如何更改此函数的 return 类型？

Question

我正在经历 matasano crypto challenges using rust, with rust-crypto for the AES implementation. I have this function to do basic ECB mode encryption (basically taken nearly verbatim from the rust-crypto repository's example):

pub fn aes_enc_ecb_128(key: &[u8], data: &[u8]) 
                       -> Result<Vec<u8>, symmetriccipher::SymmetricCipherError> {
    let mut encryptor = aes::ecb_encryptor(
            aes::KeySize::KeySize128,
            key,
            blockmodes::NoPadding);
    let mut final_result = Vec::<u8>::new();
    let mut read_buffer = buffer::RefReadBuffer::new(data);
    let mut buffer = [0; 4096];
    let mut write_buffer = buffer::RefWriteBuffer::new(&mut buffer);

    loop {
        let result = encryptor.encrypt(&mut read_buffer,
                                       &mut write_buffer,
                                       true);

        final_result.extend(write_buffer
                            .take_read_buffer()
                            .take_remaining().iter().map(|&i| i));
        match result {
            Ok(BufferResult::BufferUnderflow) => break,
            Ok(_) => {},
            Err(e) => return Err(e)
        }
    }

    Ok(final_result)
}

以上版本编译没有问题，按预期工作。但是，为了使其适合我的其余错误处理方案，我想将 return 类型更改为 Result<Vec<u8>,&'static str>。这是应用了该更改的函数：

pub fn aes_enc_ecb_128(key: &[u8], data: &[u8]) 
                       -> Result<Vec<u8>, &'static str> {
    let mut encryptor = aes::ecb_encryptor(
            aes::KeySize::KeySize128,
            key,
            blockmodes::NoPadding);
    let mut final_result = Vec::<u8>::new();
    let mut read_buffer = buffer::RefReadBuffer::new(data);
    let mut buffer = [0; 4096];
    let mut write_buffer = buffer::RefWriteBuffer::new(&mut buffer);

    loop {
        let result = encryptor.encrypt(&mut read_buffer,
                                       &mut write_buffer,
                                       true);

        final_result.extend(write_buffer
                            .take_read_buffer()
                            .take_remaining().iter().map(|&i| i));
        match result {
            Ok(BufferResult::BufferUnderflow) => break,
            Ok(_) => {},
            Err(_) => return Err("Encryption failed")
        }
    }

    Ok(final_result)
}

当我尝试编译此版本时，出现以下错误（为清楚起见删除了路径）：

error: source trait is private
         let result = encryptor.encrypt(&mut read_buffer,
                                        &mut write_buffer,
                                        true);
error: source trait is private
let r = decryptor.decrypt(&mut read_buffer, &mut write_buffer, true);
        ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

我能够更改此类型的唯一方法是将原始函数包装在这样的转换函数中：

pub fn converted_enc(key: &[u8], data: &[u8]) 
                       -> Result<Vec<u8>, &'static str> {
   match aes_enc_ecb_128(key,data) {
       Ok(v) => Ok(v),
       Err(_) => Err("Encryption failed")
   } 
}

为了让 return 值与我的 API 的其余部分相符，我应该做什么而不是上面的操作，为什么更直接的方法失败了？

我正在使用 rust/cargo 的以下版本：

rustc 1.2.0-nightly (0cc99f9cc 2015-05-17) (built 2015-05-18)
cargo 0.2.0-nightly (ac61996 2015-05-17) (built 2015-05-17)

Answer 1

我认为您遇到了编译器的错误。你的代码应该编译

您可以 use crypto::symmetriccipher::Encryptor; 作为解决方法：

pub fn aes_enc_ecb_128(key: &[u8], data: &[u8]) 
                       -> Result<Vec<u8>, &'static str> {
    use crypto::symmetriccipher::Encryptor;
    let mut encryptor = aes::ecb_encryptor(
            aes::KeySize::KeySize128,
            key,
            blockmodes::NoPadding);
    let mut final_result = Vec::<u8>::new();
    let mut read_buffer = buffer::RefReadBuffer::new(data);
    let mut buffer = [0; 4096];
    let mut write_buffer = buffer::RefWriteBuffer::new(&mut buffer);

    loop {
        let result = encryptor.encrypt(&mut read_buffer,
                                       &mut write_buffer,
                                       true);

        final_result.extend(write_buffer
                            .take_read_buffer()
                            .take_remaining().iter().map(|&i| i));
        match result {
            Ok(BufferResult::BufferUnderflow) => break,
            Ok(_) => {},
            Err(_) => return Err("Encryption failed")
        }
    }

    Ok(final_result)
}

如何更改此函数的 return 类型？

How can I change the return type of this function?

rust