C language practice questions + Answers:

One

1． set up x The value of is 15,n The value of is 2, Expression x%=(n+=3) After operation ,x The value of is （　0　）.

2． set up  int a=7,b=9,t; After executing the expression t=(a>b)?a:b after ,t The value of is （  9  ）.

3． The output of the following program segment is （ 210  322 ）.

      int a=1234; a=a&0377; printf(“%d %o\n”,a,a);

4．a The array is defined as follows , In memory order ,a All elements in the array are （  a[0], a[1], a[2]    ）.

      char a[3];

5． The following macro definitions are provided , Then execute the assignment statement  a=PP*20;（a by int Type variable ） after ,a The value of is （ 8140 ）.

      #define PR 80

      #define PP PR+403

6． stay C In the program , There are three ways to assign address values to pointer variables , They are ：（  ）、（  ） and （    ）.

answer ：int i, a[10], *p1, *p2;
（1）p1=&i; Assign a value to a pointer variable whose base type is an integer with the address of the integer variable
（2）p2=a; p2=a+3; Give the address to the pointer variable with the array name ;
（3） p1=p2; Assign values to pointer variables through pointer variables ;

7． stay C In file , The two code forms of data are （   ASCII file    ） and （    Binary    ）.

8．C Call in language （ fopen()   ） Function to open a file , call （ fclose()   ） Function close file , call （  fseek()   ） Function can realize random reading and writing of files .

9． If you have any int a[3]={10,12,30}; be a+1 yes （ &a[1] ） The address of ,*(a+2)=（ 30  ）.

Two  

1． stay Turbo C in , Integer data occupies... In memory 2 Bytes .（ v ）

2． expression 1/4+2.75 The value of is 3.（ x ）

answer ：1/4=0;0+2.75=2.75

1/4 In this expression 1 and 4 They are all literal integers Then the result is also an integer value 0

therefore The expression should be 2.75

3． There is only one type of structure .（ x ）

answer ： Structure types can be customized and declared by users

4． If the function has no return value , Then it must be invisible .（ x ）

5．C Language can only deal with text files and binary files .（ v ）

3、 ... and   choice question （20 branch ）

1． set up  int a=12;  expression a+=a-=a*=a The value of is （ C ）.

   A 12      B 144      C 0      D 132

+=、-=、*= These three operators are executed from right to left

First step perform a=a*a  a=144; 

The second step perform a=a-a  a=144-144=0; 

The third step perform a=a+a  a=0+0=0;

2． The output of the following program is （　C　）.

  int main( )

{

int a=011;

printf(“%d\n”,++a);

}

A 12     B 11        C 10      D 9

a=011 Express 8 Base number , yes 9,++a Means to add first 1, have to a=10, So the result is 10

3． The output of the following program is （ D  ）.

   #define ADD(x) (x)+(x)

   int main()

   {   

int a=4,b=6,c=7,d=ADD(a+b)*c;

printf(“d=%d”,d);

    }

   A d=70      B d=140     C d=280      D d=80

add(x) This is replaced by X+X
The formula finally becomes （a+b）+(a+b)*c

4. The correct assignment statement in the following options is （char a[5],*p=a;）（　A　）.

   A p=“abcd”;        B a=“abcd”;     C *p=“abcd”;       D *a=“abcd”;

p It's a character pointer , take “abcd” The address of p, That is to say p Point to “abcd" The first address , That is to say

Point to ‘a' The address of ;

B In the options ,a It's a pointer constant

C、D Variable on the left , On the right is the pointer

5． if k For shaping , be while Loop execution （　D　） Time .

   k=2;  while(k==0)  printf(“%d”,k);   k--;  printf(“\n”);

　 A 10       B 9    C 0     D 1

while After the first cycle of , because （k==0） Not meeting the conditions ！ So the cycle stops ,k The value of is also 2;

6． When the array name is passed to the formal parameter as an argument , The array name is processed as （　C　）.

A  The length of the array   B  The number of elements of the array   C  The first address of the array   D  The value of each element in the array

7． Two pointer variables cannot (  A  ).

A.  Add up       B  Compare      C  Subtracting the      D  Point to the same address

A. Add up , It's not that we can't , It's meaningless

B. Subtracting the , Yes. , For example, a pointer points to the beginning of the string , The end of another pointer , Subtract to get the number of strings

C. Compare , Yes. , We can compare whether they are equal , You can also compare who is big and who is small

D. Point to the same address , No need to explain , Certainly.

8． If there are the following program segments , Then the value is 6 The expression of （ D  ）.

     struct st{ int n;struct st *next;};

     static struct st a[3]={5,&a[1],7,&a[2],9,‘\0’},*p; p=&a[0];

A p++->n       B p->n++    C (*p).n++     D ++p->n

#include<stdio.h>

void main(void)

{

    structst { int n; structst* next; };

    // Type of structure st, Two members ： An integer member n, A pointer ( quote ) Type members next, This pointer is used to point to the structure type st The variable of

    structst a[3] = { 5,&a[1],7,&a[2],9,&a[0] }, * p;

    // Define an inclusion 3 Elemental st An array of struct types a, The three elements are {5,&a[1]},{7,&a[2]},{9,&a[3]}.

