Technical-C

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Technical C Question Paper

Give the output of the programs in each case unless mentioned otherwise

1. void main()
   {
     int d=5;
     printf("%f",d);
   }Ans: Undefined

2. void main()
   {
     int i;
     for(i=1;i<4,i++)                                                                                 
     switch(i)
     case 1: printf("%d",i);break;
   {
    case 2:printf("%d",i);break;
    case 3:printf("%d",i);break;
   }
    switch(i) case 4:printf("%d",i);
   }Ans: 1,2,3,4

3. void main()
   {
    char *s="\12345s\n";
    printf("%d",sizeof(s));
   }Ans: 6

4. void main()
   {
     unsigned i=1; /* unsigned char k= -1 => k=255; */
     signed j=-1; /* char k= -1 => k=65535 */
   /* unsigned or signed int k= -1 =>k=65535 */                                  
   if(i<j)
     printf("less");
   else
   if(i>j)
     printf("greater");
   else
   if(i==j)
     printf("equal");
   }Ans: less

5. void main()
   {
     float j;
     j=1000*1000;
     printf("%f",j);
   }
   
   1. 1000000
   2. Overflow
   3. Error
   4. None
   Ans: 4

6. How do you declare an array of N pointers to functions returning  pointers to functions returning pointers to characters?     Ans: The first part of this question can be answered in at least        three ways:

7. Build the declaration up incrementally, using typedefs:
           typedef char *pc;    /* pointer to char */
           typedef pc fpc();    /* function returning pointer to char */
           typedef fpc *pfpc;    /* pointer to above */
           typedef pfpc fpfpc();    /* function returning... */
           typedef fpfpc *pfpfpc;    /* pointer to... */
           pfpfpc a[N];         /* array of... */

8. Use the cdecl program, which turns English into C and vice versa:          
           cdecl> declare a as array of pointer to function returning   pointer to function returning pointer to char        char *(*(*a[])())()
   cdecl can also explain complicated declarations, help with  casts, and indicate which set of parentheses the arguments   go in (for complicated function definitions, like the one   above).  Any good book on C should explain how to read these complicated   C declarations "inside out" to understand them ("declaration mimics use"). The pointer-to-function declarations in the examples above have not included parameter type information. When the parameters have complicated types, declarations can *really* get messy. (Modern versions of cdecl can help here, too.)

9. A structure pointer is defined of the type time . With 3 fields min,sec hours having pointers to intergers.
       Write the way to initialize the 2nd element to 10.

10. In the above question an array of pointers is declared. Write the statement to initialize the 3rd element of the 2 element to 10

11. int f()
    void main()
    {
      f(1);
      f(1,2);
      f(1,2,3);
    }
    f(int i,int j,int k)
    {
      printf("%d %d %d",i,j,k);
    }What are the number of syntax errors in the above?                     
    Ans: None.

12. void main()
    {
      int i=7;
      printf("%d",i++*i++);
    }Ans: 56

13. #define one 0
    #ifdef one
    printf("one is defined ");
    #ifndef one
    printf("one is not defined ");
    Ans: "one is defined"

14. void main()
    {
      intcount=10,*temp,sum=0;
      temp=&count;
      *temp=20;
      temp=&sum;
      *temp=count;
      printf("%d %d %d ",count,*temp,sum);
    }
    Ans: 20 20 20

15. There was question in c working only on unix machine with pattern matching.

16. what is alloca()   Ans : It allocates and frees memory after use/after getting out of scope

17. main()
    {
      static i=3;
      printf("%d",i--);
      return i>0 ? main():0;
    }
    Ans: 321

18. char *foo()
    {
      char result[100]);
      strcpy(result,"anything is good");                                        
      return(result);
    }
    void main()
    {
      char *j;
      j=foo()
      printf("%s",j);
    }
    Ans: anything is good.

19. void main()
    {
      char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
      char **p;
      p=s;
      printf("%s",++*p);
      printf("%s",*p++);
      printf("%s",++*p);
    }Ans: "harma" (p->add(dharma) && (*p)->harma)
    "harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
    "ewlett-packard"