INTERVIEW QUESTIONS C BRANCHES & LOOPS IN C DETAILS

Question: void main()
{
int x=20,y=35;
x= y++ + x++;
y= ++y + ++x;
printf("
%d %d" , x,y);
}
/* please tell me the value of x and y at each and every moment. i dont need the answer but the explanation of the anamolus behavior of increment expression.*/

Answer: the answar x=57 & y=94
from first equqtion
x=y++ +x++
x=35+20=55
then X++=56
&y=36

now from second equation
y=++y + ++x
x now 57 & y=37
y=57+37=94
which is desire output

the answar x=57 & y=94
from first equqtion
x=y++ +x++
x=35+20=55
then X++=56
&y=36

now from second equation
y=++y + ++x
x now 57 & y=37
y=57+37=94
which is desire output Source: CoolInterview.com

x=57,y=55 Source: CoolInterview.com

X = 57 Y= 93 Source: CoolInterview.com

answer is 5794
x=21+36+57
y=36+58=94

the new value of x in case of y is 57 and after incremented as 58 Source: CoolInterview.com

Final values : x = 57 and y = 94

Explanation:
In the expression, x=y++ + x++;
original values of y and x are added together first and then their values are incremented, therefore since x=20 and y=35, new value of 'x' is 20+35=55.
After this expression, x=56 and y=36.
Now y= ++y + ++x; values of y and x are first incremented (x=57 & y=37)and then added.therefore finally y =37+57= 94.
Hence x=57 and y= 94. Source: CoolInterview.com

output of your source code is:-
55 93 Source: CoolInterview.com

ans. of your ques.is:-
x=35+20=55;
so x=55;
y=36;
y=37+56=93;
so
x=55
y=93

Source: CoolInterview.com

57 94 Source: CoolInterview.com

x=y+x, y=y+1=36 and x=x+1=21;
=35+20,
=55;so latest value of x is 55 and y is 36;
y=(y+1)+(x+1)=37+56=93; Source: CoolInterview.com

x=35+20=55
now the value of:-x=21,y=36
y=37+56=93
hence output will be
x=55,y=93 Source: CoolInterview.com

The answer is -
1) For the first equation -
x = y++ + x++
x = 35 + 20 = 55
2) For the second one
y = 37 + 57 = 94

This is because in the first expression the post increment operation is used. Hence here the values of x and y are first added using the initial values and then incremented.

In the second expression the pre increment operator is used. Hence here first both x and y values are incremented first and then added. Source: CoolInterview.com

The print where u have given doesn't explain the details of operation. Below I gave detail steps:

int x=20,y=35;
x= y++ + x++; // At this point x = 35 + 20 As its post increment.
//Here x= 56
y= ++y + ++x; // At this point y = 37 + 57 As its post and pre increment.
printf("
%d %d" , x,y);

Final answer is x = 57 and y = 94 Source: CoolInterview.com

answer is x=57 ,y=94 Source: CoolInterview.com

The answers are,
x=57,y=94.
1.x=20,y=35;
2.x=y++ + x++=36+21,so x=57;
3.y=++y + ++x=36+58,so y=94 Source: CoolInterview.com

o/p will be x=57 y=94.
in 1st stat. x & y gets added first
then incremanted.so,
x=20+35=55
now increment takes place
so x=56,y=36
in 2nd stat.x & y incremanted first
then added.so,
x will become 57 y will become 37.
now added so y= 57+37=94
so o/p will be x=57 y=94. Source: CoolInterview.com

in post increment the operations done before increment.....
in pre increment the operations done after increment.....
x=y++ + x++(here the operation of adding x and y completed first.... after the only the value of x and y get increased.... x=55
after assigning x value as 55 then the value of x and y get increased by one..... x=56 y=36
y= ++Y + ++x;
here the value of x and y get increased before add their value...
x=57,y=37
these 2 values added and the value will be assigned to y...
y=57+37=94.... Source: CoolInterview.com

No guys...
The answer is x=56 y=93 (I tried out!!!It works)

Explanation:
In first Statement,
x=x++ + y++;
x=(20+35)=55; Before storing the 55 to x,Increments are done.So x=20 already,it is incremented as 21.similarly y becomes 36.
After that,the evaluated value 55 replaces the 21 and Now x=55 ,y=36

In Second Statement,
y=++x + ++y;
y=(55+1)+(36+1)
Because of pre-increment X,Y are icremented before evaluating the expression.Hence x=56 and after expression y becomes 93
Source: CoolInterview.com

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