Operator's tricky Program 5

What will be output of the following program?

#include<stdio.h>

void main(){

    int x;

    x=10,20,30;

    printf("%d",x);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 10

Turbo C ++4.5: 10

Linux GCC: 10

Visual C++: 10

Explanation :

Precedence table:

Operator	Precedence	Associative
=	More than ,	Right to left
,	Least	Left to right

Since assignment operator (=) has more precedence than comma operator .So = operator will be evaluated first than comma operator. In the following expression

x = 10, 20, 30

First 10 will be assigned to x then comma operator will be evaluated.

Operator's tricky Program 4

What will be output of the following program?

#include<stdio.h>

int main(){

    int a=2,b=7,c=10;

    c=a==b;

    printf("%d",c);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 0

Turbo C ++4.5: 0

Linux GCC: 0

Visual C++: 0

Explanation:
== is relational operator which returns only two values.

0: If a == b is false

1: If a == b is true

Since

a=2

b=7

So, a == b is false hence b=0

Operator's tricky Program 3

What will be output of the following program?

#include<stdio.h>

int main(){

    int i=1;

    i=2+2*i++;

    printf("%d",i);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 5

Turbo C ++4.5: 5

Linux GCC: 5

Visual C++: 5

Explanation:
i++ i.e. when postfix increment operator is used any expression the it first assign the its value in the expression the it increments the value of variable by one. So,

i = 2 + 2 * 1

i = 4

Now i will be incremented by one so i = 4 + 1 = 5

Operator's tricky program 2

What will be output of the following program?

        

#include<stdio.h>

int main(){

    int i=5,j;

    j=++i+++i+++i;

    printf("%d %d",i,j);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 8 24

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error

Explanation:

Rule :- ++ is pre increment operator so in any arithmetic expression it first increment the value of variable by one in whole expression then starts assigning the final value of variable in the expression.

Compiler will treat this expression j = ++i+++i+++i; as

i = ++i + ++i + ++i;

Initial value of i = 5 due to three pre increment operator final value of i=8.

Now final value of i i.e. 8 will assigned to each variable as shown in the following figure:

So, j=8+8+8

j=24 and

i=8

Operator's tricky program

What will be output of the following program?

#include<stdio.h>

int main(){

    float a=0.7;d 

    if(a<0.7){

         printf("C");

    }

    else{

         printf("C++");

    }
    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: c

Turbo C ++4.5: c

Linux GCC: c

Visual C++: c

Explanation:
0.7 is double constant (Default). Its binary value is written in 64 bit.

Binary value of 0.7 = (0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 )

Now here variable a is a floating point variable while 0.7 is double constant. So variable a will contain only 32 bit value i.e.

a = 0.1011 0011 0011 0011 0011 0011 0011 0011 while
0.7 = 0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011....
It is obvious a < 0.7