-
|-
!DC
!DC
|
|
The value of the memory word defined by the LABEL field is defined to contain the specified constant.  The LABEL field is mandatory for this opcode.  The ACC is not modified.
The value of the memory word defined by the LABEL field is defined to contain the specified constant.  The LABEL field is mandatory for this opcode.  The ACC is not modified.
|-
|-
!END
!END
|  
|  
Program terminates.  LOC field is ignored.
Program terminates.  LOC field is ignored and must be empty.
|}
|}


== Sample Problems ==
== Sample Problems ==


=== Sample #1 ===
=== Problem 1 ===


After the following program is executed,  
After the following program is executed,  
Line 107: Line 113:
'''Solution:''' The ACC takes on values -2, -6, 2, -1, and -9 in that order. The last value, -9, is stored in location TEMP.
'''Solution:''' The ACC takes on values -2, -6, 2, -1, and -9 in that order. The last value, -9, is stored in location TEMP.


=== Sample #2 ===
=== Problem 2 ===


If the following program has an input value
If the following program has an input value
Line 137: Line 143:


'''Solution:''' This program loops between labels TOP and
'''Solution:''' This program loops between labels TOP and
DONE for A times.  A has an initial value of X and subsequent terms of N, then values of A-1, A-2, …,
DONE for A times.  A has an initial value of X and subsequent terms of N,
1.  Each time through the loop, X is multiplied by
then values of A-1, A-2, …, 1.  Each time through the loop,
the the current value of A. Thus, X=A * (A-1) *  
X is multiplied by the the current value of A.
(A-2) * … * 1 or X=A! or A factorial. For
Thus, X = A * (A-1) * (A-2) * … * 1 or X=A! or A factorial.
example, 5! = 5 * 4 * 3 * 2 * 1 = 120.  Since the
For example, 5! = 5 * 4 * 3 * 2 * 1 = 120.  Since the
initial value of A is the number input (i.e., N),
initial value of A is the number input (i.e. N),
the algebraic expression is X = N!.
the algebraic expression is X = N!.


== Video Resources ==
== Video Resources ==


<!--
The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
 
{|
{|
|-
|-
| <youtube width="300" height="180">URL</youtube>
| <youtube width="300" height="180">https://youtu.be/jn2TJ-sHVzY</youtube>
| [URL ''TITLE'' ('''AUTHOR''')]
| [https://youtu.be/jn2TJ-sHVzY ACSL Math: Assembly Language Programming (Quick Coding Bytes)]


DESCRIPTION
This video introduces the topic, then using an example problem, explains the methodology to solve problems that appear on ACSL contests.
|}
-->


The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
{|
|-
|-
| <youtube width="300" height="180">https://youtu.be/RUm3iHsbO3I</youtube>
| <youtube width="300" height="180">https://youtu.be/RUm3iHsbO3I</youtube>
| [https://youtu.be/RUm3iHsbO3I ''Intro to Assembly Language'' ('''CalculusNguyenify''')]
| [https://youtu.be/RUm3iHsbO3I ''Intro to Assembly Language'' ('''CalculusNguyenify''')]


A general introduction into assembly language. In particular, covers how it fits into the source code to executable image pipeline.
A general introduction into assembly language. In particular, it covers how it fits into the source code to an executable image pipeline.


|-
|-
Line 176: Line 178:
| [https://youtu.be/NEQASUsZ0g4 ''Examples'' ('''CalculusNguyenify''')]
| [https://youtu.be/NEQASUsZ0g4 ''Examples'' ('''CalculusNguyenify''')]


Walks through a couple of ACSL Assembly language programs that have been used previous contests.
Walks through a couple of ACSL Assembly language programs that have been used in previous contests.


|}
|}
<!--
{|
|-
| <youtube width="300" height="180">URL</youtube>
| [URL ''TITLE'' ('''AUTHOR''')]
DESCRIPTION
|}
-->
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Latest revision as of 10:54, 22 January 2021

Programs written in high-level languages are traditionally converted by compilers into assembly language, which is turned into machine language programs – sequences of 1’s and 0’s – by an assembler. Even today, with very good quality compilers available, there is the need for programmers to understand assembly language. First, it provides programmers with a better understanding of the compiler and its constraints. Second, on occasion, programmers find themselves needing to program directly in assembly language in order to meet constraints in execution speed or space.

