的sizeof（结构）和sizeof（联合） [英] sizeof(struct) and sizeof(union)

问题描述

struct One{
       int x[6];
       short y[12];
} a;

union Two{
       int x[6];
       short y[12];
} b;

假定int为32位，而短裤是16位。

Assume that ints are 32 bits, and that shorts are 16 bits.

一个。是什么的sizeof（一）和sizeof（B）？

a. What is the sizeof(a) and sizeof(b)?

乙。如果＆安培; A = 0x00320000 ，什么是＆放大器; a.y

℃。如果和b = 0x00320400 ，什么是＆放大器; b.y

可能有人请解释如何在联盟和结构的大小获得，以及如何得到的值的地址。据我所知，和放大器;一个和和b分别指的a和b的地址。但是，我不完全知道如何在什么＆放得出; a.y和＆安培; b.y将返回

Could someone please explain how to derive at the size of the union and struct, as well as how to get the values for the addresses. I understand that &a and &b mean the address of a and b respectively. However, i'm not exactly sure how to derive at what &a.y and &b.y will return.

推荐答案

虽然，在大多数情况下，答案是的实现定义的，为的的调试和故障排除程序员的需要知道的（可能）的答案是什么，他/她的平台：

Even though, for the most part, the answers are implementation-defined, a programmer for debugging and trouble-shooting needs to know what the (likely) answers are, on his/her platform:

一个。是什么的sizeof（一）和sizeof（B）？

a. What is the sizeof(a) and sizeof(b)?

sizeof的一个= 6 * 4 + 12 * 2 = 48，重要提示：如果您更改 12 到 13 ，这个计算很可能是错误的，因为填充通常会增加，大概2个字节，所以结构的大小会的不的是的大小之和其元素。

sizeof a = 6 * 4 + 12 * 2 = 48. Important: if you change 12 to 13, this calculation would likely be wrong, as padding would typically be added, probably 2 bytes, and so the size of the struct would not be the sum of the sizes of its elements.

sizeof的B = MAX（6 * 4,12 * 2）= 24，因为在这个联盟， X 和是被覆盖。同样，如果你改变 12 到 13 ，有可能填充。

sizeof b = max(6 * 4, 12 * 2) = 24, because in this union, x and y are overlayed. Again if you change 12 to 13, there's likely padding.

乙。如果＆安培; A = 0x00320000，什么是＆放大器; a.y

b. if &a = 0x00320000, what is &a.y?

＆安培; a.y = 0x00320000 + 6 * 4 = 0x00320018

&a.y = 0x00320000 + 6 * 4 = 0x00320018

℃。如果和b = 0x00320400，什么是与放大器; b.y

c. if &b = 0x00320400, what is &b.y?

＆安培; b.y =和b（保证）

&b.y = &b (guaranteed)

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