使函数使用 std::span 而不是旧方法

Question

我刚刚了解到我可以将 VS 2019 与 C++ 20 一起使用，我正在尝试使用它。我正在尝试使以下函数使用 std::span，因为这样 data_size 和 key_size 将是多余的。

除了 *data 和 data++ 之外，一切都很好。它应该是什么？

int rc4(rc4_context* context, const std::uint8_t* data, const std::size_t data_size, const std::uint8_t* key, const std::size_t key_size, std::uint8_t* output)
{
    std::uint32_t i, j;

    // Check parameters
    if (!context || !key)
        return ERROR_INVALID_PARAMETER;

    // Clear context
    context->i = 0;
    context->j = 0;

    // Initialize the S array with identity permutation
    for (i = 0; i < 256; i++)
    {
        context->s[i] = static_cast<std::uint8_t>(i);
    }

    // S is then processed for 256 iterations
    for (i = 0, j = 0; i < 256; i++)
    {
        // Randomize the permutations using the supplied key
        j = (j + context->s[i] + key[i % key_size]) % 256;

        // Swap the values of S[i] and S[j]
        const auto temp = context->s[i];
        context->s[i] = context->s[j];
        context->s[j] = temp;
    }

    // Restore context
    i = context->i;
    j = context->j;
    auto* s = context->s;

    // Encryption loop
    for (size_t x = 0; x < data_size; ++x)
    {
        // Adjust indices
        i = (i + 1) % 256;
        j = (j + s[i]) % 256;

        // Swap the values of S[i] and S[j]
        const auto temp = s[i];
        s[i] = s[j];
        s[j] = temp;

        // Valid input and output?
        if (data && output)
        {
            // XOR the input data with the RC4 stream
            *output = *data ^ s[(s[i] + s[j]) % 256];

            // Increment data pointers
            data++;
            output++;
        }
    }

    // Save context
    context->i = i;
    context->j = j;

    return NO_ERROR;
}

我的尝试

int rc4(rc4_context* context, const std::span<uint8_t*> data, const std::span<std::uint8_t*> key, std::uint8_t* output)
{
    // INITIALIZATION
    std::uint32_t i, j;

    // Check parameters
    if (!context || !key.empty())
        return ERROR_INVALID_PARAMETER;

    // Clear context
    context->i = 0;
    context->j = 0;

    // Initialize the S array with identity permutation
    for (i = 0; i < 256; i++)
    {
        context->s[i] = static_cast<std::uint8_t>(i);
    }

    // S is then processed for 256 iterations
    for (i = 0, j = 0; i < 256; i++)
    {
        // Randomize the permutations using the supplied key
        j = (j + context->s[i] + key[i % key.size()]) % 256;

        // Swap the values of S[i] and S[j]
        const auto temp = context->s[i];
        context->s[i] = context->s[j];
        context->s[j] = temp;
    }

    // MAIN LOGIC PART
    // Restore context
    i = context->i;
    j = context->j;
    auto* s = context->s;

    // Encryption loop
    for (size_t x = 0; x < data.size(); ++x)
    {
        // Adjust indices
        i = (i + 1) % 256;
        j = (j + s[i]) % 256;

        // Swap the values of S[i] and S[j]
        const auto temp = s[i];
        s[i] = s[j];
        s[j] = temp;

        // Valid input and output?
        if (data.empty() && output)
        {
            // XOR the input data with the RC4 stream
            *output = *data ^ s[(s[i] + s[j]) % 256];

            // Increment data pointers
            data++;
            output++;
        }
    }

    // Save context
    context->i = i;
    context->j = j;

    return NO_ERROR;
}

Answer 1

首先，(const std::uint8_t* data, const std::size_t data_size)这对参数可以用std::span<uint8_t>代替，而不是std::span<uint8_t*>。

其次，您无需费心递增 data，因为您可以将其重写为基于范围的 for 循环：

for (uint8_t elem : data) {
    // Adjust indices
    i = (i + 1) % 256;
    j = (j + s[i]) % 256;    

    // Swap the values of S[i] and S[j]
    std::swap(s[i], s[j]);

    // Valid output?
    if (output) {
        // XOR the input data with the RC4 stream
        *output++ = elem ^ s[(s[i] + s[j]) % 256];
    }   
}

使函数使用 std::span 而不是旧方法

Making a function use std::span instead of the old way

c++

std

c++20

我的尝试