C++ 中的二进制到十进制转换：10 个值后不起作用

Question

我正在尝试为 C++ 中的 hack 汇编语言创建一个反汇编程序。现在我的问题在于将 16 位二进制转换回 A 指令。在检查操作码（第一位）为 0 后，我必须将其后的其他 15 位转换回前面带有“@”的十进制。转换一直进行到 000001111111111 并输出“@1023”，但 000011111111111更高版本仅输出“@2559”。这是到目前为止的代码

#include <iostream>
#include <string>
#include <cmath>

using namespace std;

int BtoA (int n)
{
    int num = n;
    int decimal = 0;
    int base = 0;
    int tmp = num;

    while (tmp != 0)
    {
        int last_dig = tmp % 10;
        tmp /= 10;

        decimal += last_dig * pow(2, base);

        base++;
    }

    return decimal;
}

int main()
{
    cout << "Welcome to the Hack Disassembler!\n";
    int opCode;
    int aIns;

    cout << "\nIs the OP-Code 1 or 0? ";

    cin >> opCode;

    if (opCode == 0)
    {
        cout << "\nWhat are the other 15 bits? ";
        cin >> aIns;
        cout << "\nThat A Instruction is @" << BtoA(aIns) << "!\n";
    }
}

Answer 1

尝试将变量

aIns

更改为键入

std::string

。如果是

int

类型，超过

std::numeric_limits<int>::max()

就会溢出，对于32位整数来说是

2'147'483'647

。那是一个 10 位数字。当您添加第 11 位数字时，您会得到

11'111'111'111

，即溢出。

// main.cpp
#include <iostream>
#include <stdexcept>
#include <string>

int bitstring_to_int(std::string const& s)
{
    if (s.empty())
        throw std::invalid_argument(
            '"' + s + "\" is empty. "
            "Instructions must have at least 1 bit.");
    if (s.length() > 15)
        throw std::invalid_argument(
            '"' + s + "\" is too wide. "
            "Instructions can have at most 15 bits.");
    for (auto const& c : s)
        if (c != '0' && c != '1')
            throw std::invalid_argument(
                '"' + s + "\" contains invalid character(s). "
                "It should be composed entirely of '0`s and '1's.");
    int n{};
    int bit{ 1 };
    for (auto i{ s.length() }; i--; bit <<= 1u)
        if (s[i] == '1')
            n |= bit;
    return n;
}

int main()
{
    std::cout << "Welcome to the Hack Disassembler!\n";
    int opCode;
    std::string aIns;

    std::cout << "\nIs the OP-Code 1 or 0? ";

    std::cin >> opCode;

    if (opCode == 0)
    {
        std::cout << "\nWhat are the other 15 bits? ";
        std::cin >> aIns;
        std::cout << "\nThat A Instruction is @" << bitstring_to_int(aIns) << "!\n";
    }
}
// end file: main.cpp

Welcome to the Hack Disassembler!

Is the OP-Code 1 or 0? 0

What are the other 15 bits? 111111111111111

That A Instruction is @32767!