it will gives u a garbage value according to u r compiler

for this what is the input value of x.First of all we have 
to give the input value of x.
if x = 1 then answer would be
main();
int x=1,y.
y = x++ + ++x.
printf /n'%d',y.
here y = 3

for this what is the input value of x.First of all we have 
to give the input value of x.
if x = 1 then answer would be
main();
int x=1,y.
y = x++ + ++x.
printf /n'%d',y.
here y = 3

these type of expression  evaluation is dependent on
compiler

if u use turbo c and x =1;

 then answer would be
main();
int x=1,y.
y = x++ + ++x.
printf /n'%d',y.
here y = 4

if x=1 then output will be 4

x++ starts with x and ++x starts with x+1,hence if x=1 then
ans. is 3.

answer illl always be 4
value of x ill be 2+2
it willincrement only in 2nd case th ++x

If x=4 then answer is 10

Answer is 2x+2. For Ex: if x is 2 result is 2(2)+2 = 6.

suppse if x=5,
then o/p will be 13

I have worked out this.in turbo c3
#include<stdio.h>
#include<conio.h>
void main()
{     clrscr();
	int x,y;
	x=4;
	printf("%d",x++ + ++x);
	getch();
}

output:
10

#include<stdio.h>
int main()
{
        int x=1,y;
        y=x++ + ++x;
        printf("y=%d\n",y);
        return 0;
}
o/p->
y=4

if x=3
y=x++ + ++x;
y will be 8
x will be 5

calculating y part
1) x++ is 3 and x will be 4
2)++x is 5 bcz x is incremented then assigned so x=5

so y finally gives 8.