#define

What will be output if you will compile and execute the following c code?

#define x 5+2

void main(){

    int i;

    i=x*x*x;

    printf("%d",i);

}

(A)	343
(B)	27
(C)	133
(D)	Compiler error
(E)	None of above

---

Explanation:

As we know #define is token pasting preprocessor it only paste the value of micro constant in the program before the actual compilation start. If you will see intermediate file you will find:

test.c 1:

test.c 2: void main(){

test.c 3: int i;

test.c 4: i=5+2*5+2*5+2;

test.c 5: printf("%d",i);

test.c 6: }

test.c 7:

You can absorb #define only pastes the 5+2 in place of x in program. So,

i=5+2*5+2*5+2

=5+10+10+2

=27

Operator's tricky program 12

#include<stdio.h>

int main(){

     int i=5;

     int a=++i + ++i + ++i;

     printf("%d",a);

     return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 21

Turbo C ++4.5: 21

Linux GCC: 22

Visual C++: 24

Explanation:

Rule : ++ (in ++i) is pre increment operator so in any arithmetic expression it first increment the value of variable by one in whole equation up to break point then start assigning the value of variable in the equation. There are many break point operators in. For example:

(1) Declaration statement.
(2) && or operator.
(3) Comma (,) operator etc.

In the following expression:

int a=++i + ++i + ++i;

Here break point is due to declaration .It break after each increment i.e. (initial value of i=5) after first increment value 6 assign to variable i then in next increment will occur and so on.

So, a = 6 + 7 + 8;

Operator's tricky program 11

What will be output of the following program?

#include<stdio.h>

int main(){

    float a;

    (int)a= 45;

    printf("%d,a);

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: Compilation error

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error

Explanation:

After performing any operation on operand it always return some constant value.

(int) i.e. type casting operator is not exception for this. (int) a will return one constant value and we cannot assign any constant value to another constant value in c.

(int)a = 45; is equivalent to
3456 = 45 ( Here 3456 in any garbage value of int(a)).

Operator's tricky program 10

What will be output of the following program?

#include<stdio.h>

int main(){

    int a;

    a=sizeof(!5.6);

    printf("%d",a);

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: 2

Turbo C ++4.5: 2

Linux GCC: 4

Visual C++: 4

Explanation:

! is negation operator it return either integer 0 or 1.
! Any operand = 0 if operand is non zero.
! Any operand = 1 if operand is zero.
So, !5.6 = 0

Since 0 is integer number and size of integer data type is two byte.

Operator's tricky program 9

What will be output of the following program?

#include<stdio.h>

int main(){

    int x=100,y=20,z=5;

    printf("%d %d %d");

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: 5 20 100

Turbo C ++4.5: 5 20 100

Linux GCC: Garbage values

Visual C++: 5 100 20

By default x, y, z are auto type data which are stored in stack in memory. Stack is LIFO data structure. So in stack first stores 100 then 20 then 5 and program counter will point top stack i.e. 5. Default value of %d in printf is data which is present in stack. So output is revere order of declaration. So output will be 5 20 100.