Tuesday 26 March 2013

Is the following code legal?


    Is the following code legal?
struct a
    {
int x;
 struct a b;
    }
Answer:
         No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be recursive without end.

Blog Author: Vijay Kumar

Is there any difference between the two declarations,


    Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.


Blog Author: Vijay Kumar

*a ^= *b, *b ^= *a, *a ^= *b;


    main()
{
         void swap();
    int x=10,y=8;    
         swap(&x,&y);
    printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
   *a ^= *b,  *b ^= *a, *a ^= *b;
}  
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
   *a ^= *b,  *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

Blog Author: Vijay Kumar

*f2+=*f2+=a+=2.5;


    main()
{
         int a=2,*f1,*f2;
    f1=f2=&a;
         *f2+=*f2+=a+=2.5;
    printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.

Blog Author: Vijay Kumar

extern variable


    main()
{  
         extern i;
    printf("%d\n",i);
         {
             int i=20;
         printf("%d\n",i);
         }
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.

Blog Author: Vijay Kumar

switch(float)


    main()
{
         float i=1.5;
    switch(i)
         {
         case 1: printf("1");
             case 2: printf("2");
             default : printf("0");
    }
}
Answer:
Compiler Error: switch expression not integral
Explanation:
         Switch statements can be applied only to integral types.

Blog Author: Vijay Kumar

register int a=2;


    main()
{
         register int a=2;
    printf("Address of a = %d",&a);
         printf("Value of a   = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
          & (address of ) operator cannot be applied on register variables.

Blog Author: Vijay Kumar

j = j || i++ && printf("YOU CAN");


    main()
{
         int i=4,j=7;
    j = j || i++ && printf("YOU CAN");
         printf("%d %d", i, j);
}

Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.    
    false && (anything) => false where (anything) will not be evaluated.

 Blog Author: Vijay Kumar

i=i&=j&&10;


    main()
{
         int i=5,j=10;
    i=i&=j&&10;
         printf("%d %d",i,j);
}

Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

 Blog Author: Vijay Kumar

Differentiate the above declarations


    1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
    *a='F'       : illegal
             a="Hi"       : legal

2. 'const' applies to 'a'  rather than to the value of a (constant pointer to char )
    *a='F'       : legal
    a="Hi"       : illegal

3. Same as 1.

 Blog Author: Vijay Kumar

j = i, j?(i,j)?i:j:j;


    main()
{
         int i=10,j=20;
    j = i, j?(i,j)?i:j:j;
         printf("%d %d",i,j);
}

Answer:
10 10
Explanation:
         The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
         if(i,j)
            {
if(i,j)
              j = i;
         else
             j = j;           
         }
       else
         j = j;      

Blog Author: Vijay Kumar

(i>2)?i++:i--;


    void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
    printf(“%d”, i);
}
Answer:
         32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

Blog Author: Vijay Kumar

for (;i++=0;) printf(“%d”,i);


    main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}

Answer:
         Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and  hence it cannot appear on the left hand side of an assignment operation.

Blog Author: Vijay Kumar

while (strcmp(str1,str2)) (2)


    main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

Blog Author: Vijay Kumar

while (strcmp(“some”,”some\0”))


    main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
    }
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.
Blog Author: Vijay Kumar