C语言第24天，指针难题

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

int main() {
	int a[5] = { 1,2,3,4,5 };
	int* ptr = (int*)(&a + 1);
	printf("%d,%d\n", *(a + 1), *(ptr - 1));//2,5
	//a是首元素地址，a+1是第二个元素的地址
	//ptr-1是元素 5 的地址
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

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);//4,2000000
	//ptr1[-1] = *(ptr1 - 1) - 是元素4的地址
	
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

int main() {
	int a[3][2] = { (0,1),(2,3),(4,5) };
	//逗号表达式：(0,1)=1  (2,3)=3  (4,5)=5
	int* p;
	p = a[0];
	printf("%d",p[0]);//1 
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.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]);
	//FFFFFFFC,-4
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

int main() {
	int aa[2][5] = { 1,2,3,4,5,6,7,8,9,10 };
	int* ptr1 = (int*)(&aa + 1);
	int* ptr2 = (int*)(*(aa + 1));
	printf("%d,%d\n", *(ptr1 - 1), *(ptr2 - 1));//10,5
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

int main() {
	char* a[] = { "work","at","alibaba" };
	char** pa = a;
	pa++;
	printf("%s\n", *pa);//at
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

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);//st
	printf("%s\n", cpp[-1][-1] + 1);//ew
	return 0;
}