C语言——指针题目

#include <stdio.h>
int main()
{
    int a[5] = { 1, 2, 3, 4, 5 };
    int* ptr = (int*)(&a + 1);
    printf("%d,%d", *(a + 1), *(ptr - 1));
    return 0;
}

#include <stdio.h>
int main()
{
    int a[5] = { 1, 2, 3, 4, 5 };
    int* ptr = (int*)(&a + 1);
    //&a取出的是整个数组，&a+1既跳过了整个数组
    printf("%d,%d", *(a + 1), *(ptr - 1));//2,5
    //*(a+1)就是第二个元素2，*(ptr-1)就是第5个元素
    return 0;
}

#include <stdio.h>
//涉及到结构体内存对齐问题，结构体大小为20个字节
struct Test
{
  int Num;
  char* pcName;
  short sDate;
  char cha[2];
  short sBa[4];
}*p;
//假设p 的值为0x100000。 如下表表达式的值分别为多少？
int main()
{
    printf("%p\n", p + 0x1);
    printf("%p\n", (unsigned long)p + 0x1);
    printf("%p\n", (unsigned int*)p + 0x1);
    return 0;
}

#include <stdio.h>
//涉及到结构体内存对齐问题，结构体大小为20个字节
struct Test
{
  int Num;
  char* pcName;
  short sDate;
  char cha[2];
  short sBa[4];
}*p;
//假设p 的值为0x100000。 如下表表达式的值分别为多少？
int main()
{
    printf("%p\n", p + 0x1);//0x10014
    //p是结构体，结构体大小为20个字节，对于16进制来说，20相当于14
    //所以结果为0x100014
    printf("%p\n", (unsigned long)p + 0x1);//0x100001
    //p被强制转换成为了unsigned long类型，结果为0x100001
    printf("%p\n", (unsigned int*)p + 0x1);//0x100004
    //无符号整型指针+1跳过一个整型变量
    //相当于+4
    return 0;
}

#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);
      return 0;
}

#include <stdio.h>
int main()
{
      int a[4] = { 1, 2, 3, 4 };
      //假设以小端存储方式
      //01 00 00 00 02 00 00 00 03 00 00 00 04 00 00 00
      int* ptr1 = (int*)(&a + 1);
      //&a+1相当于跳过了整个数组，到达了4之后的位置，然后强制转化成int*
      //ptr1也指向那里
      int* ptr2 = (int*)((int)a + 1);
      //这里的a被强制转化成int型，整型+1就是+1，差了一个1就1个字节
      //然后又被强转为int*，所以ptr2指向第一个元素的第二个字节位置处，与第一个元素差一个字节
      //ptr2是一个整型指针
      printf("%x,%x", ptr1[-1], *ptr2);//4,2000
      //ptar[-1]可以理解为*(ptr1+(-1)),又ptr1是整型指针4个字节，-1跳到0x 00 00 00 04
      //对ptr2解引用后向后访问4个字节，又以%x打印，所以是0x 02 00 00 00
      return 0;
}

#include <stdio.h>
int main()
{
    int a[3][2] = { (0, 1), (2, 3), (4, 5) };
    int* p;
    p = a[0];
    printf("%d", p[0]);
    return 0;
}

#include <stdio.h>
int main()
{
    int a[3][2] = { (0, 1), (2, 3), (4, 5) };
    int* p;
    p = a[0];
    //a[0]代表首元素地址，就是1的地址
    printf("%d", p[0]);//1
    //p[0]可以看作*(p+0)就是1
    return 0;
}

#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]);
    return 0;
}

#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", *(ptr1 - 1), *(ptr2 - 1));
    return 0;
}

#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", *(ptr1 - 1), *(ptr2 - 1));//10,5
    return 0;
}

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

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