Linux下C语言结构体指针数组大小从24变16的原因解析

Let's break down exactly what's happening here, using your code and output as reference.

First, a key context: you're running this on a 64-bit Linux system—this dictates the size of pointers in your environment, which is the root of the size change.

1. Pointer size in 64-bit Linux #

On 64-bit Linux, any pointer type (including const struct point *) is 8 bytes long. You can confirm this from your output: the pointer values like 0XE2BF0778 are 8 hex digits, which equals 64 bits (8 bytes).

2. Original array size calculation #

Your initial array definition:

const struct point *object_directory[] = { &SW_VERSN_POINT, &SWMVERSN_POINT, 0 };

This is an array of 3 elements:

• Two valid pointers to your struct point constants
• A trailing 0, which gets implicitly converted to a const struct point * null pointer (often used as a sentinel/terminator for arrays)

Total size = number of elements × size of each element = 3 × 8 bytes = 24 bytes. This matches your first output line: 24 0XF967D778, 0xF967D788.

3. Modified array size calculation #

When you remove the trailing 0, the array becomes:

const struct point *object_directory[] = { &SW_VERSN_POINT, &SWMVERSN_POINT };

Now the array only contains 2 pointer elements.

Total size = 2 × 8 bytes = 16 bytes. Which aligns perfectly with your second output: 16 0XF967D778, 0xF967D788.

Quick sanity check of other outputs #

Your other print statements confirm the basics:

• sizeof(SW_VERSN_POINT.formatbyte1) is 4, which makes sense because int is 4 bytes on most 64-bit Linux systems.
• sizeof(struct point) is 12, which is 3 × 4 bytes (one for each int member, no padding needed here since all members are naturally aligned).

内容的提问来源于stack exchange，提问作者beparas