64位机器上的结构填充 [英] structure padding on 64bit machine

问题描述

struct A
{
    uint32_t var1;
    uint32_t var2;
    uint32_t var3;
    uint32_t var4;
    uint32_t var5;
};

在上述结构中，编译器不填充并分配 20 个字节.

In the above structure compiler does not pad and allocates 20 bytes.

现在我们有另一个结构，它包含一个 8 字节变量而不是两个 4 字节.在这种情况下，编译器填充并分配 24 字节给这个结构.

now we have another structure which contains one 8 byte variable instead of two 4 bytes.In this case compiler pads and allocates 24 bytes to this structure.

struct B
{
    uint32_t var1;
    uint32_t var2;
    uint32_t var3;
    uint64_t var5;
};

为什么会有这种行为?如果编译器将数据对齐到 8 字节边界，那么第一个结构中应该有 4 个字节的填充，并且应该在这种情况下，不要填充第二个结构.并且如果编译器将数据对齐到 4 字节边界，那么为什么在第二个结构中有 4 字节的填充?

why there is such behaviour? If compiler align the data into 8 byte boundry then there should be padding of 4 bytes in the 1st structure and should not pad the 2nd structure in such case. and also if compiler align the data into the 4 byte boundry then why there is padding of 4 byte in the 2nd structure ?

编译器:GCC平台:64位linux，x86_64

compiler: GCC Platform: 64 bit linux , x86_64

推荐答案

对齐规则(在 x86 和 x86_64 上)通常是根据变量的大小对齐.

The rule for alignment (on x86 and x86_64) is generally to align a variable on it's size.

换句话说，32 位变量在 4 个字节上对齐，64 位变量在 8 个字节上对齐，依此类推.

In other words, 32-bit variables are aligned on 4 bytes, 64-bit variables on 8 bytes, etc.

在你的第二种情况下，4 个字节的填充被添加到

In your second case, 4 bytes of padding are added between

uint32_t var3;
uint64_t var5;

让 var5 在 8 个字节上对齐.

to get var5 to align on 8 bytes.

因此，最好将数据成员从大到小排序(但由于数据的局部性、可读性等原因，这并不那么简单).

For this reason it is better to order data members from largest to smallest (but it's not that simple due to data locality, readability etc.).

这篇关于64位机器上的结构填充的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！