1. Structure alignment number

The structure occupancy diagram is from 0 Start drawing , Then the memory occupied by each variable needs to be aligned with the alignment number , Choose with several pairs min( System alignment number , Variable size )

2.

Be careful ： Now? 32 On the bit machine （X86）,C Linguistic long and int All are 4 byte , Represents long integer and integer , On the old machine 4、2 Byte differentiation .C Language only stipulates sizeof Of long》=int It is generally believed that 4.
vs Of 32 and 64 Next ,long All are 4
And you can choose the second question at a glance A, Second position , Overall structure size , To divide the compiler alignment number .
pragma pack(4)： Compilation options ,4 Byte alignment .

3. Bit segment

memories ： The bits of the bit segment are bit position . Bit segments open up space , Develop by type , The variable in the structure is int, open 4 Bytes .
This question , It's all unsigned char type , One open 1 Bytes ,8 position .
Open sequence bit ：1,（Env、Paral）、1、1, A common need 3 Bytes .

pit ：
This problem pit is replacing define Of A+B, In the code sizeof Back , Just replace , So in the end sizeof(Recode)2+3
The result is ：32+3 = 9

4.

memories ：memset():C Initializing an array or structure in language 0 The fastest way , Because it operates directly on memory .
memset(a, 0, sizeof(a)); Directly from 0 To size a Dispose of as 0

• analysis ：

1. tag Is a bit segment ：ucP1 byte , Total below 1, common 2.
2. puc Change to bit segment type , Forget the Strong rotation represents the available space
   
3. In limine , apply puc[4], Will apply for 4 byte .
   When it becomes stronger, it turns into tagPIM Give it to pst after , You can move the first two bytes , Is equivalent to pst The structure pointer points to the head of this address .
   But because of the structure , Only the first two bytes of it can be moved . The last two are 0
   to 2：0000 0010
   to 3：0000 0011
   to 4：0000 0100
   to 5：0000 0101
   then 2 Give the first one. pim1, It's one byte , Then get ：0000 0010
   to 3, to bit1, Can only reach the bottom 1, And on the right side of the figure , Is the last of the second byte , Because use low address first , Then use high , So from right to left .
   to 4, take bit2,00
   to 5, Nabit 3,101,
   Last ：0000 0010 ,0010 1001-> 2,41
   Output is 16 Base number ,2 Representatives should take at least two ：02, 29,0,0

5.

First , The size of the consortium is a larger value ,short Array , In the end 14,int yes 4, So choose 14,
But to Pay attention to the alignment number .
short The alignment number of is 2,int yes 4, Finally, the size should be aligned , So we should 16.

6.

This anonymous definition , The resulting consortium or structure , Only once .

analysis ：

1. Open up space ： Align numbers 2,1, Press the maximum ：2 byte . It also conforms to the alignment
2. to i[0] 0x39 、i[1] 0x38,i[0] In the array , It must be a low address , and i[1] It's a high address .
3. When we use short Of k When taking out ,short It's a short integer , Even if the 16 Base number , It is also presented by the left high address and the right address .
   therefore ： 3839 .
   commonly , The small end model （ Lower the address of low-order data , High order data is put to high address ）, The high address is on the left , The low address is on the right . Even if it conforms to our normal habit of looking at numbers .

7.

The default value of enumeration is 0 Start adding 1, here we are Z1 after ,A1、B1 Will add 1

8

This question VS2013, Nothing special , Follow 19、22 almost ：