海康威视校园招聘笔试题

1、10、10、4、4四个数，怎么算出24点？
(10*10-4)/4=24
2、下列表达式在32位机器编译环境下的值（）

[cpp] view plaincopyprint?

1. class?A??
2. {??
3. };??
4. ??
5. class?B??
6. {??
7. public:??
8. ????B();??
9. ????virtual?~B();??
10. };??
11. ??
12. class?C??
13. {??
14. private:??
15. #pragma?pack(4)??
16. ????int?i;??
17. ????short?j;??
18. ????float?k;??
19. ????char?l[64];??
20. ????long?m;??
21. ????char?*p;??
22. #pragma?pack()??
23. };??
24. ??
25. class?D??
26. {??
27. private:??
28. #pragma?pack(1)??
29. ????int?i;??
30. ????short?j;??
31. ????float?k;??
32. ????char?l[64];??
33. ????long?m;??
34. ????char?*p;??
35. #pragma?pack()??
36. };??
37. ??
38. int?main(void)??
39. {??
40. ????printf("%d\n",sizeof(A));??
41. ????printf("%d\n",sizeof(B));??
42. ????printf("%d\n",sizeof(C));??
43. ????printf("%d\n",sizeof(D));??
44. ????return?0;??
45. }??

class A

{

};

class B

{

public:

B();

virtual ~B();

};

class C

{

private:

#pragma pack(4)

int i;

short j;

float k;

char l[64];

long m;

char *p;

#pragma pack()

};

class D

{

private:

#pragma pack(1)

int i;

short j;

float k;

char l[64];

long m;

char *p;

#pragma pack()

};

int main(void)

{

printf("%d\n",sizeof(A));

printf("%d\n",sizeof(B));

printf("%d\n",sizeof(C));

printf("%d\n",sizeof(D));

return 0;

}

A、1、4、84、82 ? ? ?B、4、4、82、84 ? ? ?C、4、4、84、82 ? ? ?D、1、4、82、82
3、以下程序在32位机器下运行的结果是（）

[cpp] view plaincopyprint?

1. #pragma?pack(4)??
2. struct?info_t??
3. {??
4. ????unsigned?char?version;??
5. ????unsigned?char?padding;??
6. ????unsigned?char?extension;??
7. ????unsigned?char?count;??
8. ????unsigned?char?marker;??
9. ????unsigned?char?payload;??
10. ????unsigned?short?sequence;??
11. ????unsigned?int?timestamp;??
12. ????unsigned?int?ssrc;??
13. };??
14. ??
15. union?info_u??
16. {??
17. ????unsigned?char?version;??
18. ????unsigned?char?padding;??
19. ????unsigned?char?extension;??
20. ????unsigned?char?count;??
21. ????unsigned?char?marker;??
22. ????unsigned?char?payload;??
23. ????unsigned?short?sequence;??
24. ????unsigned?int?timestamp;??
25. ????unsigned?int?ssrc;??
26. };??
27. #pragma?pack()??
28. ??
29. int?main(void)??
30. {??
31. ????printf("%d\n",sizeof(info_t));??
32. ????printf("%d\n",sizeof(info_u));??
33. ????return?0;??
34. }??

#pragma pack(4)

struct info_t

{

unsigned char version;

unsigned char padding;

unsigned char extension;

unsigned char count;

unsigned char marker;

unsigned char payload;

unsigned short sequence;

unsigned int timestamp;

unsigned int ssrc;

};

union info_u

{

unsigned char version;

unsigned char padding;

unsigned char extension;

unsigned char count;

unsigned char marker;

unsigned char payload;

unsigned short sequence;

unsigned int timestamp;

unsigned int ssrc;

};

#pragma pack()

int main(void)

{

printf("%d\n",sizeof(info_t));

printf("%d\n",sizeof(info_u));

return 0;

}

A、12 ?12 ? ? ?B、12 ?4 ? ? ? C、16 ?4 ? D、16 ?12 ? ? E、16 ?1
4、以下表达式result的值是（）

[cpp] view plaincopyprint?

1. #define?VAL1(a,b)?a*b??
2. #define?VAL2(a,b)?a/b--??
3. #define?VAL3(a,b)?++a%b??
4. ??
5. int?a?=?1;??
6. int?b?=?2;??
7. int?c?=?3;??
8. int?d?=?3;??
9. int?e?=?5;??
10. ??
11. int?result?=?VAL2(a,b)/VAL1(e,b)+VAL3(c,d);??

