Structure s

{

char c；

char arr[2]；

};

s s 1；

float f = 1. 1；

sizeof(float)=4，sizeof(f)= 4；

//sizeof(struct S) is the total size of all members of the calculation type struct S, that is, Sizeof (c)+Sizeof (arr) =1+2 = 3.

sizeof(struct S)= 3；

//Because the size of a variable is equal to the size of its type, sizeof (s1) = sizeof (structures) = 3.

sizeof(s 1)= 3；

2.malloc is a library function. Malloc (size); The function of is to allocate a memory of size in the heap, and then return the first address of this memory block, but it returns a void* pointer, so it is necessary to type the return value. For example:

int * pint =(int *)malloc(sizeof(int))；

//Suppose the first address is 100, and the heap memory addresses that pointer ps can use are 100,10, 102 (because sizeof(struct S)=3, malloc allocates 3 bytes of memory).

//and ps points to the first address 100.

struct S * PS =(struct S *)malloc(sizeof(struct S))；

3. Memory allocated with malloc must be released with free (pointer variable name), which is another library function. For example:

//release the above structure through structures * PS = (structures *) malloc (size of (structures)); 3 bytes of memory allocated by the statement.

Free (PS);

Understand the above three points, your code is not difficult to understand.