    //( In each element , The first 1 Members correspond to n, The first 2 Members correspond to next) It also defines a point that can point to st Pointer to structure type variable p

    p = &a[0];

    //p Point to st Type array a Of the 1 Addresses of elements ( The first address )

    // Then the value is 6 The expression of

    //p ++ ->n //p->n Point to 5, Take the value first and increase it to 6++＝6

    //p->n ++ //p->n Point to 5, Take the value first and increase it to 6++＝6

    //(*p).n ++// ditto

    //++ p->n // Point to p->n value 5++＝6

    printf("%d\n", ++p->n);

}

/*

expression A Equate to p->n then p=p+1, The expression value is p Before self increase p->n The value of 5

expression B First calculate p->n Value , Which is equivalent to a[0].n( The value is 5), And then execute (a[0].n)++, Note that it takes value first and then increases automatically , So the value of the expression is still 5

expression C First calculate (*p).n, That's equivalent to a[0].n( The value is 5), And then execute (a[0].n)++ Just like expression B The same , The value of the expression is still 5

expression D That's right. , amount to ++(a[0].n), The value is 6

*/

9． To open an existing non empty file ＂file＂ For modification , Choose the right statement （ C ）.

A fp＝fopen("file","r");   B fp=fopen("file","w");

C fp=fopen("file","r+");   D fp=fopen("file","w+");

r Open read-only file , The file must exist .

r+ Open a read-write file , The file must exist .

w Open write only , If the file exists, the file length is clear as 0, That is, the contents of the file will disappear . If the file does not exist, create it .

w+ Open a read-write file , If the file exists, the file length is clear to zero , That is, the contents of the file will disappear . If the file does not exist, create it .

a Open write only files in an additional way . If the file does not exist , The file will be created , If the file exists , The data written will be added to the end of the file , That is, the original contents of the document will be preserved .

a+ Open a read-write file as an attachment . If the file does not exist , The file will be created , If the file exists , The written data will be added to the end of the file , That is, the original contents of the document will be preserved .

3、 ... and 、 Write the running results of the following programs .

1．main( )

{  int a[6]={10,6,23,-90,0,3},i;

     invert(a,0,5);

     for(i=0;i<6;i++)  printf(“%d,”,a[i]);

     printf(“\n”);

}

invert(int *s,int i,int j)

{  int t;

     if(i<j)

     {  invert(s,i+1,j-1);

        t=*(s+i);*(s+i)=*(s+j);*(s+j)=t;

      }

}    

Output results ：3  0  -90 23  6  10

---

 

---

invert(int* s, int i, int j)/* Defined function invert, The parameter s It's the pointer type ,i,j Is an integer */
{ // Function entrance 
	int t; // Definition t Is an integer 
	if (i < j) // If i<j
	{
		invert(s, i + 1, j - 1); /* hold s,i+1,j-1 Call the function as an argument invert, This is the vector usage of a function */
		t = *(s + i);
		*(s + i) = *(s + j);
		*(s + j) = t;
		// These three lines are pointers (s+i) With the pointer (s+j) Point to the value exchange 
	} //if End of chunk 
} // End of the function 
// This function is used to drop the items in the array , That is, the last item becomes the first item , The first item becomes the last ……
main()
{
	int a[6] = { 10,6,23,-90,0,3 }, i; // Define an array a And integer variables i
	invert(a, 0, 5); // In array a The address of 、 Numbers 0、5 Call the function as an argument invert
	for (i = 0; i < 6; i++) printf("%d", a[i]); // function invert Put the array a Of 6 Output after sorting items 
	printf("\n"); // Last output enter 
}

---

Four 、 Read the following procedure , stay      Fill in the appropriate content , Make the program complete .

seek 100～200 All prime numbers between .

     （1）   

main()

{  int m,k,i,n=0;

  for(m=101;m<=200;m+=2)

  {   if(n%10==0)    printf("\n");

           k=sqrt(m);

           for(i=  （2）  ;i<=k;i++)     if(m%i==0)      （3）  ;

           if(i==  (4)  )

      {  printf("%d ",m);n++;}

   }

}

（1）#include<stdio.h>  #include<math.h> （2）i=2  (3) break   (4)k+1

---

#include<stdio.h>

#include<math.h>

int main()

{

    int m, k, i, n = 0;

    for (m = 101; m <= 200; m += 2)

    {

        if (n % 10 == 0)

            printf("\n");

        k = sqrt(m);

        for (i = 2; i <= k; i++)

            if (m % i == 0)

                break;

        if (i == k + 1)

        {

            printf("%d ", m);

            n++;

        }

    }

}