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\title{Opgaver til Microprocessors 2}
\date{2021-03-24}
\section{Problem 4.1}
\emph{In Fig. 4-6, the B bus register is encoded in a 4-bit field, but the C bus is represented
as a bit map. Why?}
It is often wanted to save the result from the ALU in multiple destination registers.
Therefore one cannot take the shortcut with a 4-bit field, as that would only allow one save at the time.
One cannot present 1 and 2 at the same time in 4-bit field, as that would activate register 3.
\section{Problem 4.5}
{\itshape
Suppose that in the example of Fig. 4-14(a) the statement
\begin{verbatim}
k = 5;
\end{verbatim}
is added after the if statement. What would the new assembly code be? Assume that
the compiler is an optimizing compiler.
}
Well k is set either way, so one can invert the if.
\begin{verbatim}
ILOAD j
ILOAD k
IADD
BIPUSH 3
IF_ICMPEQ L1
ILOAD j
BIPUSH 1
ISUB
ISTORE j
L1: BIPUSH 5
ISTORE k
\end{verbatim}
\section{Problem 4.9}
{\itshape
How long does a 2.5-GHz Mic-1 take to execute the Java statement
\begin{verbatim}
i = j + k
\end{verbatim}
Give your answer in nanoseconds
}
First we "compile" the java statement :-).
\begin{verbatim}
ILOAD j
ILOAD k
IADD
ISTORE i
\end{verbatim}
Then we can add how many microinstructions each takes (\textbf{bold} number) multiplied with how many times it is used.
\begin{equation}
\underbrace{\mathbf 1 \cdot 4}_{MAIN} + \underbrace{\mathbf 5 \cdot 2}_{ILOAD} + \underbrace{\mathbf 3}_{IADD} + \underbrace{\mathbf 6}_{ISTORE} = 23
\end{equation}
Then we can multiply with the nanoseconds a single instruction takes
\begin{equation}
\frac 1 {2.5 \cdot 10^9} \cdot 23 = 9.2 \cdot 10^{-9}\,,
\end{equation}
which is 9.2 Nanoseconds.
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