C PUZZLE

• void main(){

      void *v;

      int integer=2;

      int *i=&integer;

      v=i;

      printf("%d",(int*)*v);

}

Answer:

Compiler Error. We cannot apply indirection on type void*.

Explanation:

Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,

Passing generic pointers to functions and returning such pointers.

As a intermediate pointer type.

Used when the exact pointer type will be known at a later point of time.

• void main(){

      int i=i++,j=j++,k=k++;

printf(“%d%d%d”,i,j,k);

}

Answer:

Garbage values.

Explanation:

An identifier is available to use in program code from the point of its declaration.

So expressions such as  i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

• void main(){

      static int i=i++, j=j++, k=k++;

printf(“i = %d j = %d k = %d”, i, j, k);

}

Answer:

i = 1 j = 1 k = 1

Explanation:

Since static variables are initialized to zero by default.

• void main(){

      while(1){

            if(printf("%d",printf("%d")))

                  break;

            else

                  continue;

      }

}

Answer:

Garbage values

Explanation:

The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf  prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

• main(){

      unsigned int i=10;

      while(i-->=0)

            printf("%u ",i);

}

Answer:

      10 9 8 7 6 5 4 3 2 1 0 65535 65534…..

Explanation:

Since i is an unsigned integer it can never become negative. So the expression i-- >=0  will always be true, leading to an infinite loop.   

• main(){

      int x,y=2,z,a;

      if(x=y%2) z=2;

      a=2;

      printf("%d %d ",z,x);

}

Answer:

Garbage-value 0

Explanation:

The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.

Thumb Rule: Check all control paths to write bug free code.

• main(){

      int a[10];

      printf("%d",*a+1-*a+3);

}

Answer:

4 

Explanation:

      *a and -*a cancels out. The result is as simple as 1 + 3 = 4 !   

• main(){

      unsigned int i=65000;

      while(i++!=0);

      printf("%d",i);

}

Answer:

 1

Explanation:

Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

 

• main(){

      int i=0;

      while(+(+i--)!=0)

            i-=i++;

      printf("%d",i);

}

Answer:

-1

Explanation:

Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is,    while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

     

• main(){

      float f=5,g=10;

      enum{i=10,j=20,k=50};

      printf("%d\n",++k);

      printf("%f\n",f<<2);

      printf("%lf\n",f%g);

      printf("%lf\n",fmod(f,g));

}

Answer:

Line no 5: Error: Lvalue required

Line no 6: Cannot apply leftshift to float

Line no 7: Cannot apply mod to float

Explanation:

Enumeration constants cannot be modified, so you cannot apply ++.Bit-wise operators and % operators cannot be applied on float values.fmod() is to find the modulus values for floats as % operator is for ints.