Linux中的结构分配在ARM中失败，但在x86中成功 [英] Structure assignment in Linux fails in ARM but succeeds in x86

问题描述

我注意到一些非常奇怪的事情. 说我已经定义了以下结构

I've noticed something really strange. say I've got the following structure defined

typedef struct
{
  uint32_t a;
  uint16_t b;
  uint32_t c;
} foo;

此结构包含在我从网络接收到的大缓冲区中.

This structure is contained in a big buffer I receive from network.

以下代码在x86中有效，但在ARM上却收到SIGBUS.

The following code works in x86, but I receive SIGBUS on ARM.

extern void * buffer;
foo my_foo;
my_foo = (( foo * ) buffer)[0];

用memcpy替换取消引用的指针可以解决此问题.

在ARM中搜索SIGBUS后，我发现了这样一个事实，即它与内存对齐方式有关.

Searching about SIGBUS in ARM pointed me to the fact that this is related to memory alignment somwhow.

有人可以解释发生了什么事吗?

Can someone explain what's going on ?

推荐答案

您自己说过:特定处理器上存在内存对齐限制，并且buffer对齐不正确，不允许从其中读取大于一个字节的内容.该任务可能被编译成三个较大实体的动作.

You said it yourself: there are memory alignment restrictions on your particular processor, and buffer is not aligned right to permit reading larger than a byte from it. The assignment is probably compiled into three moves of larger entities.

对于memcpy()，没有对齐限制，必须能够在任意两个地址之间进行复制，因此它可以执行实现该目标所需的一切.可能逐字节复制直到地址对齐为止，这是一种常见的模式.

With memcpy(), there are no alignment restrictions, it has to be able to copy between any two addresses, so it does whatever is needed to implement that. Probably copying byte-by-byte until the addresses are aligned, that's a common pattern.

顺便说一句，我发现不用数组索引就能编写代码更加清晰:

As an aside, I find it clearer to write your code without array indexing:

extern const void *buffer;
const foo my_foo = *(const foo *) buffer;

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