Helpful Information For Competitive Programming.

Assembly Language Programming



Many of us probably know that computers read instructions as 0s and 1s (binary); this is known as machine language.

Most of us have only used high-level languages that use English and common symbols for operations, which is why they are much easier to understand for us. These languages abstract away the difficulty of understanding machine language, but need a compiler to transform the code into machine code. This process makes high level languages far more inefficient than pure binary instructions, but also much, much, much simpler to write.

A few well-known examples of high-level languages are Python, Java, and C++. (C++ is a little strange because it compiles directly to assembly, circumventing many of the performance issues and making it a lower-level language.)

Assembly lies in the middle of machine languages and high-level languages. It is a low-level language type. Assembly languages are slightly more "English-like" than machine languages, as assembly langs use names to specify operations instead of numbers. However, it's still much, much harder to program in assembly languages than high-level language.

Since assembly languages are so low-level, it runs ridiculously quickly on all hardware, while simultaneously being humanly possible to write (unlike machine code). It's applied in applications where performance is absolutely critical (which is why C++ compiles to it). Knowing it can help you understand how hardware (like the CPU) works.

As a note: ACSL defines its own assembly language. Assembly can vary in between implementations, but the basic ideas are the same.

How It Works

Execution of assembly starts from the first line down until the END instruction; exceptions are branch instructions (see List of OPCODEs), which loop back to an earlier part of the program if conditions are met. There's also something called the ACC, or "accumulator", which is essentially a special variable with an initial value of 0.

One line of an assembly language program is formatted as follows:

x ADD =6

The LABEL field is optional; when paired with the DC command, it serves as a way to declare a variable. Otherwise, it essentially "marks" a line that you may want to utilize later on. OPCODEs are uppercase reserved words that cannot be used as a label. Meaning, for example, I cannot have a label called "ADD".

The LOC field can either reference a label defined in a previous line or immediate data (data not stored in a "variable"). When using immediate data, an = sign is put before the data. Only OPCODEs that do not modify the LOC field have the option of using immediate data; these are marked in the "List of OPCODEs" section down below.

List of OPCODEs

Here are the general operations:

OPCODE Modifies LOC? Description
LOAD No The value of LOC is stored into the ACC. So, if LOC is a variable, Q, with a value of 3, then ACC will store the value 3.
STORE Yes The opposite of LOAD - the value of ACC is stored into LOC.
ADD No The value of LOC is added to ACC. The sum becomes the new value of ACC. So, if LOC is =4 and ACC's value is 2, then ACC's new value will be 6.
SUB No The value of LOC is subtracted from ACC. The difference becomes the new value of ACC.
MULT No The value of LOC is multiplied by ACC's value. The product becomes the new value of ACC.
DIV No ACC's value is divided by the value of LOC. The quotient becomes the new value of ACC; decimals are rounded down to integers.
BG This branch instruction will return to the instruction labeled with LOC if ACC's value is greater than 0 (hence BG for "greater"). So, if ACC's value is 3, and LOC is TOP, then the program will move back to the instruction labeled as TOP.
BE This branch instruction will return to the instruction labeled with LOC if ACC's value is equal to $0$ (hence BE for "equal").
BL This branch instruction will return to the instruction labeled with LOC if ACC's value is less than $0$ (hence BL for "less").
BU This branch instruction will return to the instruction labeled with LOC regardless of what ACC's value is (hence BU for "unconditional").
READ Yes This essentially serves as a way to read user input into LOC. So, writing READ X will read for an input and store it into X.
PRINT No This is exactly as it sounds; it prints the value of LOC.
DC This is a way to declare a variable. The LABEL field is mandatory, as it is where the name of your variable will go. The ACC is not changed. So, writing VAR DC 2 would declare a variable, VAR, with the value of 2.
END This signals the end of your program. The LOC field must be kept empty.

Note that any read or arithmetic operations are, for some reason, defined by ACSL to be modulo 1,000,000. This probably won't crop up in competition, but it's nice to know.

Sample Problems

As you solve these problems, it can be very easy to lose track of everything! To stay organized, create a table to track the values of ACC and any other variables that may have been defined in the program.

1. After the following program is executed, what value is in location TEMP? (This is from the ACSL Wiki.)

A DC 8
B DC -2
C DC 3

This program defines 4 variables: TEMP, A, B, and C.

The value of B (which is -2) is loaded into the ACC. The value of ACC is multiplied by C, or 3, and the new value of ACC becomes -6. Then, ACC holds a value of 2 (added 8), -1 (divided by -2), and then -9 (subtracted 8). This value is then stored into TEMP; thus, TEMP has a value of -9.

2. What is printed in this program?

W DC 2

This program loops between TOP and DONE (DONE not being included) for a total of Q times. To solve through this type of problem, it may help to assign your own value to Q and work through the program with that value; in general, it's always good to choose a smaller number.

In this solution explanation, we will use the number 3. Look at the following table below to understand how the program changes over time:

Instruction ACC Q W
DC 3
DC 3 2
LOAD 3 3 2
SUB 2 3 2
STORE 2 2 2
LOAD 2 2 2
MULT 4 2 2
STORE 4 2 4
LOAD 2 2 4
SUB 1 2 4
STORE 1 1 4
LOAD 4 1 4
MULT 8 1 4
STORE 8 1 8
LOAD 1 1 8
SUB 0 1 8

Now that ACC is equal to $0$, the branch instruction BE after SUB now applies, thus taking us to DONE, where the value of W, or 8, is printed. The program then ends.

Notice the pattern with W; it was initially equal to 2, but with each loop, it doubled, becoming 4 and then 8. As we said before, we inputted a 3 for Q. 2 ^ 3 = 8; so, this program prints 2 ^ Q.

The equivalent Java code would look something like this:

int Q = 3; //not writing Scanner for this lol
int W = 2;

while (true) {
    if (Q == 0) {
    W *= 2;


Authors: Kelly Hong, Raymond Zhao