11
\$\begingroup\$

I am coding a game boy disassembler in C++. The program (simplified here) converts a vector of byte into an assembly instruction and print it.

To do so, it iterates through the vector of bytes (char) with an iterator. I noticed that with the it!=char_vect.end() condition, there is the risk that the iterator goes out from the vector. So I have to check that I do not get out of the vector at each iteration and if it occurs, I throw an error.

#include <iostream>
#include <vector> 

int disassemble(std::vector<char>::iterator& it){

    int opcode = *it;
    int opbytes = 1; //number of bytes used by the operator

    switch(opcode){
        case 0x76 : std::cout<<"HALT "<<std::endl; opbytes = 1; break;
        case 0x10 : std::cout<<"STOP $"<<(int)*(it+1)<<std::endl; opbytes =2; break; 
        //... About 250 different cases
        default : std::cout<<"Instruction not implemented yet"<<std::endl;
    }

    return opbytes;
}

int main(){

    std::vector<char> char_vect = {0x76, 0x10, 0x20}; // Fill vector of char
    std::vector<char>::iterator it = char_vect.begin();

    // First I only did that
    //while (it!=char_vect.end()){
    //  it+=disassemble(it);
    //}

    // But I have to check that I do not go out of the char_vect

    while (it!=char_vect.end()){
        try{
            it+=disassemble(it);
            //check that it do not overpass end
            if (size_t(it - char_vect.begin())>char_vect.size()){
                throw "Try to disassemble instruction outside of memory bounds";
            }
        }
        catch(const char * c){
            std::cerr << "Fatal error: " << c << std::endl;
            return 1;
        }
    }
    return 1;
}

This code works fine. Now I wonder, I have a possible exception case which is the iterator going further than vect_char.end(). Is a try / catch an appropriate way to handle this case knowing that:

  1. The reason for the iterator to go out of bounds could be either:

    • A non valid input byte sequence, for example {0x76, 0x20, 0x10} since the instruction 0x10 expects an argument. This should not append.
    • A mistake in the disassemble function. For exemple with the valid input {0x76, 0x10, 0x20}, if I code by mistake that 0x10 uses 3 opbytes instead of 2, the iterator will go out of bounds. This should not append either.
  2. If I iterated through the vect_char with an index, I would not have this issue and the code would be more compact

I never really used try/catch before and I do not know if it is intented for such situations, so I would like to know, how would you have handled this unexpected case?

\$\endgroup\$
5

3 Answers 3

12
\$\begingroup\$

I have some ideas about how you might be able to improve your program.

Avoid problems

Rather than trying to deal with the problem for every instruction, one approach is avoiding it entirely. One way to do that is to simply append a number of bytes to the end of the vector. If the maximum bytes for an instruction is \$n\$, then append \$n-1\$ bytes to the end and stop when you've advanced into the padded area.

Check before advancing

One could also pass the number of remaining bytes to the disassemble function. However, the mechanism I'd suggest would be to pass a range, e.g.:

int diassembleOne(std::vector<char>::iterator& it, std::vector<char>::iterator& end) {
    //  figure out number of bytes for this opcode
    if (std::distance(it, end) > opbytes) {
        it = end;
        // throw or return 0
    }
    // disassemble the instruction thoroughly
    std::advance(it, opbytes);
    return opbytes;
}

Use const iterators

If all the code is doing is disassembling, then it shouldn't alter the underlying vector. For that reason, I'd recommend passing a std::vector<char>::const_iterator &.

Use classes

I'd suggest using an Opcode class like this:

class Opcode {
    char code;
    short int bytes;
    std::string_view name;
    bool operator==(char opcode) const { return code == opcode; }
    int decode(std::vector<char>::const_iterator& it, std::ostream& out=std::cout) const {
        out << name; 
        ++it;
        for (int i{bytes-1}; i; --i) {
            out << static_cast<unsigned>(*it++);
        }
        out << '\n';
        return bytes;
    }
};

constexpr std::array<Opcode,2> instructions {
    { 0x10, 2, "STOP $" },
    { 0x76, 2, "HALT " },
};

