c - 计算结构校验和时跳过/避免对齐填充字节

在计算C结构的校验和时，是否有通用的方法跳过/避免对齐填充字节？

我想通过对字节求和来计算结构的校验和。问题是，该结构具有对齐填充字节，它们可以获取随机（未指定）值，并使具有相同数据的两个结构获得不同的校验和值。

注意：我主要关心的是可维护性（无需更新代码即可添加/删除/修改字段）和可重用性，而不是可移植性（该平台非常具体且不太可能更改）。

目前，我找到了一些解决方案，但是它们都有缺点：

1. Pack the struct (e.g. #pragma pack (1)). Disadvantage: I prefer to avoid packing for better performance.
2. Calculate checksum field by field. Disadvantage: The code will need to be updated when modifying the struct and requires more code (depending on the number of fields).
3. Set to zero all struct bytes before setting values. Disadvantage: I cannot fully guarantee that all structs were initially zeroed.
4. Arrange the struct fields to avoid padding and possibly add dummy fields to fill padding. Disadvantage: Not generic, the struct will need to be carefully rearranged when modifying the struct.

Is there a better generic way?

Calculating checksum example:

unsigned int calcCheckSum(MyStruct* myStruct)
{
    unsigned int checkSum = 0; 
    unsigned char* bytes = (unsigned char*)myStruct;
    unsigned int byteCount = sizeof(MyStruct);
    for(int i = 0; i < byteCount; i++)
    {
        checkSum += bytes[i];
    }
    return checkSum;
}

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