Why is the size of a C struct with short members not always a multiple of the member size?

Understanding Memory Alignment in C Structs

Memory alignment is a crucial concept in C programming as it impacts the way data is stored in memory and accessed by the processor. In the context of structs, memory alignment dictates how individual members of a struct are laid out in memory.

Consider the following struct:

<code class="c">typedef struct {
    unsigned short v1;
    unsigned short v2;
    unsigned short v3;
} myStruct;</code>

On a 32-bit machine where memory alignment is typically set to 4 bytes, one would expect the total size of this struct to be 8 bytes. However, the sizeof(myStruct) operator returns only 6 bytes.

The discrepancy arises because each member of the struct, being an unsigned short, has a size of 2 bytes. Since all members are of the same size, no padding is inserted between them. As a result, the struct occupies 6 bytes in memory.

Now, let's modify the struct:

<code class="c">typedef struct {
    unsigned short v1;
    unsigned short v2;
    unsigned short v3;
    int i;
} myStruct;</code>

In this case, the presence of an int member, which has a size of 4 bytes, changes the memory alignment of the struct. The int member must be aligned on a 4-byte boundary. Since it is preceded by 6 bytes, 2 bytes of padding are inserted between v3 and i. This brings the total size of the struct to 12 bytes, as confirmed by sizeof(myStruct).

In summary, memory alignment in C structs ensures that members are properly aligned in memory to optimize processor performance and reduce memory usage. Types are only aligned to a boundary as large as the type itself. When a larger type is introduced, it sets the alignment for the entire struct, potentially introducing padding to maintain the desired alignment.

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