C++试卷（华南理工大学）

华南理工大学期末考试 

《高级语言程序设计（I）》A卷

注意事项：    1. 考前请将密封线内各项信息填写清楚；

            2. 所有答案写在答题纸上，答在其它地方无效；

3．考试形式：闭卷；

           4. 试卷可做草稿纸，试卷必须与答题纸同时提交；

5. 本试卷共五大题，满分100分，考试时间120分钟。

Question 1 (2 points for each, 20 points for total) Choose the correct answer

1) To initialize an array of characters which one is NOT correct? (       )

      A) char s1[3]={'a','b','c'};          B) char s2[5]="abc";

       C) char s3[]={'m','n','s','r'};     D) char s4[4]="ijkl";

D需要分配5个空间

A则不需要

2)    If the array name as an argument, the function call passed  (    A ) to the parameter.

A) the starting address of the array           

B) the value of the first element

     C) all the elements values of an array

D) the number of array elements

传递（A）数组的起始地址。

3)    If we have : int x=3, y=2;  double a=2.9, b=3.5;  the value of expression (x+y)/2+(int)a%(int)b   would be（      ）。

A)  1                    B)  2                  C)   3                D)   4

D

4)    In the following statements about the overload function which one is correct? （      ）

A) overload functions must have different return types

B) overload functions must have different numbers of parameters

C) overload functions must have a different parameter list

D) overload functions’ name can be different

C

5)    In the following data types which one does not belong to the basic data types in C ++. （      ）

A)     int         B)   double         C) char      D) class     

D显而易见

6)    We have :  double fun( double l ); typedef double ft ( double ) ;

 ft * pfun[2];    pfun[0]=fun;，in the following function calls which one is NOT correct （     ）?

A）( *pfun[0] )(3.14)                      B）fun(3.14)

C）   ( pfun[0] )(3.14)                     D）( &pfun[0] )(3.14)

D

7)    in the following identifiers groups, which one is both legitimate user identifiers? (      )

A)    _0123  ssiped                B)   del-word  signed

C)      list  *jer                          D) keep%  wind

 A

8)    The operand data type on both sides of logical operators is (      )

A) can only be 0 or 1

B) can only be a positive integer or 0

C) can only be an integer or character data

D) can be any type of legitimate data

D

9)    In the following statements, which one is correct to initialize a two-dimensional array? (   B，D   )

A)   int a[2][]={{1,0,1},{5,2,3}};

B)    int a[][3]={{1,2,3},{4,5,6}};

C)    int a[2][4]={{1,2,3},{4,5},{6}};

D)   int a[][3]={{1,0},{},{1,1}};

10)  For the members of class, the implies access is ( )

(A) public;   (B)private;   (C) protected;    (D) static;

默认私有

Question 2: ( 20 points )

What is the output of the following code fragment?

Part1 (3 points)

#include  

using namespace std;

main()

{  int  x=1,y=0,a=0,b=0;

        switch(x)

        {  case  1:

                switch(y)

{   case  0:a++;  break;

                case  1:b++;  break;

}

case 2:a++; b++; break;

case 3:a++; b++;

}

cout<<"a="<

}

 

Output

a=2,b=1

Part2( 3 points)

#include

int f1(int a,int b) {return a%b*5;}

int f2(int a,int b) {return a*b;}

int f3(int(*t)(int, int),int a,int b) { return (*t)(a, b);}

void main()

{      int (*p)(int, int) ;

    p=f1 ;  cout<

    p=f2 ;  cout<

}

Output

25 56

Part3 (3 points)

#include  

using namespace std;

int fun(int n)

{      if(n==1)return 1;

      else

    return(n+fun(n-1));

}

main()

{      int x;

cin>>x; x=fun(x);

cout<

}

Output

55

When the input for x is 10,the output is：

Part4 (3 points)

#include

using namespace std;

void fun(int a,int b)

{      int k;

        k=a;a=b;b=k;

}

void main()

{      int a=4,b=7,*x,*y;

x=&a;y=&b;

fun(*x,*y);

cout<<"No.1:"<

fun(a,b);

cout<<"No.1:"<

}

No.1:4,7

No.1:4,7

Part 5( 4 points)

#include

using namespace std;

int fun(int n)

{     static int a=3;

        int t=0;

        if(n%2)

{      static int a=5; t+=a++; }

    else

{      static int a=5;  t+=a++; }

    return t+=a++;

}

void main()

{ int i,s=0;

  for(i=0;i<3;i++) s+=fun(i);

  cout<

}

Output

28

Part 6( 4 points)

#include

using namespace std;

void main()

{      char ch[2][6]={"2100","0846"},*pch[2];

        int i,j,s=0;

        for(i=0;i<2;i++)

                 pch[i]=ch[i];

        for(i=0;i<2;i++)

                 for(j=0;pch[i][j]>='0' && pch[i][j]<='9';j+=2)

                         s=10*s+pch[i][j]-'0';

        cout<

Output

2004

}

Question 3    Short answers (20 points)

