Analysis of problems related to C language pointer

1. Determine the output of the following three expressions ？

struct Test
{
 int Num;
 char *pcName;
 short sDate;
 char cha[2];
 short sBa[4];
}*p;
// hypothesis p  The value of is 0x100000.  What are the values of the expressions in the following table ？
// It is known that , Structure Test The variable size of type is 20 Bytes 
int main()
{
 printf("%p\n", p + 0x1);
 printf("%p\n", (unsigned long)p + 0x1);
 printf("%p\n", (unsigned int*)p + 0x1);
 return 0;
}

a. First analyze the concept defined according to the structure to know this p Pointer to this structure ,p+0x1(16 Base number ), That is to say p+1,p For a pointer +1 That is to skip one p Related types . So for p The address of + The bytes it occupies 20 by ：0x100014.

b. The second is the same as the first , But it needs to be added because its type is forcibly converted to （unsigned long） Is an unsigned long shape , So direct integer plus one , by ：0x100001

c. The third is the same as above , Because it is transformed into （ Pointer types ） So the pointer is four bytes, so add four directly , by ：0x100004

2. Judge the following output value

int main()
{
    int a[4] = { 1, 2, 3, 4 };
    int* ptr1 = (int*)(&a + 1);
    int* ptr2 = (int*)((int)a + 1);
    printf("%x,%x", ptr1[-1], *ptr2);//4   
    return 0;
}

a.&a because a For array name , Address it directly into an array pointer int (*)[4], When it is added by one, because it is a pointer type , So skip the length of a pointer type directly , The array pointer type length is 4*4 Bytes , Jump to the array element 4 The last one of them , Translates into int* Pointer variables exist for type ptr1 in , When it is printed at the bottom , the ptr1[-1] operation , This operation is equivalent to    *(ptr1-1), namely ptr1 This pointer moves forward one bit and dereferences , Finally, I will use %x Print 16 Base number , by ：4.

b. Drawing solution ：

3. Judge the following output value

int main()
{
    int a[5][5];
    int(*p)[4];
    p = a;
    printf("%p,%d\n", &p[4][2] - &a[4][2], &p[4][2] - &a[4][2]);//
    return 0;
}

 a. The illustration

 4. Judge the following output value .

int main()
{
    char* c[] = { "ENTER","NEW","POINT","FIRST" };
    char** cp[] = { c + 3,c + 2,c + 1,c };
    char*** cpp = cp;
    printf("%s\n", **++cpp);//point
    printf("%s\n", *-- * ++cpp + 3);//er
    printf("%s\n", *cpp[-2] + 3);//rt
    printf("%s\n", cpp[-1][-1] + 1);//ew
    return 0;
}

The illustration ：