IAS computer

logistic_guy

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Given the memory contents of the \(\displaystyle \text{IAS}\) computer shown below.

Address​
Contents​
\(\displaystyle 08\text{A}\)​
\(\displaystyle 010\text{FA}210\text{FB}\)​
\(\displaystyle 08\text{B}\)​
\(\displaystyle 010\text{FA}0\text{F}08\text{D}\)​
\(\displaystyle 08\text{C}\)​
\(\displaystyle 020\text{FA}210\text{FB}\)​

Show the assembly language code for the program, starting at address \(\displaystyle 08\text{A}\). Explain what this program does.
 
Given the memory contents of the \(\displaystyle \text{IAS}\) computer shown below.

Address​
Contents​
\(\displaystyle 08\text{A}\)​
\(\displaystyle 010\text{FA}210\text{FB}\)​
\(\displaystyle 08\text{B}\)​
\(\displaystyle 010\text{FA}0\text{F}08\text{D}\)​
\(\displaystyle 08\text{C}\)​
\(\displaystyle 020\text{FA}210\text{FB}\)​

Show the assembly language code for the program, starting at address \(\displaystyle 08\text{A}\). Explain what this program does.
Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
First, let's learn some operation codes.

\(\displaystyle \bold{1} \longrightarrow \ \textcolor{red}{\bold{load}}\)

\(\displaystyle \bold{2} \longrightarrow \ \textcolor{blue}{\bold{subtract}}\)

\(\displaystyle \bold{15} \longrightarrow \ \textcolor{green}{\bold{jump}}\)

\(\displaystyle \bold{21} \longrightarrow \ \textcolor{indigo}{\bold{store}}\)
 
The contents in the address say \(\displaystyle 08\text{A}\) are \(\displaystyle 010\text{FA}210\text{FB}\). These contents are \(\displaystyle 40\) bits. We divide them into two halves. That gives us:

\(\displaystyle 010\text{FA} \longrightarrow 20\) bits
\(\displaystyle 210\text{FB} \longrightarrow 20\) bits
 
We will also take these \(\displaystyle 20\) bits and divide it. But not into half.

\(\displaystyle 010\text{FA} \longrightarrow 20\) bits

\(\displaystyle 01\longrightarrow 8\) bits
\(\displaystyle 0\text{FA} \longrightarrow 12\) bits
 
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