#define VAL1(a,b) a*b

#define VAL2(a,b) a/b--

#define VAL3(a,b) ++a%b

int a = 1;

int b = 2;

int c = 3;

int d = 3;

int e = 5;

int result = VAL2(a,b)/VAL1(e,b)+VAL3(c,d);

A、-2 ? ? B、1 ? ? C、0 ? ? D、2
5、请写出以下程序的输出（5分）

[cpp] view plaincopyprint?

1. void?swap_1(int?a?,?int?b)??
2. {??
3. ????int?c;??
4. ????c?=?a;??
5. ????a?=?b;??
6. ????b?=?c;??
7. ????return?;??
8. }??
9. void?swap_2(int?&a?,?int?&b)??
10. {??
11. ????int?c;??
12. ????c?=?a;??
13. ????a?=?b;??
14. ????b?=?c;??
15. ????return?;??
16. }??
17. void?swap_3(int?*a?,?int?*b)??
18. {??
19. ????int?c;??
20. ????c?=?*a;??
21. ????*a?=?*b;??
22. ????*b?=?c;??
23. ????return?;??
24. }??
25. ??
26. int?main(void)??
27. {??
28. ????int?a?=?100;??
29. ????int?b?=?200;??
30. ????swap_1(a?,?b);??
31. ????printf("a?=?%d?,?b?=?%d\n",a?,?b);??
32. ????swap_2(a?,?b);??
33. ????printf("a?=?%d?,?b?=?%d\n",a?,?b);??
34. ????swap_3(&a?,?&b);??
35. ????printf("a?=?%d?,?b?=?%d\n",a?,?b);??
36. ????return?0;??
37. }??

void swap_1(int a , int b)

{

? int c;

? c = a;

? a = b;

? b = c;

? return ;

}

void swap_2(int &a , int &b)

{

? int c;

? c = a;

? a = b;

? b = c;

? return ;

}

void swap_3(int *a , int *b)

{

? int c;

? c = *a;

? *a = *b;

? *b = c;

? return ;

}

int main(void)

{

? int a = 100;

? int b = 200;

? swap_1(a , b);

? printf("a = %d , b = %d\n",a , b);

? swap_2(a , b);

? printf("a = %d , b = %d\n",a , b);

? swap_3(&a , &b);

? printf("a = %d , b = %d\n",a , b);

? return 0;

}

输出结果：
a = 100 , b = 200
a = 200 , b = 100
a = 100 , b = 200
6、下面的程序是否有问题，如有问题，请重构代码（5分）

[cpp] view plaincopyprint?

1. void?test_type(bool?b?,?const?char?*p?,?float?f)??
2. {??
3. ????if(!b)??
4. ????{??
5. ????????return?;??
6. ????}??
7. ????else?if(!p)??
8. ????{??
9. ????????return?;??
10. ????}??
11. ????else?if(!f)??
12. ????{??
13. ????????return?;??
14. ????}??
15. }??

void test_type(bool b , const char *p , float f)

{

? if(!b)

? {

??? return ;

? }

? else if(!p)

? {

??? return ;

? }

? else if(!f)

? {

??? return ;

? }

}

修改如下：

[cpp] view plaincopyprint?

1. void?test_type(bool?b?,?const?char?*p?,?float?f)??
2. {??
3. ????if(!b)??
4. ????{??
5. ????????return?;??
6. ????}??
7. ????else?if(!p)??
8. ????{??
9. ????????return?;??
10. ????}??
11. ????else?if(f?>?-1e-10?&&?f?<?1e-10)??
12. ????{??
13. ????????return?;??
14. ????}??
15. }??

void test_type(bool b , const char *p , float f)

{

if(!b)

{

? return ;

}

else if(!p)

{

? return ;

}

else if(f > -1e-10 && f < 1e-10)

{

? return ;

}

}

7、请指出以下程序有什么问题（5分）

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1. void?test_mem()??
2. {??
3. ????char?*p?=?new?char[64];??
4. ????delete?p;??
5. ????p?=?NULL;??
6. ????return?;??
7. }??

void test_mem()

{

? char *p = new char[64];

? delete p;

? p = NULL;

? return ;

}

应该修改为 delete[]p; ?p指向的是一个字符型的数组空间，原来的代码只是简单的释放了指向申请空间的指针，并没有释放申请的空间，容易造成内存崩溃。
回收用 new 分配的单个对象的内存空间的时候用 delete，回收用 new[] 分配的一组对象的内存空间的时候用 delete[]。
8、以下程序有什么问题，请指出。

[cpp] view plaincopyprint?