Part 1 (2 points)

In the function prototype int fun(int, int=0);  what does the “int=0” mean？

参数默认值是0

Part 2 (2 points)

After executing the following statements, what is the values of x and y ?

int x, y; x=y=1;    ++x||++y;

__x=2，y=1____________

Part 3 (2 points)

When the following statements are running, how many times does the loop execute?

int i=0, x=1;  do {  x++;   i++;  }  while ( !x&&i<=3 );

_1次__

Part 4  (8 points,2 points for each)

Insert the braces in the following code to produce a specified output. Please NOTE that the program may not make any other changes except braces.

if (y==8)

if (x==5)

cout << “@@@@” << endl;

else

cout << “####” <

cout << “$$$$” <

cout << “&&&&” <

1. When x=5 and y=8, The output is:

@@@@

$$$$

&&&&

1. When x=5 and y=8, The output is:

@@@@

1. When x=5 and y=8, The output is:

@@@@

&&&&

1. When x=5 and y=7, The output is:

####

$$$$

&&&&

{ } 可有可无，{ }必须有

1. if (y==8)

if (x==5)

{  cout << “@@@@” << endl;  }

else

{   cout << “####” <}

cout << “$$$$” <

cout << “&&&&” <

2） if (y==8)

if (x==5)

{   cout << “@@@@” << endl;  }

else

{  cout << “####” <

cout << “$$$$” <

cout << “&&&&” <}

1. if (y==8)

if (x==5)

{  cout << “@@@@” << endl;  }

else

{    cout << “####” <

cout << “$$$$” <}

cout << “&&&&” <

4） if (y==8)

{  if (x==5)

cout << “@@@@” << endl;  }

else

{   cout << “####” <

cout << “$$$$” <

cout << “&&&&” <}

Part 5  (6 points,1 point for each)

Find the errors and fix it in each of the following program segments. Assume the following declarations and statements:

void * sPtr = 0;

int number;

int z[5] = {1,2,3,4,5};

int * zPtr;

fix each statement.

1. // use pointer point the starting address of the array z

zPtr=z[0];

1. // use pointer to get first value of array

number = zPtr;

1. // assign the third element of array z (the value 3) to number

number = *zPtr[ 2 ];

1. // print entire array z

for( int i=0; i<=5; i++)

       cout<

1. //let the pointer point the second element of the array

zPtr=2;

1. // assign the value pointed by zPtr to number

number = zPtr;

zPtr=z / zPtr=&z[0]

_________________________________________________________

   number=zPtr[0]  /  number=*zPtr

_________________________________________________________

   Number=zPtr[2]  /  number = *(zPtr+2)

_________________________________________________________

1.  

i<5

_________________________________________________________

   zPtr=z+1   /  zPtr=&a[1]

_________________________________________________________

   Number=*zPtr

__________________________

Question 4: Fill in the blanks ( 20 Point,2 points for each blank)

1. The following program calculates values of e according to the formula 1!+3!+5!+……+n! .

#include

using namespace std;

void main()

{   long int f,s;

     int i,j,n;

     cin>>n;

     s=0,f=1;

     for(i=1;i<=n;       （1）       )

     {

              f=1;

              for(j=1; j<=i; j++)       （2）       ;

                   （3）       ;

     }

     cout<<"n="<}

1______i+=2 / i=i+2______________2_____f*=j / f=f*j__________

2．The main function get a string, then call other function to change the numbers 0～9 to lower case letter a～j；And change all the lower case letters to up case letters, then output the result in function main.

#include

using namespace std;

_____(4)_____

void main()

{   char str1[20], str2[20];

   cin>>str1;

 change(str1,str2);

 cout<

}

void change(char *s1, char *s2)

{   while(_____(5)_____)

 {     if(*s1>='0'&&*s1<='9')

                  *s2=_____(6)_____;

          else *s2=toupper(*s1);

          _____(7)_____

 }

 *s2='\0';

}

#include

#include

using namespace std;

void change(char *s1, char *s2);

int main()

{

    char str1[20], str2[20];

    cin >> str1;

    change(str1, str2);

    cout << str2 << endl;

    return 0;

}

void change(char *s1, char *s2)

{

    while (*s1)

    {

        if (*s1 >= '0' && *s1 <= '9')

            *s2 = *s1 - '0' + 'a';

        else

            *s2 = toupper(*s1);

        s1++;

        s2++;

    }

    *s2 = '\0';

}

3.    Output matrix as the Figure right of the code:

main()

{ int a[7][7];

 int i,j;

 for (i=0;i<7;i++)

   for (j=0;j<7;j++)

     { if (_____(8)___________) a[i][j]=1;

       else if (i

       else if (i>j&&i+j<6) a[i][j]=3;

       else if (____(10)____ ___) a[i][j]=4;

       else a[i][j]=5;

      }

   for (i=0;i<7;i++)

     {         for (j=0;j<7;j++)   cout<

                 cout<

     }

}

9____a[i][j]=2__________10 ____i6__________________

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