OK. Let’s start. Before I get going with the real task, I might as well get me a 1 in register R1. This is quite handy since we quite often need the constant 1. Think only of incrementing or decrementing some value. Right? I am supposed to get a value from standard input. Get one. What is it? Does it already mark the end of the input data? No. Good. Is it the end of a line? No. Good. Is it a blank character? Yes? Yes! Then we just ignore it. We don’t consider blank characters as useful information at the moment. We are currently waiting for something else to appear. What’s the next value? Again a lousy blank character. What’s next? Another Wow! Now we got an “A”. Actually, it is the number 0x41, i.e. the equivalent of an “A”. Well, to be honest, it’s not even 0x41, but rather the bit pattern 0000 0000 0100 0001. But for convenience, we denote it in hexadecimal notation and say “hex four one”. In decimal this corresponds to 4*16+1 = 49, i.e. fourtynine. Let’s store this “A” at the first location in my buffer. Well, I have not told you, but I have allocated 3 consecutive words in main memory and call it “buffer”. I use this buffer to collect important stuff I am getting from the standard input. The first location of this buffer has the index 0. Computer people like to start counting from 0. Therefore we should rather be precise and say that we put the “A” at the “zero-eth” position of the buffer. OK. On to the next value from standard input. It’s the ASCII-equivalent of the character D: 0x43. I copy it into the buffer next to the “A”. On to the next value from standard input. We’re getting another “D”. Copy it next to the two other characters in the buffer. We now have “A”, “D”, and “D” in the buffer. On to the next value from standard input. It is a blank character. Well, it seems that we got the first “word” finished. Let’s call it the first “token”. Let’s hope that I can identify this token. What I have not told you yet is that I have prepared a list of 7 3-letter tokens inside main memory. These are “HLT”, “ADD”, “SUB”, “AND”, “XOR”, “SHL”, and “SHR”. Actually, I have stored the ASCII equivalents of these tokens; and since I know that each token consists of three characters, I just put them next to each other inside main memory. In main memory I thus have “HLTADDSUBANDXORSHLSHR”. OK, what was I up to? Oh yes, I wanted to figure out whether the contents of the buffer can be found in my list of known tokens. I start by comparing the first character in the buffer with the first character in my list of tokens. Is “A” equal to “H”? No, definitely not. Apparently, the comparison with “HLT” is wrong. Let’s try with the next token in the token list. Its first character starts at location 3 (when starting to count with 0). I compare my “A” from the buffer with the character at location 3 in the token list. “A” equals “A”. Starts nicely, doesn’t it? Let’s compare the first “D” from the buffer with the character at location 4 in the token list: “D” equals “D”. Wow. No on to the third character of the buffer, the second “D”. We need to compare it with the character at location 5 in the token list. Super – with this third comparison we now know that the token in the buffer can be found in the token list. In fact it is the second token in the token list. But since we have numbered the tokens in the token list with 0, 1, 2, …, 6, we call it the token with index 1. The translation of this token into TOY machine language starts with a 1. Thus, we already know that the machine command looks something like 0x1???. Whenever I identify a token other than HLT, I know that I have to continue with reading the standard input for finding the rest of the machine instruction. What’s the next character from standard input. Its another lousy blank. Forget it. Forget it. But here comes the long awaited “R”. Let’s store it in the buffer at location 0. Read one more. Its the value 0x31. This is the ASCII equivalent of 1. Into the buffer at location 1. Read one more: a blank character. Seems that we got another token finished. We know that it is something like R0, R1, R2,…, RA, …, RF. We check the buffer. The first letter at location 0 is in fact an R. Perfect. If this were not the case, we would have stopped our activity of scanning the standard input and trying to translate the input to TOY machine language. We just would  have … what should we have done in this case? Maybe we output 0xFFFF and quit. This is not a very sophisticated error message, but anyway. TOY is only a toy. However, we have an “R” at location 0 in the buffer. What’s next to it? It’s a 1. With this we now know two hex digits of the machine instruction: 11??. Let’s go on with reading words from standard input. I a getting rather tired from telling you this story. Can you continue yourself without my help? I am sure you can. Think in terms of for-loops, while-loops, if-statements, simple assignments, additions, subtractions, shifts, or logic expressions. Write down your thoughts in detail. Learn to code your ideas in the language C. Compile this code and try it out. Test with simple test data first. Later with more complex ones. Start early to make the deadline of handing in your assignment.

I wish you all the successes you deserve with your work on assignment 1.