C pointer (interview classic topic exercise)

The first question is

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	int a[5] = {
    1,2,3,4,5};
	int* ptr = (int*)(&a + 1);//&a Get the address of the entire array ,&a+1 Skip the entire array 
			// Array pointer strongly typed to integer pointer 
	printf("%d,%d\n",*(a+1),*(ptr-1));// The addition and subtraction of the pointer depends on the pointer type ptr Is an integer pointer , therefore  ptr-1 Point to the last of the array 
	return 0;
}

The second question is

#include<stdio.h>
#include<stdlib.h>

struct Test
{
    
	int Num;
	char* pcName;
	short sDate;
	char cha[2];
	short sBa[4];
}* p;
// hypothesis P The value of is 0x100000
// Known structures test The variable size of type is 20 Bytes 
int main()
{
    
	p = (struct Test*)0x100000;
	printf("%p\n",p+0x1);//00100014 p Is a structure pointer ,p+1 Skip the entire structure 
						 //00100000+00000014=00100014
	printf("%p\n",(unsigned long)p+0x1);//00100001 p The pointer type of is converted to an integer +1 Is simply add 1
	printf("%p\n",(unsigned int*)p+0x1);//00100004 p Convert to integer pointer ,P+1 Just walk by 4 byte 
	return 0;
}

Third question

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	int a[4] = {
    1,2,3,4};
	int* ptr1 = (int*)(&a+1);//&a Get the address of the entire array ,&a+1 Skip the entire array 
				
	int* ptr2 = (int*)((int)a+1);
	printf("%x\n",ptr1[-1]);//4 ptr[-1]==*(ptr+(-1))==*(ptr-1)
				// Array pointer strongly typed to integer pointer ptr-1, Point to last element 
	printf("%x\n",*ptr2);//2 00 00 00  Pay attention to small end storage 
	return 0;
}

Fourth question

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	int a[3][2] = {
    (0,1),(2,3),(4,5)};// Note that there are parentheses inside the parentheses , It is not the use of braces and comma operators 
	//a[3][2]{(0,1),(2,3),(4,5)}=={1,3,5}=={1,3,5,0,0,0}={
    {1,3,5},
	// {0,0,0}}
	int* p = a[0];//p Is the pointer to the first row of the array 
	printf("%d\n",p[0]);//p[0]== The value of the first element in the first line 
	return 0;
}

Fifth question

#include<stdio.h>
#include<stdlib.h>
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]);
}

Sixth question

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	int aa[2][5] = {
    1,2,3,4,5,6,7,8,9,10};
	int *ptr1 = (int*)(&aa + 1);//&aa Represents the address of the entire array ,&aa+1： Skip the entire array 
	int *ptr2 = (int*)(*(aa + 1));//aa Represents the first element of the array , The first element is a one-dimensional array ;
								   //aa+1： The address of the first element in the second line 
								   //*(aa+1)  The first element in the second line 
	printf("%d\n",*(ptr1-1));//10
	printf("%d\n",*(ptr2-1));//5
	return 0;
}

Question seven

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	char* a[] = {
    "work","at","alibaba"};
	char** pa = a;
	pa++;
	printf("%s\n",*pa);
	return 0;
}

The eighth question

#include<stdio.h>
#include<stdlib.h>
int main()
{
    
	char* c[] = {
    "ANTER","NEW","POINT","FIRST"};
	char** cp[] = {
    c + 3,c + 2,c + 1,c};
	char*** cpp = cp;
	printf("%s\n",**++cpp);//POINT 
	//cpp+1 Point to c+2, After two dereferences POINT
	
	printf("%s\n", *--*++cpp + 3);//ER
	//++cpp:cpp Point to c+1
	//*(++cpp): obtain c The contents of the pointer to the second element of the pointer array 
	//--(*(++cpp)): The second element pointer -1, Get the contents of the first element pointer 
	//*(--(*(++cpp))): Dereference to get ENTER
	//*(--(*(++cpp)))+3: Point to T, Output TR
	
	printf("%s\n",*cpp[-2]+3);//ST
	// Be careful cpp So far, it still points to c+1,cpp[-2] The content of the address pointed to changes ,cpp The direction of has not changed 
	//*cpp[-2]==*(*(cpp+(-2)))  therefore *cpp[-2] Point to ,c+3, Get is FIRST
	//*cpp[-2]+3 Point to S
	
	printf("%s\n",cpp[-1][-1]+1);//EW
	// Be careful cpp So far, it still points to c+1
	//cpp[-1][-1]+1 == *(*(cpp-1)-1)+1
	//*(cpp-1) Point to c+2;
	//*(*(cpp-1)-1) Point to c[2] The content of ;
	//*(*(cpp-1)-1)+1 Point to NEW Medium N
	return 0;
}