Ç64位指针对准 [英] C 64-bit Pointer Alignment

问题描述

在64位系统上的指针仍然4字节对齐（类似于双在32位系统）？或者，他们注意到8字节对齐？

Are pointers on a 64-bit system still 4 byte aligned (similar to a double on a 32 bit system)? Or are they note 8 byte aligned?

例如，在64位系统有多大下面的数据结构：

For example, on a 64-bit system how big is the following data structure:

struct a {
    void* ptr;
    char myChar;
}

请问指针8字节对齐，导致7个字节填充的字符（总= 8 + 8 = 16）？或者将鼠标指针是4字节对齐（4字节+ 4个字节），造成3个字节的填充（总= 4 + 4 + 4 = 12）？

Would the pointer by 8 byte aligned, causing 7 bytes of padding for the character (total = 8 + 8 = 16)? Or would the pointer be 4 byte aligned (4 bytes + 4 bytes) causing 3 bytes of padding (total = 4 + 4 + 4 = 12)?

谢谢， 瑞安

推荐答案

数据对齐和包装的具体实施中，可以从编译器设置被更改，通常（或甚至编译指示）。

Data alignment and packing are implementation specific, and can be usually changed from compiler settings (or even with pragmas).

但是假设你使用默认设置，在大多数（如果不是全部）编译器的结构应该最终会被16字节的总和。其原因是因为计算机读取数据块具有其天然字的大小（这是在64位的系统的8个字节）的大小。如果它是垫它到4字节偏移，下一个结​​构不会被正确填充到64位边界。例如在一情况下，ARR [2]，则阵列的第二个元素将开始在12字节的偏移量，这是不是在机器的本机字节边界

However assuming you're using default settings, on most (if not all) compilers the structure should end up being 16 bytes total. The reason is because computers reads a data chunk with size of its native word size (which is 8 bytes in 64-bit system). If it were to pad it to 4 byte offsets, the next structure would not be properly padded to 64-bit boundary. For example in case of a arr[2], the second element of the array would start at 12-byte offset, which isn't at the native byte boundary of the machine.

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