Important Errors

    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. 

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while(+(+i--)!=0)

    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.

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while(i++!=0);

    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.

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printf("%d",prod(x+2,y-1));

    #define prod(a,b) a*b

main()

{

    int x=3,y=4;

    printf("%d",prod(x+2,y-1));

}

Answer:

10

Explanation:

    The macro expands and evaluates to as:

    x+2*y-1 => x+(2*y)-1 => 10

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printf("%d",*a+1-*a+3);

    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 !

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if(x=y%2) z=2;

    #include<conio.h>

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.

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infinite loop example

    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.  

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

    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.

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Related to previous example

    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.

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Garbage Values

    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).

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void *v;

    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,

1. Passing generic pointers to functions and returning such pointers.

2. As a intermediate pointer type.

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

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if(printf("%s\n",a))

    void main()

{

    int i;

    char a[]="\0";

    if(printf("%s\n",a))

         printf("Ok here \n");

    else

         printf("Forget it\n");

}

Answer:

 Ok here

Explanation:

Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

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char a[]="12345\0";

    void main()

{

    char a[]="12345\0";

    int i=strlen(a);

    printf("here in 3 %d\n",++i);

}

Answer:

here in 3 6

Explanation:

    The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

   

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int ret(int ret)

    void main()

{

    int k=ret(sizeof(float));

    printf("\n here value is %d",++k);

}

int ret(int ret)

{

    ret += 2.5;

    return(ret);

}

Answer:

 Here value is 7

Explanation:

    The int ret(int ret), ie., the function name and the argument name can be the same.

    Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed,  after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

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Recursion Example

    void main()

{

    static int i=5;

    if(--i){

         main();

         printf("%d ",i);

    }

}

Answer:

 0 0 0 0

Explanation:

    The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

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