我在geeksforgeeks中浏览了一篇有关结构填充的文章,https://www.geeksforgeeks.org/is-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member/ 但是我不知道为什么会这样:
int main()
{
struct C {
// sizeof(double) = 8
double z;
// sizeof(short int) = 2
short int y;
// Padding of 2 bytes
// sizeof(int) = 4
int x;
};
printf("Size of struct: %d", sizeof(struct C));
return 0;
}
我知道y(短整数)后跟x(整数),因此在y之后需要填充。但是为什么这里的填充是2?
答案 0 :(得分:1)
编译器希望将int
对齐到四个字节的边界上,并且该边界短了两个字节,因此填充被计算为两个字节:
struct C {
// offset = 0
double z;
// offset = 8
short int y;
// offset = 10
// ... padding ...
// offset = 12
int x;
// offset = 16
};
大于10的4的下一个倍数是12。
编辑:实际上,结构是按照此算法(伪代码)对齐的:
offset = 0;
alignment = 1;
for each field in structure {
offset = (offset + field.alignment - 1) / field.alignment * field.alignment;
field.offset = offset;
alignment = lcm(alignment, field.alignment);
}
structure.alignment = alignment;
structure.size = (offset + alignment - 1) / alignment * alignment;