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Thursday, June 5, 2014

C Aptitude-2

Predict the output or error(s) for the following:

27) main()
        {
        clrscr();
        }
        clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28) enum colors {BLACK,BLUE,GREEN}
        main()
        {
        printf("%d..%d..%d",BLACK,BLUE,GREEN);
        return(1);
        }
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

29) void main()
        {
          char far *farther,*farthest;
          printf("%d..%d",sizeof(farther),sizeof(farthest));
        }
Answer:
4..2  
Explanation:
the second pointer is of char type and not a far pointer

30) main()
        {
        int i=400,j=300;
        printf("%d..%d");
        }
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31) main()
        {
         char *p;
         p="Hello";
         printf("%c\n",*&*p);
        }
Answer:

Explanation:
* is a dereference operator & is a reference  operator. They can be    applied any number of times provided it is meaningful. Here  p points to  the first character in the string "Hello". *p dereferences it and so its value is H. Again  & references it to an address and * dereferences it to the value H.

32) main()
        {
         int i=1;
         while (i<=5)
        {
         printf("%d",i);
         if (i>2)
 goto here;
         i++;
        }
      }
      fun()
      {
       here:
       printf("PP");
      }
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

33) main()
        {
         static char names[5][20]=                                                                      {"pascal","ada","cobol","fortran","perl"};
         int i;
         char *t;
         t=names[3];
         names[3]=names[4];
         names[4]=t;
         for (i=0;i<=4;i++)
     printf("%s",names[i]);
        }
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

34) void main()
        {
int i=5;
printf("%d",i++ + ++i);
        }
Answer:
Output Cannot be predicted  exactly.
Explanation:
Side effects are involved in the evaluation of   i

35) void main()
        {
int i=5;
printf("%d",i+++++i);
        }
Answer:
Compiler Error 
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators. 
   
36) #include<stdio.h>
        main()
       {
        int i=1,j=2;
        switch(i)
       {
        case 1:  printf("GOOD");
        break;
        case j:  printf("BAD");
  break;
       }
       }    
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements. 

37) main()
        {
        int i;
        printf("%d",scanf("%d",&i));  // value 10 is given as input here
        }
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0.  Here 10 is given as input which should have been scanned successfully. So number of items read is 1. 

38) #define f(g,g2) g##g2
        main()
       {
        int var12=100;
        printf("%d",f(var,12));
}
Answer:
100 

39) main()
        {
        int i=0;
        for(;i++;printf("%d",i)) ;
        printf("%d",i);
        }
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40) #include<stdio.h>
        main()
        {
        char s[]={'a','b','c','\n','c','\0'};
        char *p,*str,*str1;
        p=&s[3];
        str=p;
        str1=s;
        printf("%d",++*p + ++*str1-32);
        }
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32. 
i.e. (11+98-32)=77("M");

41) #include<stdio.h>
       main()
      {
       struct xx
       {
         int x=3;
         char name[]="hello";
        };
       struct xx *s=malloc(sizeof(struct xx));
       printf("%d",s->x);
       printf("%s",s->name);
      }
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

42) #include<stdio.h>
        main()
        {
         struct xx
         {
   int x;
   struct yy
  {
    char s;
    struct xx *p;
  };
          struct yy *q;
         };
        }
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

43) main()
       {
        extern int i;
        i=20;
        printf("%d",sizeof(i));
       }
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44) main()
       {
        printf("%d", out);
       }
       int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45) main()
       {
        extern out;
        printf("%d", out);
       }
       int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
     
46) main()
       {
        show();
       }
       void show()
      {
       printf("I'm the greatest");
      }
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47) main( )
       {
         int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
         printf(“%u %u %u %d \n”,a,*a,**a,***a);
 printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
       }
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array. 

thus, for the first printf statement a, *a, **a  give address of  first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48) main( )
       {
        int a[ ] = {10,20,30,40,50},j,*p;
        for(j=0; j<5; j++)
       {
       printf(“%d” ,*a); 
       a++;
      }
      p = a;
      for(j=0; j<5; j++) 
      {
       printf(“%d ” ,*p); 
       p++;
     }
   }
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

49) main( )
       {
       static int  a[ ]   = {0,1,2,3,4};
       int  *p[ ] = {a,a+1,a+2,a+3,a+4};
       int  **ptr =  p;
       ptr++;
       printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
      *ptr++;
      printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
      *++ptr;
     printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
     ++*ptr;
     printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
     }
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
                                                                       a

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor) = 1,  **ptr is the value stored in the location pointed by  the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is  1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2. 
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3. 
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4. 

50) main( )
       {
       char  *q;
       int  j;
       for (j=0; j<3; j++) scanf(“%s” ,(q+j));
       for (j=0; j<3; j++) printf(“%c” ,*(q+j));
       for (j=0; j<3; j++) printf(“%s” ,(q+j));
      }
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE,  TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

The third input  starts filling from the location 102

This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2  = M T V
The second printf prints three strings starting from locations q, q+1, q+2
 i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.

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