DAY2-大厂面试典例题（二）-指针练习篇

大家好，我是枫晨~，拖欠了许久的指针习题终于到来，废话不多说，来看看大厂中考过哪些好玩的笔试题吧

---

一、上半节

笔试1

int main()
{
    int a[5] = { 1, 2, 3, 4, 5 };
    int *ptr = (int *)(&a + 1);
    printf( "%d,%d", *(a + 1), *(ptr - 1));
    return 0; 
}
//程序的结果是什么？

笔试2

//由于我还没介绍过结构体，这里告知结构体的大小是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？
//已知，结构体Test类型的变量大小是20个字节
int main()
{
  p = (struct Test*)0x100000;
 printf("%p\n", p + 0x1);
 printf("%p\n", (unsigned long)p + 0x1);
 printf("%p\n", (unsigned int*)p + 0x1);
 return 0; 
}

笔试3

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; 
}
//分析最后打印的结果

笔试4

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

---

上半节答案及解析：

笔试1----知识点:数组指针

答案：2,5

详解：

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

笔试2----知识点：指针类型，指针±整数

答案：100014,100001,100004

详解：

struct Test
{
 int Num;
 char *pcName;
 short sDate;
 char cha[2];
 short sBa[4];
}*p;
int main()
{
  p = (struct Test*)0x100000;
 printf("%p\n", p + 0x1);
 printf("%p\n", (unsigned long)p + 0x1);
 printf("%p\n", (unsigned int*)p + 0x1);
 return 0; 
}

笔试3—知识点:指针类型权限，大小端

答案：4,2000000

详解:

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; 
}

​ （gif图解）

笔试4 知识点：逗号表达式，二维数组

答案：1

详解：

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

---

二、下半节

笔试5

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;
}
//打印结果？

笔试6

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; 
    }
//分析打印结果

笔试7

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

笔试8

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; 
 }
//分析打印结果

---

下半节答案及解析:

笔试5 知识点：指针-指针运用，数组指针使用

答案：FFFFFFFC,-4

详解：

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; 
}

练习6

答案：10,5

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; }

笔试7

答案：“at”

详解：

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

练习8

答案：POINT
ER
ST
EW

详解：

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; }

超详细动图图解：

好啦，这次的练习有点烧脑，结合前面指针的知识来理解运用会好很多啦，希望大家收获知识，收获好offer！

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; }