在铸件C，那么解引用指针 [英] casting then dereferencing pointers in C

问题描述

在与字符在C缓冲器工作，有时它会是有用的，更有效的能够与 INT 一次-sized数据块。要做到这一点，我可以投我的char * 到为int * ，并使用该指针来代替。不过我并不完全相信，这工作我的思维方式确实。

When working with char buffers in C, sometimes it would be useful and more efficient to able to work with int-sized chunks of data at a time. To do this I can cast my char * to an int * and use that pointer instead. However I'm not entirely confident that this works the way I think it does.

例如，假设我有的char *数据，确实 *（* int32_t）的数据= -1 总是覆盖字节数据[0] ，数据[1] ，数据[2] 和数据[3] 并没有其他字节？

For example, suppose I have char *data, does *(int32_t *)data = -1 always overwrite the bytes data[0], data[1], data[2] and data[3] and no other bytes?

推荐答案

拓展我的评论。

这里有两个主要问题：

• 这违反了严格走样。


• 您可能打破对齐。

违反严格走样技术上的未定义行为即可。你被允许使用别名的char * 任何数据类型，而不是周围的其他方式。

Violating strict-aliasing is technically undefined behavior. You are allowed to alias any datatype with char*, but not the other way around.

您可以解决此问题与 F [无糖]严格走样的GCC。

You can get around the issue with f[no-]strict-aliasing on GCC.

另一个问题是对齐。在字符指针可能无法正确​​对齐。访问不对齐的数据可能会导致性能下降，甚至是硬件异常如果硬件不支持未对齐的访问。

The other issue is alignment. The char pointer might not be properly aligned. Accessing misaligned data may lead to performance degradation or even a hardware exception if the hardware doesn't support misaligned access.

如果性能是不是一个问题，全防爆方式是的memcpy（）到 INT 数组缓冲区。

If performance isn't an issue, the full-proof way is to memcpy() to an int array buffer.

在这两个问题都解决了，你的例子：

Once these two issues are resolved, your example with:

*(int32_t *)data = -1

覆盖数据[0] ，数据[1] ，数据[2] 和数据[3] 应该按预期工作如果的sizeof（int32_t）== 4 。只是要注意字节顺序...

overwriting data[0], data[1], data[2], and data[3] should work as expected if sizeof(int32_t) == 4. Just pay attention to the endianness...

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