Try these C aptitude

Replies

• 

  Neha Kochhar

  moonfalsh

  Go for these C questions and tel the answer and errors if any:
  
  1. void main()
  {
  int const * p=5;
  printf("%d",++(*p));
  }
  
  

  Answer of 1st question
  According to me it will result in an error as we cannot change the value the pointer is pointing as pointer is pointing to constant integer .
  Tell me where I am wrong?
• 

  gaurav.bhorkar

  moonfalsh

  3. main()
  {
  extern int i;
  i=20;
  printf("%d",i);
  }

  Output: 20
  
  Reason: A local variable is always preferred over global variable if their names are same.
• 

  Manish Goyal

  @ Gaurav which compiler are you using? Actually it is giving error in my system
• 

  gaurav.bhorkar

  goyal420

  @ Gaurav which compiler are you using? Actually it is giving error in my system

  I didn't compile that.
  
  You are getting error because the variable 'i' is not defined outside main ().
  Maybe my answer was wrong.
• 

  Mangesh6688

  moonfalsh

  
  3. main()
  {
  extern int i;
  i=20;
  printf("%d",i);
  }
  

  Its giving error as:
  
  "Undefined symbol _i in module"
• 

  Mangesh6688

  moonfalsh

  
  5. main()
  {
  printf("%x",-1<<4);
  }
  

  The %x format specifier represents the integer value in a hexadecimal form.
  -1 is represented as all 1's and when it is four times left shifted the least significant 4 bits are filled with 0's.
  
  Output: fff0
• 

  gaurav.bhorkar

  Mangesh6688

  Its giving error as:
  
  "Undefined symbol _i in module"

  Thats what I said. The variable should also be defined outside main ().
• 

  Mangesh6688

  moonfalsh

  
  2. main()
  {
  int c[ ]={2.8,3.4,4,6.7,5};
  int j,*p=c,*q=c;
  for(j=0;j<5;j++) {
  printf(" %d ",*c);
  ++q; }
  for(j=0;j<5;j++){
  printf(" %d ",*p);
  ++p; }
  }
  
  

  As c[] is declared as integer, c will contain {2,3,4,6,5}.
  In first 'for' loop we are printing *c, as *c contains 2 it will be printed 5 times.
  In second 'for' loop we are printing *p, initially *p contains 2 then it is incremented.
  So, 2,3,4,6,5 will be printed.
  
  Final Output: 2 2 2 2 2 2 3 4 6 5
• 

  moonfalsh

  Sorry for late reply guys.
  
  Answer for the questions are:
  
  1. Compilation error - cannot modify a constant value.
  
  2. 2222223465
  
  3. linker error- undefined symbol i
  
  4. 00131
  
  5. fffo
• 

  sherya mathur

  answers are
  ans1. compilation error
  explanation
  because p is a constant integer but here we are going to change the value of a constant integer
  ans2: the out will be
  2 2 2 2 2 2 3 4 6 5
  Explanation:
  Initially pointer c is assigned to both p and q. In the first loop,
  since
  only q is incremented and not c , the value 2 will be printed 5 times.
  In
  second loop p itself is incremented. So the values 2 3 4 6 5 will be
  printed.
  ans3: Linker Error
  explaination: Undefined symbol '_i'
  Explanation:
  extern storage class in the following declaration,
  extern int i;
  specifies to the compiler that the memory for i is allocated in some
  other
  program and that address will be given to the current program at the
  time
  of linking. But linker finds that no other variable of name i is
  available
  in any other program with memory space allocated for it. Hence a linker
  error has occurred
  ans 4:00131
  ans5: fff0