1. char*?GetMem()??
2. {??
3. ????char?p[]?=?"hello";??
4. ????return?p;??
5. }??
6. ??
7. void?test_get_mem()??
8. {??
9. ????char?*p?=?GetMem();??
10. ????printf(p);??
11. ????return?;??
12. }??

char* GetMem()

{

char p[] = "hello";

return p;

}

void test_get_mem()

{

char *p = GetMem();

printf(p);

return ;

}

GetMem函数中的p是一个在栈上的局部变量，当函数运行结束的时候，栈上的内容会自动释放的，此处返回的值有可能会成为一个野指针，会出现一个意想不到的结果。
9、请写出strcpy 和 memcpy 的区别（5分）
答：strcpy和memcpy都是标准C库函数，它们有下面的特点。
strcpy提供了字符串的复制。即strcpy只用于字符串复制，并且它不仅复制字符串内容之外，还会复制字符串的结束符。
strcpy函数的原型是：char* strcpy(char* dest, const char* src);
memcpy提供了一般内存的复制。即memcpy对于需要复制的内容没有限制，因此用途更广。
memcpy函数的原型是：void *memcpy( void *dest, const void *src, size_t count );
strcpy和memcpy主要有以下3方面的区别。
1、复制的内容不同。strcpy只能复制字符串，而memcpy可以复制任意内容，例如字符数组、整型、结构体、类等。
2、复制的方法不同。strcpy不需要指定长度，它遇到被复制字符的串结束符"\0"才结束，所以容易溢出。memcpy则是根据其第3个参数决定复制的长度。
3、用途不同。通常在复制字符串时用strcpy，而需要复制其他类型数据时则一般用memcpy。

10、请写出以下程序的输出结果

[cpp] view plaincopyprint?

1. class?Base??
2. {??
3. public:??
4. ????Base()??
5. ????{??
6. ????????printf("I?am?Base()\n");??
7. ????}??
8. ????virtual?~Base()??
9. ????{??
10. ????????printf("I?am?~Base()\n");??
11. ????}??
12. public:??
13. ????virtual?void?SayHello()??
14. ????{??
15. ????????printf("Hello?Base\n");??
16. ????}??
17. ????void?SayWorld()??
18. ????{??
19. ????????printf("World?Base\n");??
20. ????}??
21. };??
22. class?Derived?:?public?Base??
23. {??
24. public:??
25. ????Derived()??
26. ????{??
27. ????????printf("I?am?Derived()\n");??
28. ????}??
29. ????virtual?~Derived()??
30. ????{??
31. ????????printf("I?am?~Derived()\n");??
32. ????}??
33. public:??
34. ????void?SayHello();??
35. ????void?SayWorld();??
36. };??
37. ??
38. void?Derived::SayHello()??
39. {??
40. ????printf("Hello?Derived\n");??
41. }??
42. void?Derived::SayWorld()??
43. {??
44. ????printf("World?Derived\n");??
45. }??
46. ??
47. int?main(void)??
48. {??
49. ????Base?*b1?=?new?Base;??
50. ????Base?*b2?=?new?Derived;??
51. ????Derived?*d?=?new?Derived;??
52. ??
53. ????b1->SayHello();??
54. ????b1->SayWorld();??
55. ??
56. ????b2->SayHello();??
57. ????b2->SayWorld();??
58. ??
59. ????d->SayHello();??
60. ????d->SayWorld();??
61. ??
62. ????delete?d;??
63. ????delete?b2;??
64. ????delete?b1;??
65. ??
66. ????d=?NULL;??
67. ????b2?=?NULL;??
68. ????b1?=?NULL;??
69. ??
70. ????return?0;??
71. }??

class Base

{

public:

? Base()

? {

??? printf("I am Base()\n");

? }

? virtual ~Base()

? {

??? printf("I am ~Base()\n");

? }

public:

? virtual void SayHello()

? {

??? printf("Hello Base\n");

? }

? void SayWorld()

? {

??? printf("World Base\n");

? }

};

class Derived : public Base

{

public:

? Derived()

? {

??? printf("I am Derived()\n");

? }

? virtual ~Derived()

? {

??? printf("I am ~Derived()\n");

? }

public:

? void SayHello();

? void SayWorld();

};

void Derived::SayHello()

{

? printf("Hello Derived\n");

}

void Derived::SayWorld()

{

? printf("World Derived\n");

}

int main(void)

{

? Base *b1 = new Base;

? Base *b2 = new Derived;

? Derived *d = new Derived;

? b1->SayHello();

? b1->SayWorld();

? b2->SayHello();

? b2->SayWorld();

