Again Question about C

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#include <stdio.h>
#include <conio.h>
void main()
{

int x=2,y=3;
x=++y + ++x;
y=++y + ++x;
printf("%d%d",x,y);
getch();




}

The output of code is 8 and 13

i want how this comes plz help i

Posted 19-Oct-11 20:48pm

umairshoaib

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4 solutions

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P.Salini · Answer 1 · 2011-10-19T20:54:00

First study about the pre-increment and post-increment operators then you will know exactly what happening.

pre-increment will be done before execution of statement
and post-increment will be done after execution of statement

so
your result will be as below.

x=++y + ++x;

here

 ++y=4,++x=3
x=4+3=7

y=++y + ++x;

from above statement

++y=4+1=5
++x=7+1=8
y=5+8=13

so finally x=8 and y=13

sravani.v · Answer 2 · 2011-10-19T20:55:00

Solution 3

x=++y + ++x;

means

x=(1+3)+(1+2)=7 ,y=4

y=++y + ++x;

means

y=(1+4)+(1+7)=13

Posted 19-Oct-11 20:55pm

sravani.v

Updated 19-Oct-11 23:41pm

v3

Mehdi Gholam · Answer 3 · 2011-10-19T20:54:00

Solution 1

Follow the guideline given in your last question and you can figure this out.

Posted 19-Oct-11 20:54pm

Mehdi Gholam

Comments

Albert Holguin 20-Oct-11 10:41am

This isn't a solution, post as a comment in the future.

Mehdi Gholam 20-Oct-11 10:50am

This is the second question from the OP in this regard.

Albert Holguin 20-Oct-11 15:46pm

It's still not a solution... flag repeat questions as "repost"

ShacharK · Answer 4 · 2011-10-19T21:04:00

As you probably know, there are two kinds of increment (and decrement) operators. One comes before the operand - and is called prefix (++x), and the other one comes after the operand - and is called postfix (x++). Same can be said about the -- (decrement) operators.

Lets see what happens when we're applying both operators

C++

int x = 5;
int y = 5;

printf("x = %d, y = %d",++x, y++);

The answer would be: "x = 6, y = 5". Why is that?
The reason is that the prefix operator (that in this case is being applied to x), is incrementing the value of x by one, and uses the result right away. In contrary, y's current value is being used, and only afterwards increments. Look at an equivalent example:

C++

int x = 5;
int y = 5;

x = x + 1;
printf("x = %d, y = %d",x,y);
y = y + 1;

So lets translate now the example that you've presented, into a similar code.

C++

int x=2,y=3;
x=++y + ++x;
y=++y + ++x;
printf("%d %d",x,y);

becomes:

C++

int x=2,y=3;

y = y+1; // now y = 4
x = x+1; // now x = 3
x = y + x; // now x = 4 + 3 = 7

y = y+1; // now y = 5
x = x+1; // now x = 8
y = y + x; // now y = 5 + 8 = 13

printf("%d %d",x ,y); // output is 8 and 13.

Again Question about C

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