C Sample Question

Note : All the C sample programs are tested under Turbo C/C++ compilers.
It is assumed that,

• Programs run under DOS environment,
• The underlying machine is an x86 system,
• Program is compiled using Turbo C/C++ compiler.

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Following are some C sample questions.

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

1. void main() { 	int  const * p=5; 	printf("%d",++(*p)); }

   Answer:
   

   Compiler error: Cannot modify a constant value.

   Explanation:

   p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main() { 	char s[ ]="man"; 	int i; 	for(i=0;s[ i ];i++) 	printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]); }

   Answer:

   mmmm
   aaaa
   nnnn

   Explanation:

   s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main() { 	float me = 1.1; 	double you = 1.1; 	if(me==you) printf("I love U"); else 		printf("I hate U"); }

   Answer:

   I hate U

   Explanation:

   For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

   Rule of Thumb:

   Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ).

4. main() 	{ 	static int var = 5; 	printf("%d ",var--); 	if(var) 		main(); 	}

   Answer:

   5 4 3 2 1

   Explanation:

   When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. 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;	  } } 

   Answer:

   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.

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

   Answer:

   Linker Error : 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 .

7. main() { 	int i=-1,j=-1,k=0,l=2,m; 	m=i++&&j++&&k++||l++; 	printf("%d %d %d %d %d",i,j,k,l,m); }

   Answer:

   0 0 1 3 1

   Explanation :

   Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression 'i++ && j++ && k++' is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for '0 || 0' combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main() { 	char *p; 	printf("%d %d ",sizeof(*p),sizeof(p)); }

   Answer:

   1 2

   Explanation:

   The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main() { 	int i=3; 	switch(i) 	 { 	    default:printf("zero"); 	    case 1: printf("one"); 		   break; 	   case 2:printf("two"); 		  break; 	  case 3: printf("three"); 		  break; 	  }   }

   Answer :

   three

   Explanation :

   The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

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

   Answer:

   fff0

   Explanation :

   -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

   main() { 	char string[]="Hello World"; 	display(string); } void display(char *string) { 	printf("%s",string); }

   Answer:

   Compiler Error : Type mismatch in redeclaration of function display

   Explanation :

   In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

10. main() { 	int c=- -2; 	printf("c=%d",c); }

    Answer:

    c=2;

    Explanation:

    Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

    Note:
    However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

11. #define int char main() { 	int i=65; 	printf("sizeof(i)=%d",sizeof(i)); }

    Answer:

    sizeof(i)=1

    Explanation:

    Since the #define replaces the string int by the macro char

12. main() { int i=10; i=!i>14; printf("i=%d",i); }

    Answer:

    i=0

    Explanation:

    In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

13. #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:

    77

    Explanation:

    p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
    Now performing (11 + 98 - 32), we get 77("M");
    So we get the output 77 :: "M" (Ascii is 77).

    #include‹stdio.h› main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d----%d",*p,*q); }

    Answer:

    SomeGarbageValue---1

    Explanation:

    p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

14. include‹stdio.h›

Answer:

Compiler Error

Explanation:

You should not initialize variables in declaration

#include‹stdio.h› main() { struct xx { int x; struct yy { char s; 	struct xx *p; }; struct yy *q; }; }

main() { printf("\nab"); printf("\bsi"); printf("\rha"); }

main() { int i=5; printf("%d%d%d%d%d%d",i++,i--,++i,--i,i); }