Pass a pair of iterators to the dissemble function

As mentioned before, you can pass a pair of iterators to the disassemble function. Using that plus the class above:

int disassembleOne(std::vector<char>::const_iterator& it, std::vector<char>::const_iterator& end){
    auto op{std::find(instructions.begin(), instructions.end(), *it)};
    if (op == instructions.end()) {
        std::cout << "Instruction not found\n";
        it = end;
        return 0; // instruction not found or off the end of the passed array
    }
    if (std::distance(it, end) < op->bytes) {
        std::cout << "Not enough bytes left to decode " << op->name << '\n';
        it = end;
        return 0; // instruction not found or off the end of the passed array
    }
    return op->decode(it);
}

Now main becomes very simple:

int main(){
    const std::vector<char> char_vect = {0x76, 0x10, 0x20, 0x10}; // Fill vector of char
    auto end{char_vect.cend()};

    for (auto it{char_vect.cbegin()}; it != end; disassembleOne(it, end)) {
    }
}

Another way to do this would be to put more of the processing in the Opcode itself -- it would make sense that each opcode would know how to decode itself.

Be clear about caller expectations

This code, as with the original code, expects that the it being passed is a valid iterator that is not the end. It is good to document that in comments in the code.

\$\endgroup\$
2
  • \$\begingroup\$ The loop in main doesn't use the result of the call to disassembleOne. \$\endgroup\$ Commented Mar 13, 2019 at 19:17
  • \$\begingroup\$ @RolandIllig It doesn't need to use the return value because the iterator, which is passed by reference, is updated within the function. \$\endgroup\$
    – Edward
    Commented Mar 13, 2019 at 19:32
7
\$\begingroup\$

I think the big switch is a problem. Consider a more data-driven approach where each opcode is described by a struct:

struct OpCode
{
    unsigned char command_byte;
    unsigned char mask;  // if only a subset of bits determine command
    unsigned char length;
    // other members as needed - e.g. a pointer to the "print" function
};

Now, the code that reads instructions can determine whether the command is unterminated, without needing to repeat the logic for every multi-byte opcode.

I've included the mask so that we don't encode every single instruction (e.g. ld R1, R2 encodes the source and destination registers in the bits of a single-byte command; it would be tedious and error-prone to write them all separately here).

The simple length value I've shown isn't quite enough, given that the Game Boy's LR35902 processor supports the 0xCB prefix for extended Z80 instructions - we might want to handle that outside of the normal flow, and use it to switch between different instruction tables.
We get away with a simple length value here, because the only prefix instruction supported by the Game Boy's LR35902 processor is 0xCB, which is always followed by a single-byte instruction. If we were decoding Z80 instructions (with ED prefix), then we'd need something a little more sophisticated.

\$\endgroup\$
3
\$\begingroup\$

Your code's division into functions isn't very natural. The objective is to have each function performing one specific task, which should also be as orthogonal as possible to the tasks performed by the other functions. For instance, your disassemble function performs three different functions: it reads from the instruction stream, it interprets assembly code and returns the number of bytes that should be skipped to get to the next instruction. That's a not-so-coherent mix of responsibilities.

There's also a bug in your code, because it+=disassemble(it); could point beyond char_vect.end() which is in itself undefined behavior, even if you don't dereference the iterator.

I would build my disassembler around an iterator, or better even a range (if you don't mind external libraries or anticipating the next standard):

#include <algorithm>
#include <vector>
#include <iostream>
#include <range/v3/view.hpp>

using iterator = std::vector<char>::const_iterator;
struct truncated_instruction {};

// one function to interpret the instruction
void interpret(char instruction) { 
     if (instruction == 'x' throw bad_instruction();
     std::cout << (int) instruction << ' ';
}

// one function to get the number of bytes to skip
int nb_bytes(char c) { return 1; } 

class disassembled
  : public ranges::view_facade<disassembled>
{
    friend ranges::range_access;
    iterator first, last;
    char const & read() const { return *first; }
    bool equal(ranges::default_sentinel) const { return first == last; }
    void next() { 
    // one function to get to the next instruction
        auto bytes_to_skip = nb_bytes(*first); 
        if (std::distance(first, last) < bytes_to_skip) throw truncated_instruction();
        // check if there's enough space left to advance before advancing
        std::advance(first, bytes_to_skip);
    }
public:
    disassembled() = default;
    explicit disassembled(const std::vector<char>& v) : first(v.begin()), last(v.end()) {}
};

int main() {
    std::vector<char> char_vect = {0x76, 0x10, 0x20, 0x30};
    try {
        for (auto instruction : disassembled(char_vect)) interpret(instruction);
    }
    catch // ...
}
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.