ACSL chose to define its own assembly language rather than use a “real” one in order to eliminate the many sticky details associated with real languages. The basic concepts of our ACSL topic description are common to all assembly languages.

Reference Manual

Execution starts at the first line of the program and continues sequentially, except for branch instructions (BG, BE, BL, BU), until the end instruction (END) is encountered. The result of each operation is stored in a special word of memory, called the “accumulator” (ACC). The initial value of the ACC is 0. Each line of an assembly language program has the following fields:

LABEL OPCODE LOC

The LABEL field, if present, is an alphanumeric character string beginning in the first column. A label must begin with an alphabetic character(A through Z, or a through z), and labels are case-sensitive. Valid OPCODEs are listed in the chart below; they are also case-sensitive and uppercase. Opcodes are reserved words of the language and (the uppercase version) many not be used a label. The LOC field is either a reference to a label or immediate data. For example, “LOAD A” would put the contents referenced by the label “A” into the ACC; “LOAD =123” would store the value 123 in the ACC. Only those instructions that do not modify the LOC field can use the “immediate data” format. In the following chart, they are indicated by an asterisk in the first column.

OP CODE DESCRIPTION
*LOAD The contents of LOC are placed in the ACC. LOC is unchanged.
STORE The contents of the ACC are placed in the LOC. ACC is unchanged.
*ADD The contents of LOC are added to the contents of the ACC. The sum is stored in the ACC. LOC is unchanged. Addition is modulo 1,000,000.
*SUB The contents of LOC are subtracted from the contents of the ACC. The difference is stored in the ACC. LOC is unchanged. Subtraction is modulo 1,000,000.
*MULT The contents of LOC are multiplied by the contents of the ACC. The product is stored in the ACC. LOC is unchanged. Multiplication is modulo 1,000,000.
*DIV The contents of LOC are divided into the contents of the ACC. The signed integer part of the quotient is stored in the ACC. LOC is unchanged.

.

BG

Branch to the instruction labeled with LOC if ACC>0.

BE

Branch to the instruction labeled with LOC if ACC=0.

BL

Branch to the instruction labeled with LOC if ACC<0.

BU Branch to the instruction labeled with LOC.
READ

Read a signed integer (modulo 1,000,000) into LOC.

PRINT

Print the contents of LOC.

DC

The value of the memory word defined by the LABEL field is defined to contain the specified constant. The LABEL field is mandatory for this opcode. The ACC is not modified.

END

Program terminates. LOC field is ignored and must be empty.

Sample Problems

Problem 1

After the following program is executed, what value is in location TEMP?

TEMP DC 0
A DC 8
B DC -2
C DC 3
LOAD B
MULT C
ADD A
DIV B
SUB A
STORE TEMP
END

Solution: The ACC takes on values -2, -6, 2, -1, and -9 in that order. The last value, -9, is stored in location TEMP.

Problem 2

If the following program has an input value of N, what is the final value of X which is computed? Express X as an algebraic expression in terms of N.

READ X
LOAD X
TOP SUB =1
BE DONE
STORE A
MULT X
STORE X
LOAD A
BU TOP
DONE END

Solution: This program loops between labels TOP and DONE for A times. A has an initial value of X and subsequent terms of N, then values of A-1, A-2, …, 1. Each time through the loop, X is multiplied by the the current value of A. Thus, X = A * (A-1) * (A-2) * … * 1 or X=A! or A factorial. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120. Since the initial value of A is the number input (i.e. N), the algebraic expression is X = N!.

Video Resources

The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.

ACSL Math: Assembly Language Programming (Quick Coding Bytes)

This video introduces the topic, then using an example problem, explains the methodology to solve problems that appear on ACSL contests.

Intro to Assembly Language (CalculusNguyenify)

A general introduction into assembly language. In particular, it covers how it fits into the source code to an executable image pipeline.

Syntax of ACSL Assembly Language (CalculusNguyenify)

A very nice introduction to this ACSL category.

Examples (CalculusNguyenify)

Walks through a couple of ACSL Assembly language programs that have been used in previous contests.


xx