将给定大小的char数组复制到C中的整数的正确方法是什么？ [英] What's the proper way to copy a char array of a given size to an integer in C?

问题描述

假设我有一个char数组和一个关联的长度： Arr 和 Len 。不是字符串，是char数组。没有空终止符。但是我必须将数组数据复制到 int64_t 类型的整数中。这是完成的过程，出于这个问题的目的，我假设 Len 不会超过8：

Suppose I have a char array and an associated length: Arr and Len. Not a string, a char array. There is no null terminator. Yet I have to copy the array data into an integer of type int64_t. Here's how it's done, and for the purpose of this question I'm assuming Len will not exceed 8:

int64_t Word = 0;
memcpy(&Word, Arr, Len);

这实际上是执行此操作的正确方法吗？我正在复制内存，但是有没有更快的方法来内联，例如？因此， Word 可以注册？

Is this actually the proper way to do this? I am copying memory, but is there a faster way to do it inline, for example? So Word can be register?

双关语类型的问题是假定 Arr 分配了8个字节。不， Arr 最多分配了 8个字节。它可能有5个，因此将 Arr 强制转换为 int64_t * 然后取消引用它可能会尝试访问目录中的三个非法字节。

The problem with a type pun is it assumes that Arr has 8 bytes allocated. No, Arr has at most 8 bytes allocated. It could have 5, so casting Arr to a int64_t * then dereferencing it could try to access three illegal bytes at the end, resulting in segfault.

是执行我描述的 memcpy（）调用的正确方法，还是有更快的方法？还是更好的方法？

Is the proper way to do what I describe a memcpy() call, or is there a faster or better way?

推荐答案

由于您指定 Len 是最多 （8），假设使用低位字节存储，即 Arr [0]的最低有效字节是合理的。

Since you specify Len is at most (8), it's reasonable to assume little-endian storage, i.e., the least-significant byte at Arr[0].

如果 Len 固定为（8），编译器也许可以简单地通过从内存中加载值来替换 memcpy 。这也取决于平台是否可以执行未对齐的读取-如果编译器无法证明对齐-并可能涉及x86-64上的 bswap 指令，如果

If Len was fixed at (8), the compiler might be able to replace memcpy simply by loading the value from memory. That would also be dependent on whether the platform can do unaligned reads - if the compiler can't prove alignment - and may involve something like the bswap instruction on x86-64 if the architecture is big-endian.

Len 是运行时值这一事实很可能会生成对<$ c的调用$ c> memcpy 。通话本身的开销并不小。考虑所有因素，最好是使用字节算术以与字节序无关的方式处理此问题。该代码假定8位字节，这似乎与您的问题相符。

The fact that a Len is a run-time value will likely generate a call to memcpy. The overhead of the call itself is not trivial. All things considered, it's probably best just to handle this in an endian-independent way using byte arithmetic. The code assumes 8-bit bytes, which seems consistent with your question.

uint64_t Word = 0;

while (Len--)
    Word = (Word << 8) | Arr[Len];

在其他平台上，其中（CHAR_BIT> 8） ，则可以将 OR 表达式的右侧替换为（Arr [Len]& 0xff）。实际上，在具有8位（标准）字节的平台上对此进行了优化，因此您也可以添加它以提高完整性。或者只是记住这些问题。

On more exotic platforms, where (CHAR_BIT > 8), you can replace the right-hand side of the OR expression with (Arr[Len] & 0xff). In fact, this is optimised away on platforms with 8-bit (normative) bytes, so you might as well add it for completeness. Or just keep these issues in mind.

有 个带有合法C实现的平台，其中 char ，<例如，code> short ， int 是32位值。这些在嵌入式世界中很常见。

There are platforms with legal C implementations where char, short, int are 32-bit values, for example. These are quite common in the embedded world.

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