? d->SayHello();

? d->SayWorld();

? delete d;

? delete b2;

? delete b1;

? d= NULL;

? b2 = NULL;

? b1 = NULL;

? return 0;

}

输出结果：
I am Base()
I am Base()
I am Derived()
I am Base()
I am Derived()
Hello Base
World Base
Hello Derived
World Base
Hello Derived
World Derived
I am ~Derived()
I am ~Base()
I am ~Derived()
I am ~Base()
I am ~Base()


11、阅读以下程序并给出执行结果

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1. class?Bclass??
2. {??
3. public:??
4. ????Bclass(int?i?,?int?j)??
5. ????{??
6. ????????x?=?i;??
7. ????????y?=?j;??
8. ????}??
9. ????virtual?int?fun()??
10. ????{??
11. ????????return?0;??
12. ????}??
13. protected:??
14. ????int?x?,?y;??
15. };??
16. ??
17. class?lclass?:?public?Bclass??
18. {??
19. public:??
20. ????lclass(int?i?,?int?j?,?int?k)?:?Bclass(i?,?j)??
21. ????{??
22. ????????z?=?k;??
23. ????}??
24. ????int?fun()??
25. ????{??
26. ????????return?(x+y+z)/3;??
27. ????}??
28. private:??
29. ????int?z;??
30. };??
31. int?main(void)??
32. {??
33. ????lclass?obj(2,4,10);??
34. ????Bclass?p1?=?obj;??
35. ????cout<<p1.fun()<<endl;??
36. ??
37. ????Bclass?&p2?=?obj;??
38. ????cout<<p2.fun()<<endl;??
39. ????cout<<p2.Bclass::fun()<<endl;??
40. ??
41. ????Bclass?*p3?=?&obj;??
42. ????cout<<p3->fun()<<endl;??
43. ??
44. ????return?0;??
45. }??

class Bclass

{

public:

? Bclass(int i , int j)

? {

??? x = i;

??? y = j;

? }

? virtual int fun()

? {

??? return 0;

? }

protected:

? int x , y;

};

class lclass : public Bclass

{

public:

? lclass(int i , int j , int k) : Bclass(i , j)

? {

??? z = k;

? }

? int fun()

? {

??? return (x+y+z)/3;

? }

private:

? int z;

};

int main(void)

{

? lclass obj(2,4,10);

? Bclass p1 = obj;

? cout<<p1.fun()<<endl;

? Bclass &p2 = obj;

? cout<<p2.fun()<<endl;

? cout<<p2.Bclass::fun()<<endl;

? Bclass *p3 = &obj;

? cout<<p3->fun()<<endl;

? return 0;

}

输出结果：
0
5
0
5
12、如何减少频繁分配内存（malloc或者new）造成的内存碎片？（10分）


13、请写出strchr的实现（10分）
函数功能：找出在字符串str中第一次出现字符ch的位置，找到就返回该字符位置的指针（也就是返回该字符在字符串中的地址的位置），找不到就返回空指针（就是NULL）
const char* strchr(const char* str , char ch)

[cpp] view plaincopyprint?

1. const?char*?strchr(const?char*?str?,?char?ch)??
2. {??
3. ????char?*p?=?NULL;??
4. ????const?char*?s?=?str;??
5. ????for(?;?*s?!=?'\0'?;?++s)??
6. ????{??
7. ????????if(*s?==?ch)??
8. ????????{??
9. ????????????p?=?(char?*)s;??
10. ????????????break;??
11. ????????}??
12. ????}??
13. ????return?p;??
14. }??

const char* strchr(const char* str , char ch)

{

char *p = NULL;

const char* s = str;

for( ; *s != '\0' ; ++s)

{

? if(*s == ch)

{

?? p = (char *)s;

?? break;

? }

}

return p;

}

14、请写出冒泡排序法算法（20分）
void BubbleSort(int r[] , int n);

[cpp] view plaincopyprint?

1. void?BubbleSort(int?r[]?,?int?n)??
2. {??
3. ????int?i?,?j?,?temp;??
4. ????for(i?=?0?;?i?<?n?-?1?;?++i)??
5. ????{??
6. ????????for(j?=?0?;?j?<?n-i-1?;?++j)??
7. ????????{??
8. ????????????if(r[j]?>?r[j?+?1])??
9. ????????????{??
10. ????????????????temp?=?r[j];??
11. ????????????????r[j]?=?r[j?+?1];??
12. ????????????????r[j?+?1]?=?temp;??
13. ????????????}??
14. ????????}??
15. ????}??
16. }??