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Hello again.
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In this problem here, we have the memory cell starting with the address 0 until 05 and our machine in the appendix and it contains the bit patterns shown here.
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So we assume that we start with the value 0 in our program counter, so that's where the execution will start.
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And we need a to translate the instructions that are being excused.
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It into english.
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So let's look at the first instruction.
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So it's 1202.
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That's the first one in 0 .0 and 0 .01.
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And afterwards the code 3242 will be executed and finally c 000.
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And afterwards the machine holds.
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So the first instruction here, let's call this instruction 1, 2 and 3.
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The first instruction here is simply to load register 2, register 2 with the content of the memory cell number 0 2 of memory cell with the address 02.
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So that's the first one.
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And to keep track of what we're doing, now we know that register 2 will contain the value 32, which is simply the content of the memory cell 02.
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So register 2 now holds the value 32.
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That's the first one.
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The second one here is 3 to 42...